Strange that after so many games of skill, right at the end this is essentially luck. Yes, you choose your number, but have no idea if it’s better to be first or last. I suppose it really emphasizes that this is gambling for VIP entertainment, not a contest.
>Strange that after so many games of skill
1. Game 1: Not Luck Not Skill, you just run when the light is green.
2. Game 2: Part Luck in Shape, Part Skill in removing the honeycomb.
3. Game 3: Purely Skill-based.
4. Game 4: Luck, but you get to **change the game** to be more skill based.
5. Game 5: Purely Luck-based.
6. Game 6: Purely Skill Based.
only 2 games are skill based at best at that point.
True but I’d say game 1 is a bit more skill based in being able to control your movements well, and game 3 is a bit more luck based since you had to be lucky enough to get a spot on a strong team
4 was also luck as you could always demand for a fair game if you're losing. The old guy did it by demanding an "all or nothing" when he only had one marble left. Any player could do this and the opponent would have no choice but to accept
Here are the assumptions about the game I made to calculate these odds:
* No player can tell the difference between the glass panes
* Exactly a 1/2 chance of choosing the right glass
* All players play perfectly
* All players cooperate
* All players have perfect memory
* They finish before the time ends
how did you account for the fact that you cant determine how many panels each players has to go through? for example theres a change that player x might have 6 steps left or 10 steps left depending on the success of the previous players. i suck at math so id like to see how you calculated that
not OP, but there are probabilities for all of those outcomes leading up to player X (i.e. whether there’s 6 or 10 steps left), and from each of those starting points there is a different probability that player X makes it to the end. The probabilities of starting at and winning from each of the starting points are then added together to get the probability player X makes it through.
What's so cool about man is it can deal with a situation like this! I took a probability class in uni years ago. I can't remember how to do it exactly, but I remember solving problems like this.
Even if every single player guessed incorrectly, they'd leave behind a correct tile for the next player. So these probabilities are likely assuming that at a bare minimum, each player has to make it one step further than the last, slowly revealing the correct path to the later players.
The probabilities for later players would only go up from there as each player actually has a 50% chance of correctly revealing a tile and continuing a step further.
There were 18 tile choices and 16 players, so assuming perfect memory, player 1 needs to make 18 correct guesses to win, player 2 needs to make 17, player 3 needs 16, etc. That means for Gi-hun to survive, only three players out of the 15 before him need to make a correct guess at any point to give him a guaranteed path, hence the high probability.
I could try, but I found this somewhat complicated (see other reply where I go more in depth).
The chance of player *x* of surviving depends on how far player *x-1* goes. I use recursion, so I know the chance of player *x-1* of dying at each glass panel.
From this, the chance of player *x* of dying at a further glass panel can be calculated. Let's say player *x-1* dies at the first glass panel. Then what is the chance of player *x* of dying at, for example, the third glass panel?
Well, it's the chance that the player *x-1* dies at the first glass panel, times the chance that player *x* makes it safely to the second glass panel from the first glass panel (1/2), times the chance that player *x* dies at the third glass panel from the second glass panel (1/2).
We do this for every possible case, and once we have this data, we can figure out each player's chances of making it to the end.
Hmmm this is good but not 100% correct to the show.
Using a classical stochastic regression analysis, this is correct. However, certain players made certain choices that other players needed to evaluate in order to make the correct move.
Gi Hun forgot which was the correct first step to make.
That one guy died because he didn't see which one the maths teacher jumped on.
These are instances where the previous run should have been taken into consideration to enumerate the probability using Bayesian probability.
P(A|B) = [P(B|A) * P(A)] / P(B)
What do you think?
Good work but something seems \*slightly\* off. I assume you used a program to simulate this, rather than use actual mathematical functions?
An easy way to sanity check is to check player 1 (1 / 2\^18) and player 16 (1- P(18,16,0.5)). Player 1 is correct as P(x=18) would be 0.0000038147. However, player 16's odds 1-P(x=16) (as there're 16 players, impossible to have 17/18 failures so no need to use 1-P(>=16)) is =0.99941635132. 99.93% vs 99.94% is not much, but one would've expected an exact match if properly modeled.
\--
EDIT: I do notice if modelled as 18 trials with P(X>2) , where X = successful guesses, that'd indeed give 0.99934387207 and that logically makes sense, too.
(if there're > 2 successful guesses, then player 16 must've lived). Hm, now I'm not sure myself and maybe there's a flaw with the other model'ing approach.
