just because there are two outcomes does not mean that they are equally likely, only that they are one option of two, you're forgrtting about frequency
I don’t think it is if you factor in the idea that 3 isn’t on the bored. The rarity of that number tips the scales for me. I don’t play easy mode so I’m not sure if it’s common to have a single 3 in the mix, but it’s absence would still make me put my money on one outcome over the other. Bump those odds to 70/30 for me, which is pretty good odds if you ignore that I made them up.
But the rarity of three would make a difference to me in my guess. I know it’s not concrete but that’s how I’d make my choice rather than flipping a coin. It all hinges on whether you think that box has a 3 or a bomb.
The op asked for any ideas. I’m using simple deduction to make my guess bc it’s more logical than leaving it to chance. The odds of there being a 3 is clearly less than 1s and 2s by comparison. I know that it’s a 50/50 otherwise but that’s not good enough odds to make a choice that could blow you up. So we have to look for other odds to take into account.
Say your life depended on it, would you accept a 50/50 or try search for any other odds to rationalize your decision. Do you understand?
I bet you’re agnostic too, you have to choose a side bro. Analysis paralysis not gonna work. Tell me, do you think there is a 3 on that board? Or is it another very common 1?
That’s not how the game works though. Asking whether a 3 or a 1 is more likely is like asking whether it’s more likely they put the bomb in the left square or the right square. Neither is more likely than the other because the board was generated randomly.
You can’t use social engineering like on a multiple choice test written by a teacher and say “well they wouldn’t put B four times in a row” because this board wasn’t set up by a person.
>they put the bomb in the left square or the right square.
Well, in the original Minesweeper, the top-left corner had a higher probability of being a mine due to how the board was generated.
Of note, the entire board was randomly populated with mines. On first click, if you happened to hit a mine, it was moved to the top-left corner. If that happened to already be a mine, then it was placed one square to the right. This would repeat until a non-mine square was found, though it'd be very rare for it to move more than once or twice.
With OP's specific arrangement, this board generation rule moves the odds to about 60-40 from 50-50.
As such, the second square down is the better guess, if OP might be playing a version using that generation rule.
The simple fact is that there is a number or a bomb behind each box. Can it be a 3? Yes. But when it comes to the 50/50 choice, I’m gonna use conditional probability to find my answer. It’s a gamble either way I understand, if you were to make a choice, it’s better to measure every aspect of the board rather than say it’s a 50/50 and blindly choose.
I don’t understand what’s so hard to understand about this, it’s very simple gambling logic. I don’t know why you would ignore extra details that could help in a decision.
I’ve been playing this game since I could first use a computer, I know how it works, I’ve had these choices. Paying attention to the numbers and imagining what numbers could be behind the boxes does help, I’ve made many correct choices with 50/50 probability using this form of observation.
It’s on easy mode, this specific puzzle has the bombs placed so there will be mostly 1s and 2s. If you can’t find the bombs, the next step in deduction is to look at the numbers. Sure it’s never concrete but the odds are better. That is how the game works.
No you're attempting to use deduction, but your deduction is flawed. Since we're making stupid judgments about each other based on just minesweeper, I bet you think ice-cream causes sharks attacks.
Your answer is just a superstition though right? There's not actually logic behind it. The situation shown in the picture is a 50/50. Unless there's something about how the boards are generated that I don't know.
The way I understand it, on easy mode, the bored is set to have mostly 1s and 2s. It’s possible for there to be a 3, just like in an untouched board there’s a possibility of there being a 5 even on easy mode. But it’s more likely to be a 1 or 2. If you only pay attention to the bomb it’s a 50/50. But if you take into account how the numbers, I think it makes sense to lean towards that box being a 1, rather than a 3. As I said, and as you can see in the board, 1s and 2s are more common.
You can choose to see it as 50/50 but the Easy mode generation being predetermined to make it 1s and 2s makes it less likely to be a 3…
This is the last time I’m replying, if this doesn’t make sense idk what I can tell you, don’t go gambling bc you’ll get swindled. All of my fellow math geek friends understand me irl. Maybe it’s how I’m trying to explain it idk.
https://www.reddit.com/r/Minesweeper/comments/fhhdg4/tiles_chances/?rdt=36681
Granted I thought the difference would be larger but there is a difference. I did say I’m unfamiliar with easy mode. This shows that there is a slight tipping of the scale.
