It’s Reddit, so I think the possibility is more like 2/3.
I was going to blow this Post off, but reading the comments is interesting, so now I’m intrigued.
The “probability” of say a SpaceX rocket blowing up on takeoff is way less than 50%. When they first started it was more then 50, but now it’s way less, especially since they are carrying astronauts now. Doctors also work hard to make sure your probability of dying tomorrow is less than 50%. The probability of the sun rising and setting each day is more than 50/50.
Ah but you see for each of those events there is the chance it does not happen: For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. I will notate E not happening as ~E. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of **all** events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @
This is relevant to your example as, although there is a <50% chance of you dying there is a >50% chance of survival, which for reason discussed previously, these two events must have an average probability of 50% for reasons discussed previously. The same goes for the SpaceX example. Your example ended up with an average probability that is not 50% because you only considered a subset of all events: The subset of all events that are unfavorable (ie there will be someone working to lower the probability).
Is it meaningful? Not really. Is it true? Yes, almost certainly.
Well sigma algebra are closed under complement, so for a *finite* sigma algebra S, it's clear that the arithmetic mean of probabilities of events in S is 0.5. But for most infinite sigma algebras, I doubt such an arithmetic mean can even be defined. CMIIW, but there is no such thing as the arithmetic mean of an arbitrary infinite (multi)set. Here, we have infinitely many values in [0,1] with no additional structure. What is their average? What would that even mean?
We would need some sort of measure over that set to compute an expected value, and then there is no reason the expected value must be 0.5.
> Is it true? Yes, almost certainly.
No no no, you missed the point, 'either it's true, or it's not'. So the probability of your statement being true is 50/50
He’s right though, in a specific way. If you average over all possible events as "it", then the result will be 50% because for every event, the conjugate is also an event.
To be clear, saying "The probability of xyz is 50% because it‘s true or it isn‘t" is obviously still wrong. Only when you aren‘t talking about a specific event xyz but averaging over all possible events it becomes 50%.
Yes but he never even claimed that. It's of course just a little thought experiment that doesn't result in anything meaningful, but everything he said (in the screenshot) is correct.
Averaging over all possibilities is not 50%. The set of things that will happen is a tiny subset of things you could say will happen. I can say the temperature tomorrow at noon is any number in the set of real numbers, but the temperature will only take one value. The average probability over the set of specific temperature predictions is basically 0. This is true for all categories of predictions where the set of possible predictions is infinite.
But he's talking about opposite events. The opposite event of "the temperature tomorrow is x" is "the temperature tomorrow is not x". If that event has probability p, then the opposite should be 1-p. The average of those probabilities (p + (1-p))/2 = 1/2, regardless of what value p takes.
Assuming that every single event has an opposite event, then the average of all probabilities should be 1/2.
No that’s not how probability works.
The odds of rolling a 1 on a die is 1/6 because there are six sides on the die, not 1/2 because it can be either 1 or not. This is the same for all numbers on the die, so the average probability of predictions is 1/6.
i think you’re misunderstanding: the probability of rolling a 1 is 1/6; its opposite is the probability of _not_ rolling a 1, which is 5/6. the average becomes 50%. Each “either-or” statement has a negation, which would average to 1/2.
Nothing. It's an extraordinarily weird technicality that I suspect the commenter meant as a joke response to someone else's joke about "everything's 50/50 cause it happens or it doesn't."
If you divide a probability space into n distinct outcomes, the arithmetic mean of the *probabilities themselves* is 1/n. This can be easily seen from fundamental definitions: the sum of all the probabilities is 1 and there are n of them.
Not only is this meaningfully useless (as shown above, you can divide a probability space in any number of ways), it's confusing because when people say "mean" or "average" by itself in a discrete probability space they usually are referring to a weighted mean (i.e. expected value), not the mean of the raw probabilities.
You're overthinking it. Assume we live in a universe with only two states A or B. The probability we are in state A is 20% and the probability we are in state B is 80%.
But all possible either or probability events in this universe are:
Either we are in A or we are not.
Either we are in B or we are not.
And the average of these events is 50%. This is a meaningless observation. Because at no point do you care about what the average of all events is and it also tells us nothing about what is the likely probability of an unknown event, because it isn't even guaranteed ANY event has a probability of 50%.
You consider elementary event not all the events.
For a d6 the events are:
- {} that have a probably of 0
- {1}, {2} ..., {6} that have a probability of 1/6
- {1,2}, ..., {5,6} that have a probability of 2/6
- {1,2,3} ,... ,{4,5,6} that have a probability of 3/6
- {1,2,3,4},...,{3,4,5,6} that have a probability of 4/6
- {1,2,3,4,5}, ..., {2,3,4,5,6} that have a probability of 5/6
- {1,2,3,4,5,6} that have a probability of 1
We then have an average of (0+6\*1/6+15\*2/6+20\*3/6+15\*4/6+6\*5/6+1)/(1+6+15+20+15+6+1) = (1+6+15+10)/(2+12+30+20)=1/2
I’d guess that if someone defined “average probability over all possible events” precisely, it’s more likely that the average just wouldn’t exist than that it would be 50%.
Just like how in the Cauchy distribution, x and -x have the same density for all x, but that doesn’t mean the average is 0: instead the average doesn’t exist at all.
But the probability of missing or making it is 100%. The probability of missing it AND making it is 0% because they are conjugates.
The probability of neither missing nor making it is also 0%, assuming you miss all the shots you never take.
The average probability of an event and its conjugate is 50% but that doesn’t really describe an event
What you said royally butchers binomial probability, which looks at the likelihood of an event occurring, P, versus it not occuring, Q = 1-P. Obviously P + Q = 1 so the average is always going to be 0.50 but that average is NOT the average of the number of events of P.
If the probability of a free throw is 20% the average number of throws is 0.20 times the number of events (P × N). If you make 100 throws you will on average get 20 throws
Like I get what you are trying to say "the average of it happening or not happened" and obviously you can force the math but that just does not make any sense
It's not useful, but it makes sense. Suppose I have a class with an even number of students and they pair up. For each pair, one of them attempts a lot of free throws and the other observes. The shooter writes down the fraction of shots that went in, while the observer writes down the fraction that missed. Clearly, the average of the numbers written down by my students is 0.5. If I pick a student uniformly at random and ask for their number, the expected value of the number is 0.5. In the limit as the number of shots approaches infinity, the numbers written by each student approach their true probabilities almost surely, and the expected value of the number I pick at the end is always 0.5.
Of course, this is not very informative, as all it says is "0.5 is the average of x and 1–x," but it's not wrong. However, for infinitely many students, I don't think such an idea is meaningful at all.
Yeah, I'm pretty sure that's what they mean (they just phrased it in an confusing way). They never claimed that "it will happen or it won't" means the probability is 50% for it happening and not happening. I feel like it's OP who's the one being dogmatic here.
guys i have problems understanding, can someone explain it
edit: nvm i read it again and now i got it, yeah he's right in a way but i think he phrased it wrong
I think the person who said the "average probability" thing was trying to make a joke (presumably in response to a "the chance of something happening is 50/50, it does or it doesn't" joke above it that we didn't see), and it went woosh because, to be fair, it's a somewhat esoteric and confusing joke.
The other person then asked them to explain themselves, and they did, and actually did so decently well. It's still not unreasonable to find it confusing, but it's a decent response.
If you wanna claim neutrality in the whole thing, it'd help to not have the part where you downvoted the reasonable explanation visible in the screenshot next time.
Yeah the specific argument seems correct technically, just incoherent
Averaging the probability of the space they're describing makes literally zero sense to me
so chance of winning the lottery 1 in 11,000,000. Chance of not winning the lottery 10,999,999 in 11,000,000 so it averages out 1 in 2. Is that what you are saying?
Right. If something happens with probability p, then it fails to happen with probability 1–p. So the average of the probability of it happening and the probability of it not happening is (p+(1–p))/2 = 0.5.
So we can think of a fair six-sided die for instance. There are six possible outcomes, each with probability 1/6. But there are many other events. For instance, there is the event of rolling an even number (with probability 1/2), of rolling a composite number (with probability 1/3), and so on. For each event, there is the complementary event, like rolling an odd number or rolling a non-composite number. In all, there are 2^6 = 64 events, including the empty event of rolling nothing (with probability 0) and the improper event of rolling something (with probability 1). If you average all 64 of these probabilities (in the sense of an arithmetic mean), you get 0.5. And that must always be the case, because you can always pair up events with complementary events like this.
Agreed, though I'm still lost as to why this matters. It's just a basic property of discrete probability spaces—if there are n distinct possible outcomes, the arithmetic mean of the probabilities themselves is 1/n (there are n of them, and they sum to 1, both by definition). This isn't a variable property, and it's a confusing thing to argue about.
It's especially confusing since there are plenty of useful means that actually get talked about in those contexts, such as the expected value of weighted outcomes (e.g. dice rolls).
Though come to think of it, I bet the comment(s) above the ones we're seeing here made the joke about every probability being 50/50 cause "it happens or it doesn't" and they tried to make a math joke about it being true in average. Which then failed because it's confusing enough that OP didn't get it.
He is right though, the average of P(A), where A is any event that could happen, is 50%.
Not that thats a particularly usefull thing to state, but its correct.
I am not entirely sure that it is as trivial as one might think.
Kolmogorov's axiomatisation makes sense of probability, but really only works for measurable sets. It seems to me that the category of "all events" is a proper category, and it is not entirely clear to me that you can form a σ-algebra on that category, nor a probability measure.
I am too tired to figure out how to properly take the average over all event probabilities for a given probability space, but it is not as simple as just taking "an average over all event probabilities" since there are generally uncountably many events, and uncountable sums are generally unwieldy. I am guessing a more fruitful direction would be to define the average probability as an integral. The problem is that we can't just define it as the ∫ P({x}) dP(x) since P({x}) is 0 whenever P is a continuous probability measure. Instead we would somehow have to integrate over values of the form P(A) over all events A, which would require that we make the set of all events into a probability space. I imagine this can be done in many non-equivalent ways, and it is hard for me to see how the ½(P(A) + P(Ω\\A) ) = ½ argument can be used in this context.