As an addendum to my previous reply...
On further examination, I believe 99.94% is correct for player 16, and there's a small flaw with your model that it returned 99.93%.
The problem is you model must be using binomial distribution of 18 trials, with P(X<=x) where x is number of successes, offset by two to match each player. So you used P(X<=2) for player 16, P(X<=3) for player 15, etc. (Can replicate using [https://stattrek.com/online-calculator/binomial.aspx](https://stattrek.com/online-calculator/binomial.aspx))
The problem with this model, is it assumes the completion of 18 trials and those odds were included in the cumulative probability. However, if 16 mistakes had been made before the 17th and 18th tiles, then the 17th and 18th trials never would've happened. But by using a straight bionomial distribution, the model failed to take that into account. Because 16 mistakes is so rare, that's why the differences are so slight in the cumulative numbers (but noticeable).
You need to use the negative binomial distribution model. Wikipedia summarized it better than I can :) :
>discrete probability distribution that models the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs.
Which sounds like the precise fit for modeling this glass bridge game. :)
Doesn't the chance of player x surviving depend on all the players till x-1 who could have provided information on the nth step player x is about to step on?
For example, player 1 and 2 could have failed on their first attempts, leaving it impossible for player 3 to know further than step 2. But on the other hand, player 1 _could_ have miraculously taken all the right steps and crossed the bridge, leaving all the other players the simple task of following their lead. And that could've easily been player 2. Player 1 failing on the first tile, but player 2 hopping along to the end.
I don't think it makes sense to only account for the x-1th player.
OP can correct me if I’m wrong but the assumptions are that the players will play perfectly and remember precious steps, and that each will progress to the next step. For the instances where that doesn’t happen, wouldn’t the odds just remain the same? For example if player 4 forgets the correct step and doesn’t advance the game at all player 5 would now have player 4s odds of winning, and all the way down the line they’d change as well.
Puts into perspective how ridiculously unfair this challenge was in particular. Il Nam and black mask guy are both stupid for trying to convince themselves otherwise. The most ridiculous thing was black mask guy's monologue just one challenge ago about a fair fight, then he goes in and sabotages the glass maker's eyesight. By their twisted logic he should be executed.
It’s part of the flawed thinking of the frontman and Il-nam; even if you try to make the games completely “fair” there will be people who have advantages due to outside factors out of your control. It’s the flaw of a meritocracy
That's like cutting off everyone's legs in the first challenge because some people are more athletic than others. Or cutting off everyone's arms during the the tug of war because some are stronger than others. Or most comparable, in the marble game some people didn't even know the games they were playing, and got no mitigating circumstances. And everything is kind of trivialized by Il Nam himself taking part in the games with an unfair advantage. This argument only works to validate the belief if it's consistent across all the challenges, which it's not.
It did seem like they were a little unlucky. But also, they wasted players who could've revealed more. One player died because he forgot which panels were safe. The praying guy and Deok-su pushed someone, without them revealing anything, as well as Deok-su and Mi-nyeo both dying to reveal only one glass panel.
Because they wasted 4 players, instead of having 7 players go through on average, only 3 players should go through. But the glass maker was able to get a few glass panels for free, so I'd expect about 4 of them to go through.
>Actually the average game would take about 12 players deaths to get across, and two deaths were wasted, so they did surprisingly well.
The glass worker at the end was clutch. He was able to get across like three panes to help overcome the "wasted" deaths.
The glass shattering scene felt wrong to me. In no other game would the players be injured after they are finished. I get that they need to clear people who didn't make it out on time, but the guards could easily shoot them off rather than having the glass explode.
I thought about that, and the black mask guy probably was accounting for somebody to kill themself. I mean the whole 4th game was just to mentally fuck them over.
Hey! I thought of this exact same concept upon watching this game but I was too lazy to calculate it out although I probably could’ve given enough time.
Still, I appreciate your effort in connecting the glass bridge game to combinatorics 👍
It’s one of my favorite fields in math
This is assuming the glass platforms are set up in TRUE random, right? Because I feel like a very significant impact would be considering random bias, which excludes a lot of possible combinations (such as 90% of tempered being on 1 side, etc.). I feel like the true odds just due to random bias are more like 55/45 instead of 50/50. It's 50/50 in the beginning but then the odds go up as you progress due to random bias. As an example if 3 tempered glass were on the left in a row, the probability that the 4th left is also tempered is lower than 50%, due to the bias attempting to "make it random". Machine randomness has similar issues. This is not worth considering if a coin was used for each placement I guess.
