Is Veritasium WRONG? Prisoner Riddle That Seems Impossible Even If You Know The Answer

Veritasium showed that the answer to the 100 Prisoner Riddle can be 31%. This was surprising, but I think they're wrong, and that its 50%! Watch and let me know what you think!

[jamiegillett]

You've calculated the probability that any specific prisoner will find their number in a loop of n and not the probability that the maximum loop that exists is a loop of size n. e.g. an individual has a 1/10 chance of finding their number in a loop of 1, but there is a 1/10! chance that the maximum loop size is 1. This is an important distinction because if an individual prisoner finds themselves in a loop of 1 (1/10 chance) it's still possible for the group to fail whereas if the maximum loop size is 1 (1/10! chance) it's not possible for the group to fail.

From this, I think it should be fairly easy to recognize that the chance of failure as a group must be over 50%. The probability that the group will fail = (the probability prisoner A fails to find their number) + (the probability prisoner A succeeds in finding their number while somebody else fails to find theirs). The chance prisoner A fails is 50% therefore the chance the group fails is >50%.

If in doubt you could always double check with 4 boxes. With 4 boxes there's 24 possible permutations. There is 1 permutation where the max loop size is 1. There are 9 permutations where the maximum loop size is 2. There are 8 permutations where the maximum loop size is 3. There are 6 permutations where the maximum loop size is 4. If there exists a loop of 3 or 4 the groups will fail, therefore there is a (8+6)/24 chance the group will fail (which is greater than 50%).

Hope that helped. :)

[miguelmunoz]

I created a computer model of this puzzle, and my simulation shows that Veritasium's answer is correct. According to my sim, for 4 prisoners, the survival odds are ~0.417%, which agrees with the predicted figure of 5/12. With 12 prisoners, it drops to ~35%, and eventually converges on 31%. Veritasium's prediction was that it would converge on log(2), which is 30.1%. That number is a bit off.

[rgbelastung]

I think the issue with your answer is if prisoner number 1 reaches a loop of the lenght 3, that there is still a possibility that a loop with lenght 7 can exist.

[ShadowViking]

If you run a python script of this simulation the results match the ~31% probability of success that was mentioned in the original video.

[davidjames]

Just run a simple computer simulation and you will see around 31.18% is correct for 100 prisoners opening 50 boxes max each. This is one reason why computer simulation is sometimes even better than using math, cuz someone might get the math wrong, but a computer simulation this simple is hard to get wrong.

[MR-cuxn]

The math of veritasim is correct, probability and combination is awsome topic

[caseygust]

Prisoner #5 opens box #5. In this box is the number 2. So prisoner #5 opens box #2. In the second box is the number 1. So the prisoner #5 opens box #1. In this box is the number 2. This is now a closed loop without the prisoners number, is it not?

[sledgenwedge]

And if the actual published paper says something different in the rules I would be happy to rethink my position but until then I know I'm right and each man only has to pick one damn box l o freaking l

[michealkinney]

I also think you are right, but I didn't work out the math, instead I created a simple Python script and despite the amount of prisoners or sample size you use (over 100, and slightly higher, around 60%, with a smaller sample), the numbers come out anywhere from about 49% to 52%. Interestingly, if you use odd numbers, the probability is slightly lower but only marginally. I put the code on my GitHub but if I try to include a link my comments generally get removed.

[sledgenwedge]

Did anybody actually read the rules correctly? It's not a numbers thing it's a words thing!

[sledgenwedge]

He is wrong and anybody thinking it is a numbers thing is wrong, the right answer is that 100% of the men can leave free men and they only have to pick one box ask me how LOL

[Herzinfucked]

Not every loop size has a 10% Chance, maybe u should get familiar with permutations and combinatoric. For Exemplar u only have 1 way to form a 10-circle, but u have 10 ways to form a 9-circle. (U can leave each number out, each gives a 9-circle). Every combination has the same possibilty, so they cant be both at 10%