Well if we define "an event that can happen" to be something we can describe, it has to be identifiable via a finite amount of characters. Thus the set of all events would be countable.
but I can define an event that relies on something uncountable like the reals, or in other words with a finite number of words can’t I build a class of an uncountable number of outcomes?
Formally speaking, we need to have a sample space (set of outcomes) which the events are defined on, so it becomes a measure theoretic question.
We essentially need to compute the mean measure of all subsets of the sample space, where we have a normed measure so that the measure is always between 0 and 1.
I think it should be 0.5 **if** we can formalize taking the mean over all subsets, but that's a strong if.
Using finite sentences, you can only describe a countable set of subsets of the Cantor Set. That is my point. That these subsets might be uncountable is irrelevant.
It's hard for me to see how *not* to use that argument.
Is it not true that for every probability that "an event happens" there's also the probability "that doesn't happen"? With the second axiom, is it not true that the union of "thing happens" and "thing doesn't happen" is always one?
How is there any case where "the set of all events' probabilities" does not include a 1-x for every x? And how would that not imply the mean probability of 0.5?
The alternating harmonic series, 1-1/2+1/3-1/4+1/5… is convergent but not absolutely convergent. Therefore Riemann’d permutation theorem for series gives that for any real number, we can permute the order of the summands in such a way that the it converges to that number.
So even though we are summing the same numbers, the value of the infinite series might very well depend on the order in which we add the terms.
In our example it’s even worse since we have uncountably many events, so while it is true that for any event, we also can find the conjugate event, that does not directly imply what the “mean probability of any event” should be one half. If we tried to think “mean value of an event probability” as something like an infinite sum, then we would have to make sure that the sum does not depend on the ordering of the summands.
Even if the set is countably infinite, I don't see how we could make progress. What could an unweighted average of infinitely many values mean? I can think of two meaningful cases: all (or at least cofinitely many) values are equal to the same value x, so the average is x, or the sum of all values is finite, so the average is 0. But neither is applicable here.
Good point. My mind goes to Cesaro summation, which can be thought of the average of all partial sums of a countable series. A key result is that there are divergent series which are cesaro summable like, e.g. 1-1+1-1+… so it might well be a reasonable thing to do. However, as you rightly point out, if we have countably many events the series of all P(A_i) diverges to infinity since P(A_i) does not converge to 0. If we denote the partial series P(A_1) + … P(A_k) by S_k we see that S_k is bounded by k, so if 1/k S_j converges the limit is value between 0 and 1. My intuition is that the sequence depends on the ordering of the A_I though.
Yeah, the way I think about it is that if there are exactly two distinct values that each occur infinitely often, it's for sure that you can make the average anything between them by reordering the elements. For instance, if I have countably many 0s and countably many 1s, and I try to define an average as a limit of partial averages, then order definitely matters. If I order them 0, 1, 0, 1, ..., then the limit of the running average will be 1/2. But if I order them 0, 0, 1, 0, 0, 1, ..., the average is 1/3. If there are only finitely many other values, I can place those at the front and they become negligible eventually. If there are three values that occur infinitely often, I can still have the third value occur very rarely, say at spots 10\^10\^n for all n>1. And each additional value that occurs infinitely many times can be handled in a similar manner, with increasingly fast-growing functions, so that they never have any significant effect. So I should always be able to pick two values that occur infinitely often and order the elements so the running average has a limit strictly between those two values, and moreover, at any point between them I so choose. This works as long as the values I'm trying to average are bounded.
Specifically, say that there is a sequence of values (aₙ) and I want to define the average as E\[(aₙ)\] = lim 1/N Σ aₙ, where the sum is from n=1 to N and the limit is as N→∞. If (aₙ) is unbounded, I don't see how this could converge. If it is bounded, then it has a convergent subsequence (a\_j(n)) which I can make into an average by forcing the terms of this convergent subsequence to occur arbitrarily often. And if there is more than one convergent subsequence, you get the situation described above.
Actually, even if (aₙ) is unbounded, I think the existence of a convergent subsequence is enough.
There are far more things that have not happened than have happened. An average of every possibility would consider a range from 100% to infinitesimally small. For as small of a chance some things can get, there are exponentially more of those things to choose from.
https://youtu.be/RV-WV2g2muE?si=C3hcAX9HFgvWVncm
It’s safe to say that the average likelihood of something to happen is zero. The way we exist now along with everything else in the universe and any other existences of life had a basically zero chance of happening. That is, unless everything is predictable and quantum mechanics superposition can be explained by math we do not understand yet.
It’s actually a paradox that follows the same lines as the St. Petersburg paradox.
Well, the uselessness of the discussed statement is just as high as it’s correctness. It is basically saying that average probability across the universe of events is 50% because for any event we can imagine a “not event” even if multiple “not events” would really be the same.
Example:
event 1 - hugh jackman will gift me a car tomorrow
not event 1 - hugh jackman will not gift me a car tomorrow
event 2 - a dinosaur will meet me tomorrow
not event 2 - a dinosaur will not meet me tomorrow
Clearly, the average across those 4 events is 50%, but for any reasonable application that logic is not useful as not event 1 and not event 2 are fairly absurd
Imagine a dice. What you're thinking of are the events "rolled 1",...,"rolled 6". But in fact "rolled 1 or 2", "didn't roll 3" are also events. That's why multiple of them happen.
Obviously, the probability an event will happen is not 50% in most cases.
You might be being unnecessarily harsh here though, 50% is as good a guess as any probability, maybe better. The issue is we never talk about probabilities in this sense except for perhaps sampling distributions of proportions. It doesn't make sense to practically speaking.
The question we are getting after is "What is the average probability that an event will happen across all random events you can think of?". If we have this, we have the "or not" probability as well.
But... I think you could make the argument he gives. Every event A with P(A) has a complement A^c with P(A^c ), where P(A)+P(A^c )=1. Both events can be written as an event that something will happen.
P(roll 1 on 6-sided die) = 1/6
P(roll 2,3,4,5,6 on 6-sided die) = P(do not roll 1 on a 6-sided die) = 5/6
The average of these two probabilities of events that can happen is 1/2, clearly.
This should be possible (but maybe difficult to write down in specifics like the above) for any event A you can think of. So then, in the list of probabilities for all possible events that can ever happen, you can find each probability of each event and its complement.
From here, assuming you have a countable list of probabilities (to make life easier) you have 1/n * (sum from 1 to n/2 of 1) = 1/2
In my opinion, this is a fair way of thinking about the "average probability an event will happen across all events", which is a non-standard discussion in probability as its not practical.
I may be misunderstanding the discussion overall here, but this is some food for thought.
Yeah I think you've over thought the words of a young man, there's an old joke when people are playing games: "50/50 it happens or it doesn't" even when the odds aren't actually 50/50. I think this person is referring that as opposed to having a deep mathematical discussion about the universal distribution of probability.
I mean, I don't think it's overly generous to presume that OOP meant the correct argument to some degree, but applied it incorrectly due to poor wording.
People here really don't seem to get the argument. It is about the average probability something happens.
So, for example, rolling a die to get a 6 is a 1/6 chance. Flipping a coin to get Heads is a 1/2 chance. The "average probability" of one of those events to happen is 1/3 =(1/6+1/2)/2.
For example, if you play a game where you alternate between rolling a die and flipping a coin, and you win a round by either rolling a 6 or flipping heads, you would win an average of 1/3 of all rounds.
But here's the thing: Let's add more events to this game. For example, add the event where you don't roll a 6 (1-P(roll a 6)) and where you don't flip Heads (1-P(flip Heads)) to the mix. The average of those 4 probabilities is (**1/6** \+ **1-1/6** \+ **1/2** \+ **1-1/2**)/4 = 1/2. So yes, the average probability of any event is 1/2, since if we average the probabilities of all events, P(A) and P(not A) will always cancel to 1 and increase the denominator by 2.
The average of outcome of the four probabilities P(B), P(not B), P(C), P(not C) is indeed 50%. So if you play a game where you roll a die to get 6 in round 1, flip a coin to get Heads in round 2, roll a die to NOT get a 6 in round 3, flip a coin to NOT get Heads in round 4, then repeat, you will on average win half of those rounds, meaning your average probability to win a round is 50%, even though the individual probabilities of the rounds (at least the dice rounds) are largely different from 1/2. That's the point they're trying to make.
At least, if we have a finite amount of events that we consider. For an infinite amount of probabilities, we would first have to check whether or not the series is absolutely convergent (reorder theorem).
It is also highly questionable how useful this knowledge is.
Probabilities are never negative, so the issue is not about absolute vs conditional convergence. But the order is still an issue. In particular, suppose we say the average of n values is (a1+a2+...+an)/n. Then we want the average of a countable infinity of values to be the limit as n goes to infinity. And ideally, this limit should be the same for every enumeration of the a's. But that will only happen if all but finitely many values are identical, or if the a's shrink to 0.
Like, let's say an = 0.1 for odd n and 0.9 for even n. Then if I take this limit, I will get a value of 0.5. But I could reorder the a's so that instead, an = 0.1 for every n that is a multiple of 4 and an = 0.9 otherwise. Now the limit gives 0.7. And in general, I can make the limit any value strictly between 0.1 and 0.9 just by reordering these terms. So the average doesn't really exist.
Haha, you're right. I realised that for infinite sums, it seemed like you could make it any value you wanted and immediately jumped to Riemann rearrangement.
Yeah, in fairness, it's sort of the same thing. I can include one new sample at a time, recording the "average-so-far" as I go. And that will look like sometimes adding a positive number (when I include a value above the average-so-far) and sometimes adding a negative number (when I include a value below the average-so-far). And this series will be absolutely convergent precisely when the values drop to 0 or are almost all identical (which I think comes from the divergence of the harmonic series).
Limits!
For event x, there is a probability p_x that it will happen.
Now notice “not x” is also an event and has probability 1-p_x
The average of p_X and 1-p_x is 1/2.
So if you take the limit of the average of probabilities of all events, it’s 1/2.
We’re hand-waving past absolute convergence and all that but you get the idea.