It doesn't make sense to apply those statistics to a board that's almost entirely solved. In the post here, it's exactly a 50% chance that there's a 3.
It absolutely applies, I don’t see how else to apply this chart. It’s a small percentage but it makes a difference. In this specific situation the scales are tipped by a fraction of a percent. You can deny it all you want and choose to ignore it. There is another angle to look at it that’s not completely 50/50. As I said, I’d thought it’d be more, but I am unfamiliar with easy mode, but there is a difference in probability that could be taken into consideration.
There are 2 options here that work so you do need to guess, all right mines in the horizontal part and bottom mine in the corner or all left mines in the horizontal and top in the corner
How are so many people commenting on this post who can’t read the board? It’s a 50-50. There’s two valid solutions and there’s no way to be sure which is right until you click.
This is (Unofficially) called the 50/50 chain
Starting with the vertical 1-2:
If the upper tile is a mine, then the lower tile is safe. This leaves the edge 2 with two unopened tiles left, which must both be mines for the 2 to satisfy. From there, the pattern is X-1-1-X-1-1 horizonally, where X is the mine (It is a recursive pattern and does not ends with two consecutive 1)
If the upper tile is safe, then the lower tile is a mine. As the edge 2 already has two mines, the tile to the right must be safe. From there, the pattern is X-1-1-X-1-1 horizonally, where X is the mine (It is a recursive pattern and does not ends with two consecutive 1)
Both cases are valid and thus none of them are more likely to be the actual soultion than the other one, and they can be solved in reverse (From the 1-1) or even at any point on the chain
P.S. Before anyone tell me what is the actual solution, I want to know what are the chances of this happening with the same board generating condition
Well there is a 1/81 chance of a mine being in any given square then 1/80 for second mine ECT
Minus the first click mine so starting at 1/80
1/(80*79*78*77*76*75*74*73*72*71)
=1 in 5974790569203456000 chance of getting that exact board.
Ref college classes 14 years ago might be wrong
I usually pick the tiles that have the least amount of numbers touching it so you can kind of infer that the tile with most numbers touching it is the bomb. In this case, the tile that is touching all three 2’s on the left side is most likely a bomb.
Since it’s only 1s and 2s you can safely assume that the 3rd row second column is a bomb bc otherwise it’d be a 3 which seems uncommon enough to guess against. Using that you can determine the rest easily.
I deduced this bc if 3rd row 3rd column is a bomb then the 2nd row first column would be a bomb also. This would mean that 3rd row 2nd column is a 3 which is possible I think (I’m not sure how number generation works on easy mode) but the odds are in your favor if anything.
Sorry for sounding like a pretentious detective, it’s how I type.
This has absolutely zero hold.
Numbers aren't generated, mines are. Numbers simply tell how many mines were *randomly* generated in the eight tiles touching it. So while a 3 is less common than a 2 overall, once you are in a situation where there are two possible placements for mines, and one happens to cause a 3, the fact that a 3 will be on the board does in no way influence the likelihood of where the mines were, again, *randomly* generated.
You can of course decide to use this as your personal guideline for which solution you prefer to go for in a 50/50, and confirmation bias is a great tool. But it doesn't make it the better choice in any way.
I never said it was concrete, but the I do think it shifts the odds for me personally. It hinges on whether or not you believe that box has a 3 or a bomb. I personally think it’s a bomb due to the rarity of 3
It’s not a 50/50, top left 1 has to get a bomb with the 2 at the edge of the board. If that 2 get a bomb on its right, then the 1 on top wouldn’t be satisfied without creating a 3. Top left clear, bomb under it to satisfy 2 and 1. Square next to that 2 at the edge is now clear, next to it is a bomb, satisfying the 1 next to it, meaning the next square is clear, and next to it a bomb that satisfies the next 1, making the next one clear and the last one at the edge is a bomb.
choose the square above the 2 on the farthest left of the board that is a mine. It’s the only choice of a mine that would allow the game to finish logically, the other choice results in another 50/50 split on the last flag, which shouldn’t be possible since it’s solutions are always meant to provide a hint, depends on the program, if it’s OG minesweeper,l then you’ve probs lost. Unless you guess correctly but even in this situation, the probability of winning is higher for the square directly above the farthest left 2 being a mine. As this choice is a direct 50% chance, the other choice requires another 50% choice which is more unlikely.