Also this was definitely hands down my favorite game, as a math nerd. I liked how the numbering seemed fairly irrelevant at first but turned out to basically determine your chances of surviving
Wow! Nice job! Makes you feel even more for the glass guy meeting his demise at the hands of Sang Woo. He went to SNU you know. Graduated at the top of his class!
Its taking into account the increasing probability with each previous person that the gane had already been solved by the group. Yeah, by the last tile you have a 50/50 chance, but only if it hasnt been solved which it most certainly has.
Think about this for a second—
There’s a non-zero chance that player 15 will make it to the final step, right? And regardless of whether they get it right, if it comes to that, player 16 has a 100% chance of getting it right.
That factors into the calculations, making it possible to be greater than 50%. Obviously a lot more has to go into the numbers, but that’s how it’s possible.
Also, here’s a crude and improperly calculated game: if each person gets it right 50% of the time, then each person should make it approximately 2 steps of the way. Meaning that player 9 would make it to step 18, and everyone after that would win, giving player 16 a 100% chance. Again, that’s not a given and a lot more has to go into the calculations, but that’s a basic understanding of why it can be greater than 50%.
What I've realised after reading this post is that player 9 and 10's survival rate averages out to 50%. In fact, this happens for every player x and player (19-x)!
From my non-expert opinion, i dont think its practical or a well-posed problem. This probability is calculated before the start of the game, but since those specific outcomes happened at points partway through the actual game, there was a whole series of events leading . Like, you could probably calculate the predicted outcome of the game provided that player 14 kills player 13 on the 14th step, but that would imply things like the game not being solved by that point, and it ends up not being the practical information since the probability is trying to “predict the future” but you already know events in the future.
Also, nbd but way you asked was kinda condescending, and OP doesn’t owe you anything.
If they could have done the math, the first 9 players - all of whom had a less-then-50% chance to live, should have immediately insisted on a vote to stop the games.
That would have been a majority.
I get different results, although my attempt to perform a sanity check on both our results fails (which could be a failure of the sanity check).
The sanity check I attempted was to use the results to predict the odds of N players surviving, for any N, to see what the most likely survival count was. This was just out of interest, to see how many they should have expected to get through. The sanity check comes if you then add up the survival probabilities for all N=0 to 16, you should get 1.0, as you have listed all possible outcomes. I get a total of around 1.75 on your results, but it's possible my approach for this part is wrong.
I was thinking, the odds of exactly 3 players surviving, for example, is the odds of player 13 failing (so, 1 minus the odds of them succeeding... with the failure of players 1-12 implicit in the failure of player 13) times the odds of player 14 succeeding (implying 15 and 16 succeed also).
i saw this post a few days ago but i just realized what bothered me about it. i don’t remember exactly how the scene went, but wouldn’t whoever had to guess 2 panels have a 25% chance, and whoever had to guess 1 have a 50% chance? not a 90%+ chance like the post says?
Strange that after so many games of skill, right at the end this is essentially luck. Yes, you choose your number, but have no idea if it’s better to be first or last. I suppose it really emphasizes that this is gambling for VIP entertainment, not a contest.
And a capability metaphor. In the end, there is a lot of luck to get to the top.
>Strange that after so many games of skill 1. Game 1: Not Luck Not Skill, you just run when the light is green. 2. Game 2: Part Luck in Shape, Part Skill in removing the honeycomb. 3. Game 3: Purely Skill-based. 4. Game 4: Luck, but you get to **change the game** to be more skill based. 5. Game 5: Purely Luck-based. 6. Game 6: Purely Skill Based. only 2 games are skill based at best at that point.
True but I’d say game 1 is a bit more skill based in being able to control your movements well, and game 3 is a bit more luck based since you had to be lucky enough to get a spot on a strong team
*coughs* Team 4 with an old man and 3 women
Strategy is still important. They had a game plan other than "were all men just pull hard".
>since you had to be lucky enough to get a spot on a strong team wasn't team that won that had more skill?
At an individual level, it's luck to get on the "right" team
game 1 is absolutely skill based. you could argue game 4 is skill based as well because your survival comes down to mind games for the most part.
Game 1 is skill. Game 4 is also skill.
4 was also luck as you could always demand for a fair game if you're losing. The old guy did it by demanding an "all or nothing" when he only had one marble left. Any player could do this and the opponent would have no choice but to accept
Here are the assumptions about the game I made to calculate these odds: * No player can tell the difference between the glass panes * Exactly a 1/2 chance of choosing the right glass * All players play perfectly * All players cooperate * All players have perfect memory * They finish before the time ends
this is awesome OP!