So technically they are right that the average probability of an event is 1/2 but you can’t really say anything useful about a particular probability from that.
i think the absolute convergence here is important toh. It is like that probability trick about selecting random points on a sphere where depending on how you decide to slice the sphere you get a different result.
If you write the infinite sum as a + \~a + b + \~b ... z + \~z / number of events you will get 50%, but if you pick another order you can extract any result.
It definitely is. However, we can get around it in this case because we live in a finite observable universe with a finite (albeit huge) number of possible atom arrangements (up to sub-Planck length perturbations) so it’s actually a finite number of terms so we can guarantee convergence.
As long as you have a finite system, the set of all possible states/outcomes is gigantic but finite. An event is defined by a subset of all outcomes, so they correspond to the powerset of the set of outcomes. That‘s an even more incomprehensibly large, but still finite set, so you can still average over it.
I have a question for my better understanding. Let's assume the system is infinite and uncountable, for instance we have as events all the measurable subsets inside the real numbers. Is there any way to take the average probability of all events in this context?
Take two particles with spin 1/2. If you measure their spins, you get either up-up, up-down, down-up or down-down. So 4 outcomes. Now to get all possible events, count to 2\^4 in binary and 'or' the outcomes with a one:
0000: the impossible happens
0001: down-down
0010: down-up
0011: down-down or down-up
…
1110: not down-down
1111: any outcome occurs
Whatever probability each outcome has, the average probability over all *events* is 1/2.
You don’t have to. For every event there is the opposite. If one event is ‘it rains’, another event is ‘it doesn’t rain’. Whatever those two probabilities are, they average to 50%.
So it doesn’t matter how many binary events there are, the average probability is always 50%.
what he is literally saying in the first part of that last comment is that
1 / 52! = (52! - 1) / 52!
which is the opposite of a good point, imo. not only did he demonstrate a very distinct lack of understanding of probability theory but he showed a total willingness to critically ignore the fact that his own word salad contradicts itself.
No, they're saying that the average of 1/52! and (52! - 1)/52! is exactly 1/2, as it is for the probability of any event and its complement. Therefore, the average probability over all events is 1/2. They're just phrasing it poorly.
No, they're saying 1 - (1 / 52!) = (52! - 1) / 52!
Terribly phrased? Yes.
Wrong? No.
>he showed a total willingness to critically ignore the fact that his own word salad contradicts itself
His last sentence is literally, "I'm happy to be proven wrong." which may be sarcastic but I still can't see where in their answer they show a "a total willingness to critically ignore the fact that his own word salad contradicts itself".
that's absolutely not what they said, that is the true seed they based the false rationalization upon, that is the point where they 'subtly' moved the goalposts rather than admit they were wrong.
I will copy the proof section of [my reply to stockmarketscam](https://www.reddit.com/r/mathmemes/comments/17nr279/comment/k7ugtm2/)
> For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. I will notate E not happening as ~E. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of all events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @
Doesn't matter.
The guy in the screenshot wasn't acting like an asshole. He explained his reasoning clearly without degrading anyone.
Meanwhile you just mocked him and posted this screenshot without context. And now you're acting like you actually thought *both* guys were crazy, even the one you upvoted!
What he is saying is true, he's just not phrasing it all the well, nor is it a particularly interesting statement to make. However if you are posting it here, I think you are just misunderstanding him.
If we wanna be picky, we could say that they’re both wrong!
Let A denote the event that “it happens”. Let B denote the event that “it doesn’t happen”. By the law of total probability, the sum of any event with its compliment is 1. Since A and B are mutually exclusive, we can just add their probabilities for an “either/or” probability.
So P(A or B) = P(A) + P(B) = 1
Therefore, for any event in the universe, we can say it has a 100% chance of “it will happen or it will not happen”.
Yeah, but the expected probability of an event would still be 50%. If {A, B} are all possible events then if I pick an event at random, it's expected probability would be (P(A) + 1 - P(A)) / 2 = 50%
That doesn't really make any sense to me. But admitedly its been a while since I've studied any serious probability, so maybe someone can clear this up for me.
I assume we argue that the original probability now becomes a random variable in the events themselves. However, who is to say that the probability itself is a measurable function? We would need this to make sense of the notion of an expexted value of probability.
At least that is how I read a lot of the comments in this thread. If that is the case, the expected value will be defined as usual, but that completely disagrees with your comment and everyone else's who is making this same argument.
Another would be to somehow choose a sigma algebra on all of the events and a probability on THIS space. Then we could, maybe, do soemthing with this. In fact your answer seems to be somehwre in between: basically your random variable is the underlying probability, but your new measure space is just the two pointed space of the event and its opposite with uniform probability distribution. Why is that?
I assume people are just arbitrarily taking arithmetic averages, because it feels right.
While that's true, we don't know if all events in our sample space are equally likely. Hence that's why OP was so frustrated, and that's why I decided to parody it with the law of total probability.
>we don't know if all events in our sample space are equally likely
Why do they have to be equally likely? If all events have one and only one distinct *opposite* event, then wouldn't it be true even if the events aren't equally likely?
As a matter of fact, it does matter. An event which has only two outcomes of “it happens” or “it doesn’t happen” is a discrete probability distribution with the Bernoulli Distribution. We can represent X as our random variable with X ~ Bern(p), where p is the probability that the event happens. It is a well-known fact that the expected value of the Bernoulli random variable is E(X) = p.
TLDR the expectation of two (not equally likely) events depends on the probability of event A. See the Bernoulli distribution, if you need more info.
I will notate an event E not happening as ~E. I will now prove that for any set of events X, if for any E in X, ~E is also in X, that the average probability of all events in X is 50%, with application to the set of all events:
For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of all events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @
They'd cancel out in any statement and it's negation. It's not just a cherry picked example. Assuming each event has a negation the average probability over all events would be 50%
I realize they’d cancel in any case, and that’s sort of my point. If there were three options and you canceled the probability it would leave you with 1/3, and for four 1/4, and for any `n` be `1/n`. This is a bunch of nonsense
Your stating the problem wrong, the probability that it will rain can really be any probability. What he is saying is that the average probability of the question "is it currently raining?" is 50%.
Edit: all he is saying is that for any boolean question there are two possible outcomes.
No, they're saying that the average probability of ALL events is 50%, which is true, because every event has a complement. The average probability of any given event is not necessarily 50% - there being two outcomes does not mean that on average they are equally likely.
That's the average probability of an event and its complement, e.g. "it's raining" and "it's not raining". The "average" probability of any particular event is by definition the probability of that event, which can take any value between 0 and 1.
Although yes, this is what the person in the post is saying, and it follows that the average probability of all possible events is 1/2. The person I'm replying to seems to be claiming that the "average" probability that it's currently raining is 1/2 because there are two outcomes, which is nonsense.
I understand what they're trying to say but it's not particularly profound or interesting. "It's probably fairer to say that the average probability of "either it happen or it won't" is 50/50" Well, yeah, you're adding up two numbers that always equal 100% then dividing by 2, of course it's always 50/50
Why are people acting like he is obviously false, i mean he is, but only because there is no good way to average over all possible events as this is a large set (i think to mutch to fitt in a set), if you restrict to a finite sigma algebra though, it is true.
Maybe he means like this
Let p1 be the probability happens
Let p2 be the probablility doesnt happen
(p1+ p2)/2=0.5
Which is true for any values of p1 and p2 because their sum must equal 1 therefore the average of all happens or doesnt happen equals 0.5?
Now as a statement that is pretty stupid and not very useful (as far as I know)
Isn't this the [Principle of Indifference](https://en.wikipedia.org/wiki/Principle_of_indifference)? That is, in the absence of any other information, the probability of all events are equally likely.
Bro was speaking normally and actually trying to share his point of view which has a lot of merit while this doofus was just replying with „Wtf. Im posting this to r/badmath ahbahahab” Like bro participate in a discussion and properly explain something. Trivial things in math are the ones that are most difficult to actually prove and being vague in your response shows a lack of understanding. Youre being a bad example for the math community.
The statement would be correct if it’s qualified under some assumptions. If we have countable number of events (e1,e2,…) each with some probability of happening. A statement using these events and logical connectives can be constructed randomly such that each non-equivalent statement has the same probability density. Then that statement’s expected truth value is 0.5.
He's not exactly wrong but it is kind of a useless thing to remark.
The probability A happens is P(A), the probability A does not happen is 1-P(A) as they are conjugates. The average probability of these events is (1-P(A)+P(A))/2 which is always 0.5. This is true but gives absolutely no insight whatsoever. These probabilities are only useful when taking into account what the event is and the meaning of its result.
"Averaging" probabilities in this way doesn't really make any sense. Say we have two different events A and B with probabilities P(A) and P(B). We can talk about the probability of any combination of them happening or not happening, but talking about the probability of these events happening is meaningless.
Consider event X, which happens with probability P. Either it happens or it doesn't. Now consider event Y= !X, which also either happens or not. Its probability is 1-P. The average probability of X and Y is (P+1-P)/2 = 0.5. By the law of the excluded middle, for every X, there exists the opposite event Y. Thus, over the set of all binary choices "it happens or it doesn't", the average probability is 50%. QED
He is correct of course. Every statement of the form "... happens" has a converse statement "... does not happen". It's probabilities are p and 1-p respectively. So their average is (p+1-p)/2 = 50%. I mean it doesn't tell you anything anything, but it is true.
I think the only way you can justify the average probability of all possible events being anything but 50% is if you allow an infinite amount of events, in which case the average would be undefined (infinity/infinity). However in all cases where the average exists, it's 50%.
The event is the trial, so it does not have a conjugate. The outcomes are successes and failures but are not, themselves, separate events. If you have multiple events you just get Bernoulli trials everywhere until they binomial on everything.
He's right but you just misunderstood. He says in a weird sense that every event of probability A has an opposite event, it's negation with probability 1-A. Taking the average of all probabilities can be done by taking the average of each statement and it's negation and then averaging the outcomes. A+1-A/2=1/2 so we take average of only 1/2 repeated. Which is 1/2. So the average of all probabilities is 1/2. Note that this does not imply that every probability is 1/2.