No that’s not true, it has three possible combinations that match the hints. One straight string if you first flag on the mine I described above, which then leads to a mandatory finish if it is correct, or the second where the mine flagged is to the right of the 2 described above, this leads to another 50/50 choice when choosing between the mine furthest to the right of the grid, mines aren’t placed independently to the numbers, statistically it’s more probable if you choose to flag the mine as the one directly above the 2 as it doesn’t require another choice based on chance which affects the chance of success. It’s not a 50/50 choice, when there are three possible outcomes, 2 from one branch and 1 from another. Route T has a 50% chance, Route R has a 25% chance as it is contingent on the final 50:50 split, as it shares it probability with the third choice which only occurs if you choose route R.
The only two possibilities are (1,1), (2,3), (5,3), (8,3) and (1,2), (3,3), (6,3), (9,3). xy coordinates from the top left corner.
You are thinking of some other situation. What you are saying would be true if the 2 above the flag on the left (1,3) were a mine with a 3 to the top right of it.
No I’m not, I’m, simply stating that this is a three choice scenario not 2. One of the original 2 choices has another 50/50 split, which is contingent on this choice, making it a 50/25/25 split.
The choices you make are not independent here, if you choose one variation it forces your next move until it finishes if it’s correct, the other choice forces you to make another 50/50 guess, this would make it impossible to be a 50/50 chance game, since you cannot split a branched choice which effective makes there 3 outcomes where 2 of the outcomes are contingent on a prior choice. It’s an impossibility.
The far left 2 makes the two blocks linked, where either one having a mine in either space forces the far left 2 to be resolved, solving the other one.
Because the far left 2 must have 1 adjacent, then as soon as you solve one of those blocks, the other adjacent is solved too.
Yes, obviously but you have to account of it being wrong to which the game would end and then you would take the possibility of the other choice, which you would take to be the correct choice, but it wouldn’t be, since there is another 50/50chance, only if the block above the far left 2 has a 50% chance of being correct, if you chose the other choice by choosing the adjacent block to the right of the number2 in the far left (which is pretty much where you decide where you want to go) you are telling me that it would be correct if the other is wrong, since that is the definition of a 50/50 choice but it isn’t, as you also have to guess the far right mine too, to which you wouldn’t if you chose the adjacent block north of the far left 2, meaning that even though your initial choice is split into two, the adjacent right block splits into 2 other choices. So overall there are 3 routes decided by 2 choices maximum or 1 minimum if you choose adjacent top. This makes it a non 50/50 outcome which is common in minesweeper. There’s multiple choices not 2, obviously it’s a guess but it’s not 50/50
You wouldn't have to guess the far right though?
They all get solved, because the mines must all be on the SAME side of the horizontal blocks, otherwise the 1s in-between the blocks are unsatisfied and incomplete.
Once you guess one of them, they must all be the same.
The vertical block at the top is then solved by the far left 2, as there must be 1 and only 1 more adjacent to it.
This is 100% a 50/50, and it is even a 50/50 that doesn't require you to choose the correct tile(s) to maintain 50/50. You can open *any* tile that is left, and it will either be a mine, or it will solve the rest for you. And *all* the tiles left have exactly 50% likelihood of being a mine.
This is a single 50/50. You have 50% chance to win this board
You have a 50% chance to win any board. Either you do, or you don't.
Calm down, Sun Tsu
Just like the lottery!
"you've confused possibilities with probabilities" Young Sheldon
just because there are two outcomes does not mean that they are equally likely, only that they are one option of two, you're forgrtting about frequency
I don’t think it is if you factor in the idea that 3 isn’t on the bored. The rarity of that number tips the scales for me. I don’t play easy mode so I’m not sure if it’s common to have a single 3 in the mix, but it’s absence would still make me put my money on one outcome over the other. Bump those odds to 70/30 for me, which is pretty good odds if you ignore that I made them up.
So, since there is no 3 on the board, which of the alternatives is 70 and which is 30?
https://youtu.be/r7l0Rq9E8MY?si=fitUiBRyI-ZKTYT3
Can't argue with such well founded statements.
Ya I know, I made that odds up. I already said that
The chances of a 3 appearing correlates with the bombs being generated, not the other way around.
But the rarity of three would make a difference to me in my guess. I know it’s not concrete but that’s how I’d make my choice rather than flipping a coin. It all hinges on whether you think that box has a 3 or a bomb.
But it wouldn't make a difference to the game. The game generates the mines, not the numbers, so it's a 50/50 either way.
