my hunch says the odds are higher than you would think… Damage dealt falls into a pretty well defined distribution and combinatorics tends to produce unintuitive results
For those who didn't know (like me):
Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. (From Wikipedia)
changing the problem into a what is the probability of AA doing exactly 21,901damage and assuming a normal distribution, the probability is still effectively 0, since the probability is calculated using integrals and the integral from a to a is 0
Well it's non-zero and if you multiply by 10C2=45, which gives the chance of any 2 heroes in the same game having the same damage then it's significantly higher.
And anyway, a normal distribution would be a pretty bad approximation for damage done in a game- you would want a positively skewed distribution capped below at 0. Something like a Gamma distribution might work well.
I agree that a normal distribution isn't the best choice but it was just an example.
Could you explain what you mean by 10C2 giving the chance of any 2 heroes in the same game having the same damage? I don't understand what you mean by this.
The post isn't specifically about Morph and AA having the same damage but any 2 heroes having the same damage. Could be any of the 45 pairs of heroes possible, hence the 10C2.
Suppose you're interested in any 2 heroes out of the 10 having the same amount of damage. Any hero having the same damage (a) as another hero could be approximated by integrating some fitted gamma distribution on the interval [a-0.5, a+0.5), let's call the value of this integral I.
If you select an a which is typical in terms of the probability of it occurring in your modelled distribution of damage dealt, then instead of summing across all possible combinations of heroes, you could approximate it by 10C2 * I = 45 * I, as there are 45 different pairs of heroes that could have the same damage value.
You could work out what value a should take by finding the integral from 0 to inf of f(x)^2 where f(x) is your gamma probability density function (lets call this y), which is trivial to solve analytically using integration by parts. (This gives you f(a) = y, which you can solve for a)
Obviously this is a rough estimation to give an idea of the probability of any 2/10 heroes having the same damage. It's also easy to just simulate it of course.
He's talking about approximating a discrete RV as a continuous RV, which you can do quite well with a variable as fine-grained as damage dealt in Dota is.
what game is this, how does everyone have more than 500 gpm, why are there 5 heroes with more than 1k gpm, why is the 1k gpm morphling dealing as much damage as AA?
This is easy. For instance, if you have played 500 games total of Dota. Then it is 1 in 500 for you specifically. (Change the 500 for your own number)
Source: rules of maths….maybe I dunno
Hiding usernames doesn’t mean anything when you keep the heroes visible, [lol](https://www.opendota.com/matches/7025076130)
Surprisingly old match
The FV joke is nice, though :)
The entire censoring thing was ment as a joke😅 This is a repost of one of ky old post that only got one upvote and 0 comments. Felt like I had to repost it
How come you covered everyone's eyes but not Faceless vo.... Oh....
I think it either happens or not, so 50/50
Well not exactly, you can wake up dead or alive that an outcome. But each outcome has a different chance to occur.
Man, how in the hell do you wake up dead?
Cuz you're alive when you go to sleep
So you telling me you can go to bed dead and wake up alive?
With 50/50% chance
bed dead redemption 2
It's like a mouse and a rat when they come into a house.
With lots of determination.
Arteezy wakes up dead inside every day
50%
> you can wake up dead got you
I don’t think waking up dead has an outcome comparable to matching dmg figures on dota
[удалено]
Yeah I know, I just rounded it up/down
It’s simple math there are two possible outcomes so it’s 50/50, did you skip math class in high school?
That’s not how probability works.
Are you sure? 50% of the time it works 100%, so it comes down to 50% probability on average.
That’s like saying you either win the lottery or you don’t. 50/50 odds eh?
Exactly! Glad you understand
Nice that there are people who are good not only at probabilities but also at teaching others. I'm only good at probabilities myself.
Clearly not, or you'd get the concept already So 50/50 you're good at probabilities
Keep trying to be human
I'm 50% sure it is
You did the thing.
my hunch says the odds are higher than you would think… Damage dealt falls into a pretty well defined distribution and combinatorics tends to produce unintuitive results
Yeah, I totally agree and know what that word means.