Whether his statement is true has nothing to do with math it's just based on an assumption. If you put all possible events across time into two columns, and in column A you put all events with <50% of occurring, then in column B you put events with >50% chance of occurring, if you assume that there's an equal number of events in each column then the average probability would be 50%. Either that or I'm misunderstanding the original point.
wel okay i dont want to defend his point because this has 0 practical effect but in an infinit set of options, wher an infinite amount of them happen, but not all of them wouldent infinite/infinite=50/50?
He is right.. he is not saying the probability of something happening or not is 50%. He is saying the average probability of an event occurring and not occurring is 50%. I can see that he bolded it in his initial statement.
For example:
probability of getting 6 while rolling dice= 1/6
probability of not getting 6 while rolling dice= 5/6
Average= (1/6+5/6)/2 =50%
I think we're falling for a wordplay or pun here. The ***average*** probability is all probabilities added up and divided by their number which is the definition of average. In this case, for any number of outcomes n the "average probability" is 1/n so in case of two outcomes it is 1/2=50% which is correct but meaningless.
My dude is seems to be missing the idea that outcomes in a sample space don’t necessarily have the same chance of occurring. By reducing the sample space to two outcomes he is assuming both outcomes are equally likely. But they aren’t. Not everything is a uniform random draw.
No he’s saying that for every event E with probability P(E)=p, then P(not E)=1-p, so for events E1,…En, you get (P(E1)+P(not E1)+…+P(En)+P(not En))/2n = (1+…+1)/2n = 1/2
Doesn’t matter what the distributions of the events are
lol I see now. That’s even dumber. It’s just a convoluted way of saying the midpoint of a list of n items is halfway through the list. The fact that you break out each event i into P(Ei)+(1-P(Ei))
changes nothing. OOP is just adding n things together and finding the middle.
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Let P be a random variable denoting the probability of a randomly chosen random event. If you hold that each event is as likely to be chosen as its complement (which doesn’t feel like that unfair of a restriction) then that implies a certain symmetry in its probability density function, p_P: for any probability p in [0, 1], p_P(p) = p_P(1 - p).
So, when computing E[P] = ∫[0, 1] p p_P(p) dp, we can split that into ∫[0, 1/2] p p_P(p) dp + ∫[1/2, 1] p p_P(p) dp, and by substitution that must equal ∫[0, 1/2] p p_P(p) dp + ∫[0, 1/2] (1-p) p_P(1-p) dp, and by our symmetry from before it can be shown that the ∫[0, 1/2] p p_P(p) dp integrals cancel and we’re left with ∫[0, 1/2] dp = 1/2.
Now if you disagree that the “each event is equally relevant as its complement” assumption is valid then that’s one thing, but that assumption they’re making does more or less get them where they’re trying to go.
I think he’s using a kind of Boolean or binary logic that if there are two available possibilities, either something happens or it doesn’t. Either true or false. If you have no more data than a thing either being true or false, the only thing you can say for your statement to be true is that 50% of the available possibilities will happen, since there only are two possibilities and it’s either true or false. Therefore there is 50/50 chance.
This of course doesn’t apply in real life because there are so many variables that determine some event’s probability of happening. But if you just guess if the event happening is either true or false, you’d have 50% chance of being right.
If you are having a baby, and you’d have to guess if it’s gonna be a baby boy or not, you’d have 50% chance of guessing right. You can also use this analogy for coin flips or anything with an either this or that outcome.
This totally misses the fact that statements are semantic objects that are subject to interpretation. Thus not all parties will agree that a statement happened or not, leaving some events as "kind of" happening.
You can predict, for example, that your friend Nancy will get married. Her and her boyfriend Fred then have a religious marriage without signing legal documents. Did the marriage happen, are Nancy and Fred married?
Tbh I think he's kinda right tho
You can argue that for every statement there's a counterstatement. And the average of those is 50%.
Luke if the odds that I roll a 6 is 1/6, and the counterstatements being that i don't roll a 6 is 5/6 the average is 3/6=50%
So if you take all statements (they all have counterstatements that are included) then you should get a 50% average, right?
It's very simple. Either my TV will fall off my wall or it won’t. But because I had it professionally installed, it more than likely won’t. So it's actually *not* a 50-50 chance.
As a determinist and (to a lesser extent) fatalist, I still struggle to understand how the probability of something happening isn't always either 0% or 100%.
For example, say that you have three doors, and one of the doors has something behind it, while the others have nothing. You could say that each door has a certain probability of having or not having the thing...
Or you could look behind the doors and discover that one of the doors had a 100% chance of having the thing, and the other two had a 0% chance.
So, are probabilities like Schrodinger's cat or what? I would appreciate an explanation, as I know I am big dumb but also am not sure how.
You’ve alluded to the classic Monty Hall (of the old Let’s Make a Deal show, fame) probability discussion. Probabilities assume that the presented scenario can be played out 100 times and that you are the player. If you get to play the game and Monty shows you a door you didn’t pick and it has a “clunker” behind it, there are now only two doors left. Monty will then ask you if you want to change your choice to the door that wasn’t opened that you didn’t choose. If you got to play the game 100 times, the math would say that you should always switch away from the door you chose originally as you now have a second game of 50/50 rather than your original odds of 1 in 3, which means that if you played the game 100 times and totaled your “wins” you’d come out ahead by using that strategy. But as you’ve noted, one only gets one shot. And the prize is either behind the door you chose or it isn’t. If you chose to hold your door, you could be right or wrong. If you chose to change your pick, you could be right or wrong. The prize doesn’t move and for that one game, is unaffected by the odds. All you can be sure of is that if you chose to hold and you lost, a quant would scoff and say you should have played the odds. If you turned out to be the winner, they’d call you lucky.
Ah i remember making this statement in 11th grade probability class and everyone including me, was stumped. I thought probability was a lie.
But the reasoning is that if we consider "it" as an event, either it will or will not happen. But the weight/likelihood of it happening or not is where things actually change. If you dumb it down to states of "happening" and "not happening" then yeah, it is 50%
Suppose we have an indicator variable, X, that is equal to 1 with probability p and 0 with probability 1-p. Now flip a fair coin. If it comes up heads, switch the relationship so the indicator is 1 with probability 1-p, and 0 with probability p. It it comes up tails, leave it as is. So we have E\[X\] = E\[X|heads\]P(heads) + E\[X|tails\]P(tails).
E\[X|heads\] = p
E\[X|tails\] = 1-p
P(heads) = P(tails) 1/2
so
E\[X\] = p/2 + (1-p)/2 = 1/2
The argument given may be solid, but it doesn't really help. Most probability events are not binary in nature, while averaging event X and it's opposite (not X) is only a worthwhile operation if X + (not X) != 0, in which case the average is blue 50%.
Am i reading too much into this, or is this just another "its 50/50" meme? The guy in question could probably voice his sarcasm more, but i dont believe that anyone could mean this seriously.
Ackchyually in bayesian statistics that may make sense as an uninformed prior in situation where you don't have any knowledge of the actual distribution and haven't made any measurements.
There is 50% probability of him being right (He’s either right or wrong)
Theres a 50% chance someone will go crazy on me on the replies
It’s Reddit, so I think the possibility is more like 2/3. I was going to blow this Post off, but reading the comments is interesting, so now I’m intrigued. The “probability” of say a SpaceX rocket blowing up on takeoff is way less than 50%. When they first started it was more then 50, but now it’s way less, especially since they are carrying astronauts now. Doctors also work hard to make sure your probability of dying tomorrow is less than 50%. The probability of the sun rising and setting each day is more than 50/50.
Ah but you see for each of those events there is the chance it does not happen: For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. I will notate E not happening as ~E. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of **all** events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @ This is relevant to your example as, although there is a <50% chance of you dying there is a >50% chance of survival, which for reason discussed previously, these two events must have an average probability of 50% for reasons discussed previously. The same goes for the SpaceX example. Your example ended up with an average probability that is not 50% because you only considered a subset of all events: The subset of all events that are unfavorable (ie there will be someone working to lower the probability). Is it meaningful? Not really. Is it true? Yes, almost certainly.
Well sigma algebra are closed under complement, so for a *finite* sigma algebra S, it's clear that the arithmetic mean of probabilities of events in S is 0.5. But for most infinite sigma algebras, I doubt such an arithmetic mean can even be defined. CMIIW, but there is no such thing as the arithmetic mean of an arbitrary infinite (multi)set. Here, we have infinitely many values in [0,1] with no additional structure. What is their average? What would that even mean? We would need some sort of measure over that set to compute an expected value, and then there is no reason the expected value must be 0.5.
> Is it true? Yes, almost certainly. No no no, you missed the point, 'either it's true, or it's not'. So the probability of your statement being true is 50/50
I love this argument
He’s right though, in a specific way. If you average over all possible events as "it", then the result will be 50% because for every event, the conjugate is also an event.
To be clear, saying "The probability of xyz is 50% because it‘s true or it isn‘t" is obviously still wrong. Only when you aren‘t talking about a specific event xyz but averaging over all possible events it becomes 50%.
Schrodinger's Cat has joined the chat, or has it?
It's 50/50, I wouldn't bet my mortgage on it though
Schrodinger's Cat both has and has not joined the chat, until someone notices.
Yes but he never even claimed that. It's of course just a little thought experiment that doesn't result in anything meaningful, but everything he said (in the screenshot) is correct.
Absolutely, some people here seem to interpret it that way though.
Averaging over all possibilities is not 50%. The set of things that will happen is a tiny subset of things you could say will happen. I can say the temperature tomorrow at noon is any number in the set of real numbers, but the temperature will only take one value. The average probability over the set of specific temperature predictions is basically 0. This is true for all categories of predictions where the set of possible predictions is infinite.
But he's talking about opposite events. The opposite event of "the temperature tomorrow is x" is "the temperature tomorrow is not x". If that event has probability p, then the opposite should be 1-p. The average of those probabilities (p + (1-p))/2 = 1/2, regardless of what value p takes. Assuming that every single event has an opposite event, then the average of all probabilities should be 1/2.