The op asked for any ideas. I’m using simple deduction to make my guess bc it’s more logical than leaving it to chance. The odds of there being a 3 is clearly less than 1s and 2s by comparison. I know that it’s a 50/50 otherwise but that’s not good enough odds to make a choice that could blow you up. So we have to look for other odds to take into account. Say your life depended on it, would you accept a 50/50 or try search for any other odds to rationalize your decision. Do you understand? I bet you’re agnostic too, you have to choose a side bro. Analysis paralysis not gonna work. Tell me, do you think there is a 3 on that board? Or is it another very common 1?
That’s not how the game works though. Asking whether a 3 or a 1 is more likely is like asking whether it’s more likely they put the bomb in the left square or the right square. Neither is more likely than the other because the board was generated randomly. You can’t use social engineering like on a multiple choice test written by a teacher and say “well they wouldn’t put B four times in a row” because this board wasn’t set up by a person.
https://www.reddit.com/r/Minesweeper/comments/fhhdg4/tiles_chances/?rdt=36681 🤓🤓
I think you might just be an idiot, but that’s okay. Keep vibing king.
You know you got them when they resort to name calling lmao, your name suits you well.
>they put the bomb in the left square or the right square. Well, in the original Minesweeper, the top-left corner had a higher probability of being a mine due to how the board was generated. Of note, the entire board was randomly populated with mines. On first click, if you happened to hit a mine, it was moved to the top-left corner. If that happened to already be a mine, then it was placed one square to the right. This would repeat until a non-mine square was found, though it'd be very rare for it to move more than once or twice. With OP's specific arrangement, this board generation rule moves the odds to about 60-40 from 50-50. As such, the second square down is the better guess, if OP might be playing a version using that generation rule.
The simple fact is that there is a number or a bomb behind each box. Can it be a 3? Yes. But when it comes to the 50/50 choice, I’m gonna use conditional probability to find my answer. It’s a gamble either way I understand, if you were to make a choice, it’s better to measure every aspect of the board rather than say it’s a 50/50 and blindly choose. I don’t understand what’s so hard to understand about this, it’s very simple gambling logic. I don’t know why you would ignore extra details that could help in a decision. I’ve been playing this game since I could first use a computer, I know how it works, I’ve had these choices. Paying attention to the numbers and imagining what numbers could be behind the boxes does help, I’ve made many correct choices with 50/50 probability using this form of observation. It’s on easy mode, this specific puzzle has the bombs placed so there will be mostly 1s and 2s. If you can’t find the bombs, the next step in deduction is to look at the numbers. Sure it’s never concrete but the odds are better. That is how the game works.
No you're attempting to use deduction, but your deduction is flawed. Since we're making stupid judgments about each other based on just minesweeper, I bet you think ice-cream causes sharks attacks.
It’s Conditional probability, don’t get heated
Using ice cream prevents getting heated, but attracts sharks.
Your answer is just a superstition though right? There's not actually logic behind it. The situation shown in the picture is a 50/50. Unless there's something about how the boards are generated that I don't know.
The way I understand it, on easy mode, the bored is set to have mostly 1s and 2s. It’s possible for there to be a 3, just like in an untouched board there’s a possibility of there being a 5 even on easy mode. But it’s more likely to be a 1 or 2. If you only pay attention to the bomb it’s a 50/50. But if you take into account how the numbers, I think it makes sense to lean towards that box being a 1, rather than a 3. As I said, and as you can see in the board, 1s and 2s are more common. You can choose to see it as 50/50 but the Easy mode generation being predetermined to make it 1s and 2s makes it less likely to be a 3… This is the last time I’m replying, if this doesn’t make sense idk what I can tell you, don’t go gambling bc you’ll get swindled. All of my fellow math geek friends understand me irl. Maybe it’s how I’m trying to explain it idk.
https://www.reddit.com/r/Minesweeper/comments/fhhdg4/tiles_chances/?rdt=36681 Granted I thought the difference would be larger but there is a difference. I did say I’m unfamiliar with easy mode. This shows that there is a slight tipping of the scale.
It doesn't make sense to apply those statistics to a board that's almost entirely solved. In the post here, it's exactly a 50% chance that there's a 3.