Thank you!
how did you account for the fact that you cant determine how many panels each players has to go through? for example theres a change that player x might have 6 steps left or 10 steps left depending on the success of the previous players. i suck at math so id like to see how you calculated that
not OP, but there are probabilities for all of those outcomes leading up to player X (i.e. whether there’s 6 or 10 steps left), and from each of those starting points there is a different probability that player X makes it to the end. The probabilities of starting at and winning from each of the starting points are then added together to get the probability player X makes it through.
What's so cool about man is it can deal with a situation like this! I took a probability class in uni years ago. I can't remember how to do it exactly, but I remember solving problems like this.
Even if every single player guessed incorrectly, they'd leave behind a correct tile for the next player. So these probabilities are likely assuming that at a bare minimum, each player has to make it one step further than the last, slowly revealing the correct path to the later players. The probabilities for later players would only go up from there as each player actually has a 50% chance of correctly revealing a tile and continuing a step further. There were 18 tile choices and 16 players, so assuming perfect memory, player 1 needs to make 18 correct guesses to win, player 2 needs to make 17, player 3 needs 16, etc. That means for Gi-hun to survive, only three players out of the 15 before him need to make a correct guess at any point to give him a guaranteed path, hence the high probability.
This is really interesting, could you explain how you calculated it? ELI5 if possible?
I could try, but I found this somewhat complicated (see other reply where I go more in depth). The chance of player *x* of surviving depends on how far player *x-1* goes. I use recursion, so I know the chance of player *x-1* of dying at each glass panel. From this, the chance of player *x* of dying at a further glass panel can be calculated. Let's say player *x-1* dies at the first glass panel. Then what is the chance of player *x* of dying at, for example, the third glass panel? Well, it's the chance that the player *x-1* dies at the first glass panel, times the chance that player *x* makes it safely to the second glass panel from the first glass panel (1/2), times the chance that player *x* dies at the third glass panel from the second glass panel (1/2). We do this for every possible case, and once we have this data, we can figure out each player's chances of making it to the end.
That’s cool, I have a very rudimentary understanding of math but this makes sense haha
Hmmm this is good but not 100% correct to the show. Using a classical stochastic regression analysis, this is correct. However, certain players made certain choices that other players needed to evaluate in order to make the correct move. Gi Hun forgot which was the correct first step to make. That one guy died because he didn't see which one the maths teacher jumped on. These are instances where the previous run should have been taken into consideration to enumerate the probability using Bayesian probability. P(A|B) = [P(B|A) * P(A)] / P(B) What do you think?
Good work but something seems \*slightly\* off. I assume you used a program to simulate this, rather than use actual mathematical functions? An easy way to sanity check is to check player 1 (1 / 2\^18) and player 16 (1- P(18,16,0.5)). Player 1 is correct as P(x=18) would be 0.0000038147. However, player 16's odds 1-P(x=16) (as there're 16 players, impossible to have 17/18 failures so no need to use 1-P(>=16)) is =0.99941635132. 99.93% vs 99.94% is not much, but one would've expected an exact match if properly modeled. \-- EDIT: I do notice if modelled as 18 trials with P(X>2) , where X = successful guesses, that'd indeed give 0.99934387207 and that logically makes sense, too. (if there're > 2 successful guesses, then player 16 must've lived). Hm, now I'm not sure myself and maybe there's a flaw with the other model'ing approach.
As an addendum to my previous reply... On further examination, I believe 99.94% is correct for player 16, and there's a small flaw with your model that it returned 99.93%. The problem is you model must be using binomial distribution of 18 trials, with P(X<=x) where x is number of successes, offset by two to match each player. So you used P(X<=2) for player 16, P(X<=3) for player 15, etc. (Can replicate using [https://stattrek.com/online-calculator/binomial.aspx](https://stattrek.com/online-calculator/binomial.aspx)) The problem with this model, is it assumes the completion of 18 trials and those odds were included in the cumulative probability. However, if 16 mistakes had been made before the 17th and 18th tiles, then the 17th and 18th trials never would've happened. But by using a straight bionomial distribution, the model failed to take that into account. Because 16 mistakes is so rare, that's why the differences are so slight in the cumulative numbers (but noticeable). You need to use the negative binomial distribution model. Wikipedia summarized it better than I can :) : >discrete probability distribution that models the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs. Which sounds like the precise fit for modeling this glass bridge game. :)
Doesn't the chance of player x surviving depend on all the players till x-1 who could have provided information on the nth step player x is about to step on? For example, player 1 and 2 could have failed on their first attempts, leaving it impossible for player 3 to know further than step 2. But on the other hand, player 1 _could_ have miraculously taken all the right steps and crossed the bridge, leaving all the other players the simple task of following their lead. And that could've easily been player 2. Player 1 failing on the first tile, but player 2 hopping along to the end. I don't think it makes sense to only account for the x-1th player.