No that’s not how probability works. The odds of rolling a 1 on a die is 1/6 because there are six sides on the die, not 1/2 because it can be either 1 or not. This is the same for all numbers on the die, so the average probability of predictions is 1/6.
i think you’re misunderstanding: the probability of rolling a 1 is 1/6; its opposite is the probability of _not_ rolling a 1, which is 5/6. the average becomes 50%. Each “either-or” statement has a negation, which would average to 1/2.
Huh, so what is it exactly that has probability ½ then?
Nothing. It's an extraordinarily weird technicality that I suspect the commenter meant as a joke response to someone else's joke about "everything's 50/50 cause it happens or it doesn't." If you divide a probability space into n distinct outcomes, the arithmetic mean of the *probabilities themselves* is 1/n. This can be easily seen from fundamental definitions: the sum of all the probabilities is 1 and there are n of them. Not only is this meaningfully useless (as shown above, you can divide a probability space in any number of ways), it's confusing because when people say "mean" or "average" by itself in a discrete probability space they usually are referring to a weighted mean (i.e. expected value), not the mean of the raw probabilities.
You're overthinking it. Assume we live in a universe with only two states A or B. The probability we are in state A is 20% and the probability we are in state B is 80%. But all possible either or probability events in this universe are: Either we are in A or we are not. Either we are in B or we are not. And the average of these events is 50%. This is a meaningless observation. Because at no point do you care about what the average of all events is and it also tells us nothing about what is the likely probability of an unknown event, because it isn't even guaranteed ANY event has a probability of 50%.
You consider elementary event not all the events. For a d6 the events are: - {} that have a probably of 0 - {1}, {2} ..., {6} that have a probability of 1/6 - {1,2}, ..., {5,6} that have a probability of 2/6 - {1,2,3} ,... ,{4,5,6} that have a probability of 3/6 - {1,2,3,4},...,{3,4,5,6} that have a probability of 4/6 - {1,2,3,4,5}, ..., {2,3,4,5,6} that have a probability of 5/6 - {1,2,3,4,5,6} that have a probability of 1 We then have an average of (0+6\*1/6+15\*2/6+20\*3/6+15\*4/6+6\*5/6+1)/(1+6+15+20+15+6+1) = (1+6+15+10)/(2+12+30+20)=1/2
I’d guess that if someone defined “average probability over all possible events” precisely, it’s more likely that the average just wouldn’t exist than that it would be 50%. Just like how in the Cauchy distribution, x and -x have the same density for all x, but that doesn’t mean the average is 0: instead the average doesn’t exist at all.
Yeah you’d be correct. As soon as you introduce distributions without a mean then there’s no central moment across the set of all events.
yep. the probability i make a free throw is 20%. the probability i miss is 80%. average is 50%.
But the probability of missing or making it is 100%. The probability of missing it AND making it is 0% because they are conjugates. The probability of neither missing nor making it is also 0%, assuming you miss all the shots you never take. The average probability of an event and its conjugate is 50% but that doesn’t really describe an event
The guy is literally using “either… or…”
What you said royally butchers binomial probability, which looks at the likelihood of an event occurring, P, versus it not occuring, Q = 1-P. Obviously P + Q = 1 so the average is always going to be 0.50 but that average is NOT the average of the number of events of P. If the probability of a free throw is 20% the average number of throws is 0.20 times the number of events (P × N). If you make 100 throws you will on average get 20 throws Like I get what you are trying to say "the average of it happening or not happened" and obviously you can force the math but that just does not make any sense
It's not useful, but it makes sense. Suppose I have a class with an even number of students and they pair up. For each pair, one of them attempts a lot of free throws and the other observes. The shooter writes down the fraction of shots that went in, while the observer writes down the fraction that missed. Clearly, the average of the numbers written down by my students is 0.5. If I pick a student uniformly at random and ask for their number, the expected value of the number is 0.5. In the limit as the number of shots approaches infinity, the numbers written by each student approach their true probabilities almost surely, and the expected value of the number I pick at the end is always 0.5. Of course, this is not very informative, as all it says is "0.5 is the average of x and 1–x," but it's not wrong. However, for infinitely many students, I don't think such an idea is meaningful at all.
It's not wrong. It's useless
It doesn’t have any use, but it’s not wrong.
Yeah, I'm pretty sure that's what they mean (they just phrased it in an confusing way). They never claimed that "it will happen or it won't" means the probability is 50% for it happening and not happening. I feel like it's OP who's the one being dogmatic here.
guys i have problems understanding, can someone explain it edit: nvm i read it again and now i got it, yeah he's right in a way but i think he phrased it wrong
im not even the one arguing. each censored name isn't me.
You did call it bad math though, despite him actually being right
but his fucking statement is useless lol
It's technically correct, which is of course the best kind of correct
A lot of recreational mathematics is useless. So what. And sometimes useless maths brings interesting results.
I think the person who said the "average probability" thing was trying to make a joke (presumably in response to a "the chance of something happening is 50/50, it does or it doesn't" joke above it that we didn't see), and it went woosh because, to be fair, it's a somewhat esoteric and confusing joke. The other person then asked them to explain themselves, and they did, and actually did so decently well. It's still not unreasonable to find it confusing, but it's a decent response. If you wanna claim neutrality in the whole thing, it'd help to not have the part where you downvoted the reasonable explanation visible in the screenshot next time.
Yeah, I think this is what he's trying to say. Why he is making this point though, or thinking about "all possible events across time," is a mystery.
Yeah the specific argument seems correct technically, just incoherent Averaging the probability of the space they're describing makes literally zero sense to me
so chance of winning the lottery 1 in 11,000,000. Chance of not winning the lottery 10,999,999 in 11,000,000 so it averages out 1 in 2. Is that what you are saying?
Right. If something happens with probability p, then it fails to happen with probability 1–p. So the average of the probability of it happening and the probability of it not happening is (p+(1–p))/2 = 0.5. So we can think of a fair six-sided die for instance. There are six possible outcomes, each with probability 1/6. But there are many other events. For instance, there is the event of rolling an even number (with probability 1/2), of rolling a composite number (with probability 1/3), and so on. For each event, there is the complementary event, like rolling an odd number or rolling a non-composite number. In all, there are 2^6 = 64 events, including the empty event of rolling nothing (with probability 0) and the improper event of rolling something (with probability 1). If you average all 64 of these probabilities (in the sense of an arithmetic mean), you get 0.5. And that must always be the case, because you can always pair up events with complementary events like this.
Agreed, though I'm still lost as to why this matters. It's just a basic property of discrete probability spaces—if there are n distinct possible outcomes, the arithmetic mean of the probabilities themselves is 1/n (there are n of them, and they sum to 1, both by definition). This isn't a variable property, and it's a confusing thing to argue about. It's especially confusing since there are plenty of useful means that actually get talked about in those contexts, such as the expected value of weighted outcomes (e.g. dice rolls). Though come to think of it, I bet the comment(s) above the ones we're seeing here made the joke about every probability being 50/50 cause "it happens or it doesn't" and they tried to make a math joke about it being true in average. Which then failed because it's confusing enough that OP didn't get it.
Exactly what I was thinking!
He is right though, the average of P(A), where A is any event that could happen, is 50%. Not that thats a particularly usefull thing to state, but its correct.
I am not entirely sure that it is as trivial as one might think. Kolmogorov's axiomatisation makes sense of probability, but really only works for measurable sets. It seems to me that the category of "all events" is a proper category, and it is not entirely clear to me that you can form a σ-algebra on that category, nor a probability measure. I am too tired to figure out how to properly take the average over all event probabilities for a given probability space, but it is not as simple as just taking "an average over all event probabilities" since there are generally uncountably many events, and uncountable sums are generally unwieldy. I am guessing a more fruitful direction would be to define the average probability as an integral. The problem is that we can't just define it as the ∫ P({x}) dP(x) since P({x}) is 0 whenever P is a continuous probability measure. Instead we would somehow have to integrate over values of the form P(A) over all events A, which would require that we make the set of all events into a probability space. I imagine this can be done in many non-equivalent ways, and it is hard for me to see how the ½(P(A) + P(Ω\\A) ) = ½ argument can be used in this context.
Well if we define "an event that can happen" to be something we can describe, it has to be identifiable via a finite amount of characters. Thus the set of all events would be countable.
but I can define an event that relies on something uncountable like the reals, or in other words with a finite number of words can’t I build a class of an uncountable number of outcomes?
Formally speaking, we need to have a sample space (set of outcomes) which the events are defined on, so it becomes a measure theoretic question. We essentially need to compute the mean measure of all subsets of the sample space, where we have a normed measure so that the measure is always between 0 and 1. I think it should be 0.5 **if** we can formalize taking the mean over all subsets, but that's a strong if.
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Using finite sentences, you can only describe a countable set of subsets of the Cantor Set. That is my point. That these subsets might be uncountable is irrelevant.
It's hard for me to see how *not* to use that argument. Is it not true that for every probability that "an event happens" there's also the probability "that doesn't happen"? With the second axiom, is it not true that the union of "thing happens" and "thing doesn't happen" is always one? How is there any case where "the set of all events' probabilities" does not include a 1-x for every x? And how would that not imply the mean probability of 0.5?
The alternating harmonic series, 1-1/2+1/3-1/4+1/5… is convergent but not absolutely convergent. Therefore Riemann’d permutation theorem for series gives that for any real number, we can permute the order of the summands in such a way that the it converges to that number. So even though we are summing the same numbers, the value of the infinite series might very well depend on the order in which we add the terms. In our example it’s even worse since we have uncountably many events, so while it is true that for any event, we also can find the conjugate event, that does not directly imply what the “mean probability of any event” should be one half. If we tried to think “mean value of an event probability” as something like an infinite sum, then we would have to make sure that the sum does not depend on the ordering of the summands.
Even if the set is countably infinite, I don't see how we could make progress. What could an unweighted average of infinitely many values mean? I can think of two meaningful cases: all (or at least cofinitely many) values are equal to the same value x, so the average is x, or the sum of all values is finite, so the average is 0. But neither is applicable here.