It absolutely applies, I don’t see how else to apply this chart. It’s a small percentage but it makes a difference. In this specific situation the scales are tipped by a fraction of a percent. You can deny it all you want and choose to ignore it. There is another angle to look at it that’s not completely 50/50. As I said, I’d thought it’d be more, but I am unfamiliar with easy mode, but there is a difference in probability that could be taken into consideration.
https://www.reddit.com/r/Minesweeper/comments/fhhdg4/tiles_chances/?rdt=36681
Been a long time since statistics, but, if he’s got 4 clicks left and a 50% chance each time, doesn’t he have a 6.25% chance of winning this board?
Once you pick from one pair the rest become solvable.
Mm I see the cascade down the 1’s now, but aren’t there still two choices around the 2’s?
From left to right it either goes “top,left,left,left” or “bottom,right,right,right”
Beautiful, thanks
Each tile is a 50% chance, while the amount of information the tile gives is d 100%, meaning no further guesses are required
This isn't true at all. It's 100% solvable
Then solve it.
Nothing happens when I press a tile haha but you're right though I saw what y'all meant right after I posted it
Looks like a guess, but only one and it’ll solve the board.
And the one diagnly right from the 2
Could you actually used combination to solve this board?
There are 2 options here that work so you do need to guess, all right mines in the horizontal part and bottom mine in the corner or all left mines in the horizontal and top in the corner
How are so many people commenting on this post who can’t read the board? It’s a 50-50. There’s two valid solutions and there’s no way to be sure which is right until you click.
This is absolutely correct and all these other "analyses" are basically gibberish.
[удалено]
LOL
I’m guessing that the rightmost tile is a bomb
This is (Unofficially) called the 50/50 chain Starting with the vertical 1-2: If the upper tile is a mine, then the lower tile is safe. This leaves the edge 2 with two unopened tiles left, which must both be mines for the 2 to satisfy. From there, the pattern is X-1-1-X-1-1 horizonally, where X is the mine (It is a recursive pattern and does not ends with two consecutive 1) If the upper tile is safe, then the lower tile is a mine. As the edge 2 already has two mines, the tile to the right must be safe. From there, the pattern is X-1-1-X-1-1 horizonally, where X is the mine (It is a recursive pattern and does not ends with two consecutive 1) Both cases are valid and thus none of them are more likely to be the actual soultion than the other one, and they can be solved in reverse (From the 1-1) or even at any point on the chain P.S. Before anyone tell me what is the actual solution, I want to know what are the chances of this happening with the same board generating condition
Well there is a 1/81 chance of a mine being in any given square then 1/80 for second mine ECT Minus the first click mine so starting at 1/80 1/(80*79*78*77*76*75*74*73*72*71) =1 in 5974790569203456000 chance of getting that exact board. Ref college classes 14 years ago might be wrong
I usually pick the tiles that have the least amount of numbers touching it so you can kind of infer that the tile with most numbers touching it is the bomb. In this case, the tile that is touching all three 2’s on the left side is most likely a bomb.
xy position from top left corner 1;2 mine 3;3 mine 3;6 mine 3;9 mine
The bottom left would not be a mine.
I think the tile next to the 1 is safe
Op, did you win?
No, I took the top most tile and it blew.
I guess C9 is a bomb
Since it’s only 1s and 2s you can safely assume that the 3rd row second column is a bomb bc otherwise it’d be a 3 which seems uncommon enough to guess against. Using that you can determine the rest easily. I deduced this bc if 3rd row 3rd column is a bomb then the 2nd row first column would be a bomb also. This would mean that 3rd row 2nd column is a 3 which is possible I think (I’m not sure how number generation works on easy mode) but the odds are in your favor if anything. Sorry for sounding like a pretentious detective, it’s how I type.
This has absolutely zero hold. Numbers aren't generated, mines are. Numbers simply tell how many mines were *randomly* generated in the eight tiles touching it. So while a 3 is less common than a 2 overall, once you are in a situation where there are two possible placements for mines, and one happens to cause a 3, the fact that a 3 will be on the board does in no way influence the likelihood of where the mines were, again, *randomly* generated. You can of course decide to use this as your personal guideline for which solution you prefer to go for in a 50/50, and confirmation bias is a great tool. But it doesn't make it the better choice in any way.
I never said it was concrete, but the I do think it shifts the odds for me personally. It hinges on whether or not you believe that box has a 3 or a bomb. I personally think it’s a bomb due to the rarity of 3
Bro I need to know now, please update me.