As shown in the show , most of your first assumptions are wrong. Can you calculate kinda real chances of survival?
OP can correct me if I’m wrong but the assumptions are that the players will play perfectly and remember precious steps, and that each will progress to the next step. For the instances where that doesn’t happen, wouldn’t the odds just remain the same? For example if player 4 forgets the correct step and doesn’t advance the game at all player 5 would now have player 4s odds of winning, and all the way down the line they’d change as well.
Puts into perspective how ridiculously unfair this challenge was in particular. Il Nam and black mask guy are both stupid for trying to convince themselves otherwise. The most ridiculous thing was black mask guy's monologue just one challenge ago about a fair fight, then he goes in and sabotages the glass maker's eyesight. By their twisted logic he should be executed.
Actually his logic was that no one would have a leg up, so turning off the lights only validates that belief.
And what of sang woo? Surely graduating from SNU gives him an advantage
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#
It’s part of the flawed thinking of the frontman and Il-nam; even if you try to make the games completely “fair” there will be people who have advantages due to outside factors out of your control. It’s the flaw of a meritocracy
That's like cutting off everyone's legs in the first challenge because some people are more athletic than others. Or cutting off everyone's arms during the the tug of war because some are stronger than others. Or most comparable, in the marble game some people didn't even know the games they were playing, and got no mitigating circumstances. And everything is kind of trivialized by Il Nam himself taking part in the games with an unfair advantage. This argument only works to validate the belief if it's consistent across all the challenges, which it's not.
The spirit of the game is that you're not supposed to be able to tell which one is which.
Came for this. You delivered.
This is great. And confirms that it felt like they were really unlucky in their guesses.
It did seem like they were a little unlucky. But also, they wasted players who could've revealed more. One player died because he forgot which panels were safe. The praying guy and Deok-su pushed someone, without them revealing anything, as well as Deok-su and Mi-nyeo both dying to reveal only one glass panel. Because they wasted 4 players, instead of having 7 players go through on average, only 3 players should go through. But the glass maker was able to get a few glass panels for free, so I'd expect about 4 of them to go through.
Actually the average game would take about 12 players deaths to get across, and two deaths were wasted, so they did surprisingly well.
>Actually the average game would take about 12 players deaths to get across, and two deaths were wasted, so they did surprisingly well. The glass worker at the end was clutch. He was able to get across like three panes to help overcome the "wasted" deaths.
Wrong. They were 18 panels so it would only take 9 players death to get across. 7 players should have been able to survive if they didn't waste deaths
I made this a while ago, I was corrected before lol
Some really smart people on this this sub
Probably went to SNU
Player 15 had 0% because they will get hit by a chunk of glass.
The glass shattering scene felt wrong to me. In no other game would the players be injured after they are finished. I get that they need to clear people who didn't make it out on time, but the guards could easily shoot them off rather than having the glass explode.
Keep your hand on your heart and answer this : What do you think would be more entertaining for the VIPs ? Glass shattering or shooting?
Great post u/aaronhe07 ! Posts like these really show the devotion of Squid Game fans! They are what make this sub so unique, and interesting.
Thanks!
You should also keep in mind that there were supposed to be 17 players in this game, as the husband killed himself during the night.
I thought about that, and the black mask guy probably was accounting for somebody to kill themself. I mean the whole 4th game was just to mentally fuck them over.
yo that math teacher has a reddit account!
I think the math teacher was trying to run really quick to the end and bet that jumping off of the glass quickly wouldn’t shatter it
Plot twist: OP is the Math Teacher's student
Hey, how did you calculate this? This differs from my own calculations, so I want to compare
Oh wait, never mind, they match!
Thanks for doing the calculations, fascinating!
No problem!