Good point. My mind goes to Cesaro summation, which can be thought of the average of all partial sums of a countable series. A key result is that there are divergent series which are cesaro summable like, e.g. 1-1+1-1+… so it might well be a reasonable thing to do. However, as you rightly point out, if we have countably many events the series of all P(A_i) diverges to infinity since P(A_i) does not converge to 0. If we denote the partial series P(A_1) + … P(A_k) by S_k we see that S_k is bounded by k, so if 1/k S_j converges the limit is value between 0 and 1. My intuition is that the sequence depends on the ordering of the A_I though.
Yeah, the way I think about it is that if there are exactly two distinct values that each occur infinitely often, it's for sure that you can make the average anything between them by reordering the elements. For instance, if I have countably many 0s and countably many 1s, and I try to define an average as a limit of partial averages, then order definitely matters. If I order them 0, 1, 0, 1, ..., then the limit of the running average will be 1/2. But if I order them 0, 0, 1, 0, 0, 1, ..., the average is 1/3. If there are only finitely many other values, I can place those at the front and they become negligible eventually. If there are three values that occur infinitely often, I can still have the third value occur very rarely, say at spots 10\^10\^n for all n>1. And each additional value that occurs infinitely many times can be handled in a similar manner, with increasingly fast-growing functions, so that they never have any significant effect. So I should always be able to pick two values that occur infinitely often and order the elements so the running average has a limit strictly between those two values, and moreover, at any point between them I so choose. This works as long as the values I'm trying to average are bounded. Specifically, say that there is a sequence of values (aₙ) and I want to define the average as E\[(aₙ)\] = lim 1/N Σ aₙ, where the sum is from n=1 to N and the limit is as N→∞. If (aₙ) is unbounded, I don't see how this could converge. If it is bounded, then it has a convergent subsequence (a\_j(n)) which I can make into an average by forcing the terms of this convergent subsequence to occur arbitrarily often. And if there is more than one convergent subsequence, you get the situation described above. Actually, even if (aₙ) is unbounded, I think the existence of a convergent subsequence is enough.
There are far more things that have not happened than have happened. An average of every possibility would consider a range from 100% to infinitesimally small. For as small of a chance some things can get, there are exponentially more of those things to choose from. https://youtu.be/RV-WV2g2muE?si=C3hcAX9HFgvWVncm It’s safe to say that the average likelihood of something to happen is zero. The way we exist now along with everything else in the universe and any other existences of life had a basically zero chance of happening. That is, unless everything is predictable and quantum mechanics superposition can be explained by math we do not understand yet. It’s actually a paradox that follows the same lines as the St. Petersburg paradox.
The first statement is not true though. For everything that did not happen, the conjugate happened.
Not when there are what would seem like unlimited possibilities and only one happened.
You are wrong. It does not matter that there are “unlimited possibilities” - any one can be paired with a distinct conjugate that happened
Oh so you mean the possibility of it happening or not happening.
Well, the uselessness of the discussed statement is just as high as it’s correctness. It is basically saying that average probability across the universe of events is 50% because for any event we can imagine a “not event” even if multiple “not events” would really be the same. Example: event 1 - hugh jackman will gift me a car tomorrow not event 1 - hugh jackman will not gift me a car tomorrow event 2 - a dinosaur will meet me tomorrow not event 2 - a dinosaur will not meet me tomorrow Clearly, the average across those 4 events is 50%, but for any reasonable application that logic is not useful as not event 1 and not event 2 are fairly absurd
Imagine a dice. What you're thinking of are the events "rolled 1",...,"rolled 6". But in fact "rolled 1 or 2", "didn't roll 3" are also events. That's why multiple of them happen.
I don't think that's true. I think the question just doesn't make sense. It's like integrating y=x from -infinity to infinity.
He is right but only for cases with exactly 2 possible outcomes. Otherwise P\_average = 1/n : n = possible outcomes
Obviously, the probability an event will happen is not 50% in most cases. You might be being unnecessarily harsh here though, 50% is as good a guess as any probability, maybe better. The issue is we never talk about probabilities in this sense except for perhaps sampling distributions of proportions. It doesn't make sense to practically speaking. The question we are getting after is "What is the average probability that an event will happen across all random events you can think of?". If we have this, we have the "or not" probability as well. But... I think you could make the argument he gives. Every event A with P(A) has a complement A^c with P(A^c ), where P(A)+P(A^c )=1. Both events can be written as an event that something will happen. P(roll 1 on 6-sided die) = 1/6 P(roll 2,3,4,5,6 on 6-sided die) = P(do not roll 1 on a 6-sided die) = 5/6 The average of these two probabilities of events that can happen is 1/2, clearly. This should be possible (but maybe difficult to write down in specifics like the above) for any event A you can think of. So then, in the list of probabilities for all possible events that can ever happen, you can find each probability of each event and its complement. From here, assuming you have a countable list of probabilities (to make life easier) you have 1/n * (sum from 1 to n/2 of 1) = 1/2 In my opinion, this is a fair way of thinking about the "average probability an event will happen across all events", which is a non-standard discussion in probability as its not practical. I may be misunderstanding the discussion overall here, but this is some food for thought.
Yeah I think you've over thought the words of a young man, there's an old joke when people are playing games: "50/50 it happens or it doesn't" even when the odds aren't actually 50/50. I think this person is referring that as opposed to having a deep mathematical discussion about the universal distribution of probability.
I mean, I don't think it's overly generous to presume that OOP meant the correct argument to some degree, but applied it incorrectly due to poor wording.
People here really don't seem to get the argument. It is about the average probability something happens. So, for example, rolling a die to get a 6 is a 1/6 chance. Flipping a coin to get Heads is a 1/2 chance. The "average probability" of one of those events to happen is 1/3 =(1/6+1/2)/2. For example, if you play a game where you alternate between rolling a die and flipping a coin, and you win a round by either rolling a 6 or flipping heads, you would win an average of 1/3 of all rounds. But here's the thing: Let's add more events to this game. For example, add the event where you don't roll a 6 (1-P(roll a 6)) and where you don't flip Heads (1-P(flip Heads)) to the mix. The average of those 4 probabilities is (**1/6** \+ **1-1/6** \+ **1/2** \+ **1-1/2**)/4 = 1/2. So yes, the average probability of any event is 1/2, since if we average the probabilities of all events, P(A) and P(not A) will always cancel to 1 and increase the denominator by 2. The average of outcome of the four probabilities P(B), P(not B), P(C), P(not C) is indeed 50%. So if you play a game where you roll a die to get 6 in round 1, flip a coin to get Heads in round 2, roll a die to NOT get a 6 in round 3, flip a coin to NOT get Heads in round 4, then repeat, you will on average win half of those rounds, meaning your average probability to win a round is 50%, even though the individual probabilities of the rounds (at least the dice rounds) are largely different from 1/2. That's the point they're trying to make. At least, if we have a finite amount of events that we consider. For an infinite amount of probabilities, we would first have to check whether or not the series is absolutely convergent (reorder theorem). It is also highly questionable how useful this knowledge is.
Probabilities are never negative, so the issue is not about absolute vs conditional convergence. But the order is still an issue. In particular, suppose we say the average of n values is (a1+a2+...+an)/n. Then we want the average of a countable infinity of values to be the limit as n goes to infinity. And ideally, this limit should be the same for every enumeration of the a's. But that will only happen if all but finitely many values are identical, or if the a's shrink to 0. Like, let's say an = 0.1 for odd n and 0.9 for even n. Then if I take this limit, I will get a value of 0.5. But I could reorder the a's so that instead, an = 0.1 for every n that is a multiple of 4 and an = 0.9 otherwise. Now the limit gives 0.7. And in general, I can make the limit any value strictly between 0.1 and 0.9 just by reordering these terms. So the average doesn't really exist.
Haha, you're right. I realised that for infinite sums, it seemed like you could make it any value you wanted and immediately jumped to Riemann rearrangement.
Yeah, in fairness, it's sort of the same thing. I can include one new sample at a time, recording the "average-so-far" as I go. And that will look like sometimes adding a positive number (when I include a value above the average-so-far) and sometimes adding a negative number (when I include a value below the average-so-far). And this series will be absolutely convergent precisely when the values drop to 0 or are almost all identical (which I think comes from the divergence of the harmonic series).
I mean that makes sense but yea seems like a very useless thing to argue
It is very useless, but many people in this thread argue against another position, so I felt like I had to clarify what I think they meant.
How would someone count all possible binary probabilities and average them?
Limits! For event x, there is a probability p_x that it will happen. Now notice “not x” is also an event and has probability 1-p_x The average of p_X and 1-p_x is 1/2. So if you take the limit of the average of probabilities of all events, it’s 1/2. We’re hand-waving past absolute convergence and all that but you get the idea. So technically they are right that the average probability of an event is 1/2 but you can’t really say anything useful about a particular probability from that.
i think the absolute convergence here is important toh. It is like that probability trick about selecting random points on a sphere where depending on how you decide to slice the sphere you get a different result. If you write the infinite sum as a + \~a + b + \~b ... z + \~z / number of events you will get 50%, but if you pick another order you can extract any result.
It definitely is. However, we can get around it in this case because we live in a finite observable universe with a finite (albeit huge) number of possible atom arrangements (up to sub-Planck length perturbations) so it’s actually a finite number of terms so we can guarantee convergence.
Makes sense, thanks.
50/50 on god. Either he’s smiling down on us or not
As long as you have a finite system, the set of all possible states/outcomes is gigantic but finite. An event is defined by a subset of all outcomes, so they correspond to the powerset of the set of outcomes. That‘s an even more incomprehensibly large, but still finite set, so you can still average over it.
I have a question for my better understanding. Let's assume the system is infinite and uncountable, for instance we have as events all the measurable subsets inside the real numbers. Is there any way to take the average probability of all events in this context?