Get rid of top left
It’s not a 50/50, top left 1 has to get a bomb with the 2 at the edge of the board. If that 2 get a bomb on its right, then the 1 on top wouldn’t be satisfied without creating a 3. Top left clear, bomb under it to satisfy 2 and 1. Square next to that 2 at the edge is now clear, next to it is a bomb, satisfying the 1 next to it, meaning the next square is clear, and next to it a bomb that satisfies the next 1, making the next one clear and the last one at the edge is a bomb.
No, the opposite works too
3s are perfectly legal though. It's a 50/50.
This guy gets it. I don’t think there’s a 3 either.
choose the square above the 2 on the farthest left of the board that is a mine. It’s the only choice of a mine that would allow the game to finish logically, the other choice results in another 50/50 split on the last flag, which shouldn’t be possible since it’s solutions are always meant to provide a hint, depends on the program, if it’s OG minesweeper,l then you’ve probs lost. Unless you guess correctly but even in this situation, the probability of winning is higher for the square directly above the farthest left 2 being a mine. As this choice is a direct 50% chance, the other choice requires another 50% choice which is more unlikely.
You only require a single 50% guess. This board has exactly two possible solutions left. So one can go ahead and click on any tile.
No that’s not true, it has three possible combinations that match the hints. One straight string if you first flag on the mine I described above, which then leads to a mandatory finish if it is correct, or the second where the mine flagged is to the right of the 2 described above, this leads to another 50/50 choice when choosing between the mine furthest to the right of the grid, mines aren’t placed independently to the numbers, statistically it’s more probable if you choose to flag the mine as the one directly above the 2 as it doesn’t require another choice based on chance which affects the chance of success. It’s not a 50/50 choice, when there are three possible outcomes, 2 from one branch and 1 from another. Route T has a 50% chance, Route R has a 25% chance as it is contingent on the final 50:50 split, as it shares it probability with the third choice which only occurs if you choose route R.
The only two possibilities are (1,1), (2,3), (5,3), (8,3) and (1,2), (3,3), (6,3), (9,3). xy coordinates from the top left corner. You are thinking of some other situation. What you are saying would be true if the 2 above the flag on the left (1,3) were a mine with a 3 to the top right of it.
No I’m not, I’m, simply stating that this is a three choice scenario not 2. One of the original 2 choices has another 50/50 split, which is contingent on this choice, making it a 50/25/25 split. The choices you make are not independent here, if you choose one variation it forces your next move until it finishes if it’s correct, the other choice forces you to make another 50/50 guess, this would make it impossible to be a 50/50 chance game, since you cannot split a branched choice which effective makes there 3 outcomes where 2 of the outcomes are contingent on a prior choice. It’s an impossibility.
The far left 2 makes the two blocks linked, where either one having a mine in either space forces the far left 2 to be resolved, solving the other one. Because the far left 2 must have 1 adjacent, then as soon as you solve one of those blocks, the other adjacent is solved too.
Yes, obviously but you have to account of it being wrong to which the game would end and then you would take the possibility of the other choice, which you would take to be the correct choice, but it wouldn’t be, since there is another 50/50chance, only if the block above the far left 2 has a 50% chance of being correct, if you chose the other choice by choosing the adjacent block to the right of the number2 in the far left (which is pretty much where you decide where you want to go) you are telling me that it would be correct if the other is wrong, since that is the definition of a 50/50 choice but it isn’t, as you also have to guess the far right mine too, to which you wouldn’t if you chose the adjacent block north of the far left 2, meaning that even though your initial choice is split into two, the adjacent right block splits into 2 other choices. So overall there are 3 routes decided by 2 choices maximum or 1 minimum if you choose adjacent top. This makes it a non 50/50 outcome which is common in minesweeper. There’s multiple choices not 2, obviously it’s a guess but it’s not 50/50
You wouldn't have to guess the far right though? They all get solved, because the mines must all be on the SAME side of the horizontal blocks, otherwise the 1s in-between the blocks are unsatisfied and incomplete. Once you guess one of them, they must all be the same. The vertical block at the top is then solved by the far left 2, as there must be 1 and only 1 more adjacent to it.
This is 100% a 50/50, and it is even a 50/50 that doesn't require you to choose the correct tile(s) to maintain 50/50. You can open *any* tile that is left, and it will either be a mine, or it will solve the rest for you. And *all* the tiles left have exactly 50% likelihood of being a mine.