In a game with 0.01 chance of survival, there are 100 participants, a specific player survives twice what are the odds? Sorry ahead if my question is very beginner-ish but I'm confused with this example. Your help is appreciated
Let's say we have a survival game where 100 players participate each time and only one of them survives.
Let's say on one occasion Ali survived the game. We cannot exclude that Ali survived by chance because this is a post hoc result and the chance of at least one player surviving is 100% (Edit:i.e.high)
I guess that if we predicted Ali's win before the game starts, then Ali's survival will be significant (because his chance of survival is 1/100 - Assuming p<0.05 for significance).
My confusion is what if we didn't make any predictions beforehand and we just happen to find out that in 2 consequent games the same exact person (Ali) has survived. Does that make Ali's survival significant or we still can't exclude the chance with 2 consequent games?  How many consequent games needed for a player to survive to conclude that his survival is significant (p<0.05 in a post hoc manner)?
Thanks a lot in advance

The probability of a given player surviving twice (assuming survival is purely by chance and the two games are independent) is $.01^2 = .0001$, or 1:9999 odds. The probability of there being one player who survives twice is $100*.0001 = .01$, or 1:99 odds, because there are 100 ways for that event to happen (i.e., it could be any of the players).
If you were testing the null hypothesis of non-clairvoyance, i.e., the null hypothesis that you cannot predict who survive the first game, then correctly predicting the first game's survivor would cause you to reject the null hypothesis at the .05 level, indicating clairvoyance. Similarly, if you were testing the null hypothesis that survival is purely by chance (i.e., survival is independent across the two games), then observing anyone survive twice would cause you to reject the null hypothesis at the .05 level.