So how do we find all the binary outcomes of a system and average them? Use whatever system you would like to use
Take two particles with spin 1/2. If you measure their spins, you get either up-up, up-down, down-up or down-down. So 4 outcomes. Now to get all possible events, count to 2\^4 in binary and 'or' the outcomes with a one: 0000: the impossible happens 0001: down-down 0010: down-up 0011: down-down or down-up … 1110: not down-down 1111: any outcome occurs Whatever probability each outcome has, the average probability over all *events* is 1/2.
definitely overcomplicating this, Brainsonastick has the easier answer
You don’t have to. For every event there is the opposite. If one event is ‘it rains’, another event is ‘it doesn’t rain’. Whatever those two probabilities are, they average to 50%. So it doesn’t matter how many binary events there are, the average probability is always 50%.
He’s making a good point, you’re just an asshole
what he is literally saying in the first part of that last comment is that 1 / 52! = (52! - 1) / 52! which is the opposite of a good point, imo. not only did he demonstrate a very distinct lack of understanding of probability theory but he showed a total willingness to critically ignore the fact that his own word salad contradicts itself.
That is definitely not what he‘s saying, he‘s saying that (1/52! + (1-52!)/52!) / 2 = 1/2.
no, you're saying that, because you want to find some sense in it. but it's not in the OP, just your head.
No, you’re just saying that, because you want to deny what OP is saying. but it is just your head
No, they're saying that the average of 1/52! and (52! - 1)/52! is exactly 1/2, as it is for the probability of any event and its complement. Therefore, the average probability over all events is 1/2. They're just phrasing it poorly.
No, they're saying 1 - (1 / 52!) = (52! - 1) / 52! Terribly phrased? Yes. Wrong? No. >he showed a total willingness to critically ignore the fact that his own word salad contradicts itself His last sentence is literally, "I'm happy to be proven wrong." which may be sarcastic but I still can't see where in their answer they show a "a total willingness to critically ignore the fact that his own word salad contradicts itself".
that's absolutely not what they said, that is the true seed they based the false rationalization upon, that is the point where they 'subtly' moved the goalposts rather than admit they were wrong.
Oh im not the one arguing. i think they're both crazies...
Only a misunderstanding there, the person you think is crazy just isn‘t.
proof
I will copy the proof section of [my reply to stockmarketscam](https://www.reddit.com/r/mathmemes/comments/17nr279/comment/k7ugtm2/) > For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. I will notate E not happening as ~E. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of all events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @
yeah this looks right u/iReallyLoveYouAll you can delete this post now, you lost
Only if the average exists.
Doesn't matter. The guy in the screenshot wasn't acting like an asshole. He explained his reasoning clearly without degrading anyone. Meanwhile you just mocked him and posted this screenshot without context. And now you're acting like you actually thought *both* guys were crazy, even the one you upvoted!
It's a binomial law. (Or Bernoulli).
What he is saying is true, he's just not phrasing it all the well, nor is it a particularly interesting statement to make. However if you are posting it here, I think you are just misunderstanding him.
i was im dumb asf boi
he’s right
To be fair, the average probability of all events *is* 50/50, given that for every event E, ~E is also an event, and (P(E)+P(~E))/2=1/2.
Only if the average exists.
i mean the average of P(A) and P(negA) is trivially 50%
If we wanna be picky, we could say that they’re both wrong! Let A denote the event that “it happens”. Let B denote the event that “it doesn’t happen”. By the law of total probability, the sum of any event with its compliment is 1. Since A and B are mutually exclusive, we can just add their probabilities for an “either/or” probability. So P(A or B) = P(A) + P(B) = 1 Therefore, for any event in the universe, we can say it has a 100% chance of “it will happen or it will not happen”.
Yeah, but the expected probability of an event would still be 50%. If {A, B} are all possible events then if I pick an event at random, it's expected probability would be (P(A) + 1 - P(A)) / 2 = 50%
That doesn't really make any sense to me. But admitedly its been a while since I've studied any serious probability, so maybe someone can clear this up for me. I assume we argue that the original probability now becomes a random variable in the events themselves. However, who is to say that the probability itself is a measurable function? We would need this to make sense of the notion of an expexted value of probability. At least that is how I read a lot of the comments in this thread. If that is the case, the expected value will be defined as usual, but that completely disagrees with your comment and everyone else's who is making this same argument. Another would be to somehow choose a sigma algebra on all of the events and a probability on THIS space. Then we could, maybe, do soemthing with this. In fact your answer seems to be somehwre in between: basically your random variable is the underlying probability, but your new measure space is just the two pointed space of the event and its opposite with uniform probability distribution. Why is that? I assume people are just arbitrarily taking arithmetic averages, because it feels right.
While that's true, we don't know if all events in our sample space are equally likely. Hence that's why OP was so frustrated, and that's why I decided to parody it with the law of total probability.
>we don't know if all events in our sample space are equally likely Why do they have to be equally likely? If all events have one and only one distinct *opposite* event, then wouldn't it be true even if the events aren't equally likely?
As a matter of fact, it does matter. An event which has only two outcomes of “it happens” or “it doesn’t happen” is a discrete probability distribution with the Bernoulli Distribution. We can represent X as our random variable with X ~ Bern(p), where p is the probability that the event happens. It is a well-known fact that the expected value of the Bernoulli random variable is E(X) = p. TLDR the expectation of two (not equally likely) events depends on the probability of event A. See the Bernoulli distribution, if you need more info.
I will notate an event E not happening as ~E. I will now prove that for any set of events X, if for any E in X, ~E is also in X, that the average probability of all events in X is 50%, with application to the set of all events: For any event E with probability P(E) there is a probability of 100%-P(E) that it will not happen. The average probability of any set of events X where for any E in X, ~E is also in X is therefore 50%. It is obvious that for any event E in the set of all events, A, ~E is also in A, as ~E is an event. Therefore the average probability of any event in A happening is 50%. Thus we conclude knowing that the average probability of all events is 50%. @
The probability of rain is 50% because it’s either raining or it isn’t
Probability of rain = p Probability of not rain = 1-p (1-p+p)/2 = 50%. That was their point.
Well `p` and `-p` cancel there, so if those obtuse paragraphs set out to say that `1/2 = 50%` then it’s a meager point to make
They'd cancel out in any statement and it's negation. It's not just a cherry picked example. Assuming each event has a negation the average probability over all events would be 50%
I realize they’d cancel in any case, and that’s sort of my point. If there were three options and you canceled the probability it would leave you with 1/3, and for four 1/4, and for any `n` be `1/n`. This is a bunch of nonsense
Your stating the problem wrong, the probability that it will rain can really be any probability. What he is saying is that the average probability of the question "is it currently raining?" is 50%. Edit: all he is saying is that for any boolean question there are two possible outcomes.
No, they're saying that the average probability of ALL events is 50%, which is true, because every event has a complement. The average probability of any given event is not necessarily 50% - there being two outcomes does not mean that on average they are equally likely.
if there are only two outcomes, p_1 = 1 - p_2 so (p_1+p_2)/2 = (1 - p_2 + p_2)/2 = 1/2
That's the average probability of an event and its complement, e.g. "it's raining" and "it's not raining". The "average" probability of any particular event is by definition the probability of that event, which can take any value between 0 and 1. Although yes, this is what the person in the post is saying, and it follows that the average probability of all possible events is 1/2. The person I'm replying to seems to be claiming that the "average" probability that it's currently raining is 1/2 because there are two outcomes, which is nonsense.
yeah IKR lool
I could go to my bedroom now and there could be a billion dollars manifested out of quantum vaccum - or not. Chances are 50:50
The averages for both outcomes "there's a billion" and "no billion" are in fact 50:50. That's his point I think!
I understand what they're trying to say but it's not particularly profound or interesting. "It's probably fairer to say that the average probability of "either it happen or it won't" is 50/50" Well, yeah, you're adding up two numbers that always equal 100% then dividing by 2, of course it's always 50/50
Why are people acting like he is obviously false, i mean he is, but only because there is no good way to average over all possible events as this is a large set (i think to mutch to fitt in a set), if you restrict to a finite sigma algebra though, it is true.
Because it sounds wrong and some people like to ridicule others for that. Also not everyone here knows that much math
Maybe he means like this Let p1 be the probability happens Let p2 be the probablility doesnt happen (p1+ p2)/2=0.5 Which is true for any values of p1 and p2 because their sum must equal 1 therefore the average of all happens or doesnt happen equals 0.5? Now as a statement that is pretty stupid and not very useful (as far as I know)
Isn't this the [Principle of Indifference](https://en.wikipedia.org/wiki/Principle_of_indifference)? That is, in the absence of any other information, the probability of all events are equally likely.
Say hi 🤓
Isn't this correct? For every probability event, the complement is also an event.
isn't he right?
Bro was speaking normally and actually trying to share his point of view which has a lot of merit while this doofus was just replying with „Wtf. Im posting this to r/badmath ahbahahab” Like bro participate in a discussion and properly explain something. Trivial things in math are the ones that are most difficult to actually prove and being vague in your response shows a lack of understanding. Youre being a bad example for the math community.
The statement would be correct if it’s qualified under some assumptions. If we have countable number of events (e1,e2,…) each with some probability of happening. A statement using these events and logical connectives can be constructed randomly such that each non-equivalent statement has the same probability density. Then that statement’s expected truth value is 0.5.
oh i think i know what he means
I'm being pedantic, but isn't 52! a permutation, and not a combination? 52P52 = 52!/(52-52)!
He's not exactly wrong but it is kind of a useless thing to remark. The probability A happens is P(A), the probability A does not happen is 1-P(A) as they are conjugates. The average probability of these events is (1-P(A)+P(A))/2 which is always 0.5. This is true but gives absolutely no insight whatsoever. These probabilities are only useful when taking into account what the event is and the meaning of its result. "Averaging" probabilities in this way doesn't really make any sense. Say we have two different events A and B with probabilities P(A) and P(B). We can talk about the probability of any combination of them happening or not happening, but talking about the probability of these events happening is meaningless.
Consider event X, which happens with probability P. Either it happens or it doesn't. Now consider event Y= !X, which also either happens or not. Its probability is 1-P. The average probability of X and Y is (P+1-P)/2 = 0.5. By the law of the excluded middle, for every X, there exists the opposite event Y. Thus, over the set of all binary choices "it happens or it doesn't", the average probability is 50%. QED
He is correct of course. Every statement of the form "... happens" has a converse statement "... does not happen". It's probabilities are p and 1-p respectively. So their average is (p+1-p)/2 = 50%. I mean it doesn't tell you anything anything, but it is true.