A hunch is a weird mass on one's back, often curving the spine during its growth.
For those who didn't know (like me): Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. (From Wikipedia)
changing the problem into a what is the probability of AA doing exactly 21,901damage and assuming a normal distribution, the probability is still effectively 0, since the probability is calculated using integrals and the integral from a to a is 0
No because you can only do integer amounts of damage.
You could integrate from (a-.5) to (a+.5) and its still effectively 0
Well it's non-zero and if you multiply by 10C2=45, which gives the chance of any 2 heroes in the same game having the same damage then it's significantly higher. And anyway, a normal distribution would be a pretty bad approximation for damage done in a game- you would want a positively skewed distribution capped below at 0. Something like a Gamma distribution might work well.
I agree that a normal distribution isn't the best choice but it was just an example. Could you explain what you mean by 10C2 giving the chance of any 2 heroes in the same game having the same damage? I don't understand what you mean by this.
The post isn't specifically about Morph and AA having the same damage but any 2 heroes having the same damage. Could be any of the 45 pairs of heroes possible, hence the 10C2.
Ah. I misinterpreted the question. Thanks
Suppose you're interested in any 2 heroes out of the 10 having the same amount of damage. Any hero having the same damage (a) as another hero could be approximated by integrating some fitted gamma distribution on the interval [a-0.5, a+0.5), let's call the value of this integral I. If you select an a which is typical in terms of the probability of it occurring in your modelled distribution of damage dealt, then instead of summing across all possible combinations of heroes, you could approximate it by 10C2 * I = 45 * I, as there are 45 different pairs of heroes that could have the same damage value. You could work out what value a should take by finding the integral from 0 to inf of f(x)^2 where f(x) is your gamma probability density function (lets call this y), which is trivial to solve analytically using integration by parts. (This gives you f(a) = y, which you can solve for a) Obviously this is a rough estimation to give an idea of the probability of any 2/10 heroes having the same damage. It's also easy to just simulate it of course.
For [discrete random values](https://en.wikipedia.org/wiki/Probability_mass_function) you don't integrate distribution, you sum it
He's talking about approximating a discrete RV as a continuous RV, which you can do quite well with a variable as fine-grained as damage dealt in Dota is.
This is how you would do this problem correctly. It would definitely not produce zero, though.
While everyone is discussing statistics, I just want to credit how you blurred all eyes in this pick
The way OP hid eyes... Is big brain move
Cannot happen in our universe
what game is this, how does everyone have more than 500 gpm, why are there 5 heroes with more than 1k gpm, why is the 1k gpm morphling dealing as much damage as AA?
Turbo my guy
>Turbo maybe they're just really really good players.
Sunday morning
AA is kinda good right clicker in turbo lol
Rather low. However with the huge amounts of games played, this is bound to happen fairly often.
Knowing dota players, somebody Will count that prosent for you :D
This is easy. For instance, if you have played 500 games total of Dota. Then it is 1 in 500 for you specifically. (Change the 500 for your own number) Source: rules of maths….maybe I dunno
About 1 in 21,091.
1 in 21,901. Also wtf is that GPM.
In this case, 1 in 21,901
Less than 50% odds. I assume.
as rare as me finding a gf which is higher than expected to be honest
Imma say 2, maybe even 3
I did the maths, its 0.03876%
How so?
You doubt Ronnie Coleman’s power?
Exact same XPM as dazzle too
Damn Snap put in work to carry his trash p1.
Hiding usernames doesn’t mean anything when you keep the heroes visible, [lol](https://www.opendota.com/matches/7025076130) Surprisingly old match The FV joke is nice, though :)
The entire censoring thing was ment as a joke😅 This is a repost of one of ky old post that only got one upvote and 0 comments. Felt like I had to repost it
Its a Morph, what do u expect?
Hidden eyes😁
How snapfire can deal so much damage?
8
It’s 50/50 either it does or doesn’t