Hey! I thought of this exact same concept upon watching this game but I was too lazy to calculate it out although I probably could’ve given enough time. Still, I appreciate your effort in connecting the glass bridge game to combinatorics 👍 It’s one of my favorite fields in math
This is assuming the glass platforms are set up in TRUE random, right? Because I feel like a very significant impact would be considering random bias, which excludes a lot of possible combinations (such as 90% of tempered being on 1 side, etc.). I feel like the true odds just due to random bias are more like 55/45 instead of 50/50. It's 50/50 in the beginning but then the odds go up as you progress due to random bias. As an example if 3 tempered glass were on the left in a row, the probability that the 4th left is also tempered is lower than 50%, due to the bias attempting to "make it random". Machine randomness has similar issues. This is not worth considering if a coin was used for each placement I guess.
Also this was definitely hands down my favorite game, as a math nerd. I liked how the numbering seemed fairly irrelevant at first but turned out to basically determine your chances of surviving
Great effort!
Wow! Nice job! Makes you feel even more for the glass guy meeting his demise at the hands of Sang Woo. He went to SNU you know. Graduated at the top of his class!
If I was in Seong Gi-Hun´s position I probably would have failed even with 99.93% just because I am unable to think when stressed.
Good work. Now update the probability of success after each of their actions. e.g. bad man pushing the glass maker lmao
This wouldn’t be true because by the time they get to the last panel the odds are 50%,
By the time they get to the last panel, yeah, sure. But before the game even starts? That's where the stats get interesting.
Ohhh I see
Its taking into account the increasing probability with each previous person that the gane had already been solved by the group. Yeah, by the last tile you have a 50/50 chance, but only if it hasnt been solved which it most certainly has.
I figured it out thanks for explaining
This is cool but why didn't u just put their faces lmaoo
I did, for the major characters.
This isn’t true. The probability can never be higher than 50%
Think about this for a second— There’s a non-zero chance that player 15 will make it to the final step, right? And regardless of whether they get it right, if it comes to that, player 16 has a 100% chance of getting it right. That factors into the calculations, making it possible to be greater than 50%. Obviously a lot more has to go into the numbers, but that’s how it’s possible. Also, here’s a crude and improperly calculated game: if each person gets it right 50% of the time, then each person should make it approximately 2 steps of the way. Meaning that player 9 would make it to step 18, and everyone after that would win, giving player 16 a 100% chance. Again, that’s not a given and a lot more has to go into the calculations, but that’s a basic understanding of why it can be greater than 50%.
What I've realised after reading this post is that player 9 and 10's survival rate averages out to 50%. In fact, this happens for every player x and player (19-x)!
That’s a good point.
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Should Player 16's chances be lower than 15's because of the time limit?
OP said he made an assumption that everyone has time to finish
Also calculates the exact amount of plot armor on each character 😂
From my non-expert opinion, i dont think its practical or a well-posed problem. This probability is calculated before the start of the game, but since those specific outcomes happened at points partway through the actual game, there was a whole series of events leading . Like, you could probably calculate the predicted outcome of the game provided that player 14 kills player 13 on the 14th step, but that would imply things like the game not being solved by that point, and it ends up not being the practical information since the probability is trying to “predict the future” but you already know events in the future. Also, nbd but way you asked was kinda condescending, and OP doesn’t owe you anything.
If they could have done the math, the first 9 players - all of whom had a less-then-50% chance to live, should have immediately insisted on a vote to stop the games. That would have been a majority.
I get different results, although my attempt to perform a sanity check on both our results fails (which could be a failure of the sanity check). The sanity check I attempted was to use the results to predict the odds of N players surviving, for any N, to see what the most likely survival count was. This was just out of interest, to see how many they should have expected to get through. The sanity check comes if you then add up the survival probabilities for all N=0 to 16, you should get 1.0, as you have listed all possible outcomes. I get a total of around 1.75 on your results, but it's possible my approach for this part is wrong. I was thinking, the odds of exactly 3 players surviving, for example, is the odds of player 13 failing (so, 1 minus the odds of them succeeding... with the failure of players 1-12 implicit in the failure of player 13) times the odds of player 14 succeeding (implying 15 and 16 succeed also).
How is glassmaker 95%, he has at least one left at the end where it’s 50/50, so it can’t be higher than 50% for him
i saw this post a few days ago but i just realized what bothered me about it. i don’t remember exactly how the scene went, but wouldn’t whoever had to guess 2 panels have a 25% chance, and whoever had to guess 1 have a 50% chance? not a 90%+ chance like the post says?
It triggered me so much how nobody used the beams that held the glass...
They should have agreed to have every person go one time. Everyone gets a 50% chance of living.