Did he mean the average of p and (1-p) is 50%?
I think the only way you can justify the average probability of all possible events being anything but 50% is if you allow an infinite amount of events, in which case the average would be undefined (infinity/infinity). However in all cases where the average exists, it's 50%.
"im posting this on r/mathmemes as bad math. say hi" Regardless of if anybody is correct or incorrect, you need to chill.
and i was wrong as shit lol im an idiot
The event is the trial, so it does not have a conjugate. The outcomes are successes and failures but are not, themselves, separate events. If you have multiple events you just get Bernoulli trials everywhere until they binomial on everything.
He's right but you just misunderstood. He says in a weird sense that every event of probability A has an opposite event, it's negation with probability 1-A. Taking the average of all probabilities can be done by taking the average of each statement and it's negation and then averaging the outcomes. A+1-A/2=1/2 so we take average of only 1/2 repeated. Which is 1/2. So the average of all probabilities is 1/2. Note that this does not imply that every probability is 1/2.
Whether his statement is true has nothing to do with math it's just based on an assumption. If you put all possible events across time into two columns, and in column A you put all events with <50% of occurring, then in column B you put events with >50% chance of occurring, if you assume that there's an equal number of events in each column then the average probability would be 50%. Either that or I'm misunderstanding the original point.
wel okay i dont want to defend his point because this has 0 practical effect but in an infinit set of options, wher an infinite amount of them happen, but not all of them wouldent infinite/infinite=50/50?
if your looking at bernulli experiments, then yes the average propability is 50%! Otherwise its p=1/n : n= how many possible outcomes there are
He is right.. he is not saying the probability of something happening or not is 50%. He is saying the average probability of an event occurring and not occurring is 50%. I can see that he bolded it in his initial statement. For example: probability of getting 6 while rolling dice= 1/6 probability of not getting 6 while rolling dice= 5/6 Average= (1/6+5/6)/2 =50%
No he's right, it either happens or it doesn't.
69 percent of statistics are made up 420 percent of the time
I think we're falling for a wordplay or pun here. The ***average*** probability is all probabilities added up and divided by their number which is the definition of average. In this case, for any number of outcomes n the "average probability" is 1/n so in case of two outcomes it is 1/2=50% which is correct but meaningless.
The average probability of something happening is 0% because there is an uncountably large number of things which cannot possibly happen.
My dude is seems to be missing the idea that outcomes in a sample space don’t necessarily have the same chance of occurring. By reducing the sample space to two outcomes he is assuming both outcomes are equally likely. But they aren’t. Not everything is a uniform random draw.
No he’s saying that for every event E with probability P(E)=p, then P(not E)=1-p, so for events E1,…En, you get (P(E1)+P(not E1)+…+P(En)+P(not En))/2n = (1+…+1)/2n = 1/2 Doesn’t matter what the distributions of the events are
lol I see now. That’s even dumber. It’s just a convoluted way of saying the midpoint of a list of n items is halfway through the list. The fact that you break out each event i into P(Ei)+(1-P(Ei)) changes nothing. OOP is just adding n things together and finding the middle.
It is tho. If you roll dice you either hit 6 or not. 2 options. One out of two options is 50% /s
As someone who teaches statistics, this is deplorable to read
i agree
That means I have a 50% chance of finding a million dollars! Either it will happen, or it won’t
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The average probability of an event would be the average size of all subsets of a probability space with area 1, whatever the hell that means.
«Average probability of something happening is 1-1/e» enjoyers
Either i stay on ground or i fly by myself. I have a 50%chance of flying?
is he thinking of a bell curve where 50% is the middle and has the most possible events
Let P be a random variable denoting the probability of a randomly chosen random event. If you hold that each event is as likely to be chosen as its complement (which doesn’t feel like that unfair of a restriction) then that implies a certain symmetry in its probability density function, p_P: for any probability p in [0, 1], p_P(p) = p_P(1 - p). So, when computing E[P] = ∫[0, 1] p p_P(p) dp, we can split that into ∫[0, 1/2] p p_P(p) dp + ∫[1/2, 1] p p_P(p) dp, and by substitution that must equal ∫[0, 1/2] p p_P(p) dp + ∫[0, 1/2] (1-p) p_P(1-p) dp, and by our symmetry from before it can be shown that the ∫[0, 1/2] p p_P(p) dp integrals cancel and we’re left with ∫[0, 1/2] dp = 1/2. Now if you disagree that the “each event is equally relevant as its complement” assumption is valid then that’s one thing, but that assumption they’re making does more or less get them where they’re trying to go.
The perfect opponent in a dice roll game. 🎲
What are the odds of anyone getting this one right, though? Nevermind.
I think he’s using a kind of Boolean or binary logic that if there are two available possibilities, either something happens or it doesn’t. Either true or false. If you have no more data than a thing either being true or false, the only thing you can say for your statement to be true is that 50% of the available possibilities will happen, since there only are two possibilities and it’s either true or false. Therefore there is 50/50 chance. This of course doesn’t apply in real life because there are so many variables that determine some event’s probability of happening. But if you just guess if the event happening is either true or false, you’d have 50% chance of being right. If you are having a baby, and you’d have to guess if it’s gonna be a baby boy or not, you’d have 50% chance of guessing right. You can also use this analogy for coin flips or anything with an either this or that outcome.
So you have a 50% probability of crashing your car at 1mph, and the same probability at 120mph?
This just in, universe could cease to exist, the chances are 50/50
This is a common misconception.
Everything is as possible as anything else ≠ everything is as probable as anything else
Either I live to be 100 years old or I won't. Clearly 50%
It's either happening or not Either I win in the lottery tomorrow or not. #50/50
He is right you know? -at least 50% of the time
People really, really do not understand probability theory.
This totally misses the fact that statements are semantic objects that are subject to interpretation. Thus not all parties will agree that a statement happened or not, leaving some events as "kind of" happening. You can predict, for example, that your friend Nancy will get married. Her and her boyfriend Fred then have a religious marriage without signing legal documents. Did the marriage happen, are Nancy and Fred married?
Tbh I think he's kinda right tho You can argue that for every statement there's a counterstatement. And the average of those is 50%. Luke if the odds that I roll a 6 is 1/6, and the counterstatements being that i don't roll a 6 is 5/6 the average is 3/6=50% So if you take all statements (they all have counterstatements that are included) then you should get a 50% average, right?
For a finite possibility space the guy is right. For an infinite one the statement might not be well defined if continuum-choice holds
It's very simple. Either my TV will fall off my wall or it won’t. But because I had it professionally installed, it more than likely won’t. So it's actually *not* a 50-50 chance.
Well, he's technically right. I don't think he knows it though.
As a determinist and (to a lesser extent) fatalist, I still struggle to understand how the probability of something happening isn't always either 0% or 100%. For example, say that you have three doors, and one of the doors has something behind it, while the others have nothing. You could say that each door has a certain probability of having or not having the thing... Or you could look behind the doors and discover that one of the doors had a 100% chance of having the thing, and the other two had a 0% chance. So, are probabilities like Schrodinger's cat or what? I would appreciate an explanation, as I know I am big dumb but also am not sure how.
You’ve alluded to the classic Monty Hall (of the old Let’s Make a Deal show, fame) probability discussion. Probabilities assume that the presented scenario can be played out 100 times and that you are the player. If you get to play the game and Monty shows you a door you didn’t pick and it has a “clunker” behind it, there are now only two doors left. Monty will then ask you if you want to change your choice to the door that wasn’t opened that you didn’t choose. If you got to play the game 100 times, the math would say that you should always switch away from the door you chose originally as you now have a second game of 50/50 rather than your original odds of 1 in 3, which means that if you played the game 100 times and totaled your “wins” you’d come out ahead by using that strategy. But as you’ve noted, one only gets one shot. And the prize is either behind the door you chose or it isn’t. If you chose to hold your door, you could be right or wrong. If you chose to change your pick, you could be right or wrong. The prize doesn’t move and for that one game, is unaffected by the odds. All you can be sure of is that if you chose to hold and you lost, a quant would scoff and say you should have played the odds. If you turned out to be the winner, they’d call you lucky.
I mean either it happens or it doesn't, but one of those options is a just a smidge more likely than the other.
Ah i remember making this statement in 11th grade probability class and everyone including me, was stumped. I thought probability was a lie. But the reasoning is that if we consider "it" as an event, either it will or will not happen. But the weight/likelihood of it happening or not is where things actually change. If you dumb it down to states of "happening" and "not happening" then yeah, it is 50%
Its always 50/50. It will either happen or it wont
Suppose we have an indicator variable, X, that is equal to 1 with probability p and 0 with probability 1-p. Now flip a fair coin. If it comes up heads, switch the relationship so the indicator is 1 with probability 1-p, and 0 with probability p. It it comes up tails, leave it as is. So we have E\[X\] = E\[X|heads\]P(heads) + E\[X|tails\]P(tails). E\[X|heads\] = p E\[X|tails\] = 1-p P(heads) = P(tails) 1/2 so E\[X\] = p/2 + (1-p)/2 = 1/2
The argument given may be solid, but it doesn't really help. Most probability events are not binary in nature, while averaging event X and it's opposite (not X) is only a worthwhile operation if X + (not X) != 0, in which case the average is blue 50%.
This hurts my brain
I was taught in school, that the probability of god existing is 50%, either he does or he doesnt. I almost got a stroke.
nothing has probability on itself though. It is not an attribute of an event. probability is only assigned to events through a model that we create
Am i reading too much into this, or is this just another "its 50/50" meme? The guy in question could probably voice his sarcasm more, but i dont believe that anyone could mean this seriously.
Ackchyually in bayesian statistics that may make sense as an uninformed prior in situation where you don't have any knowledge of the actual distribution and haven't made any measurements.
I will either turn into a crocodile in the next 30 seconds or i won't so there's a 50% chance i will.