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

  • $\begingroup$ I'm not convinced that these matter, but does anyone have to survive, and can multiple people survive? $\endgroup$
    – Dave
    May 19 at 14:12
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    $\begingroup$ Then why do you say that "the chance of at least one player surviving is 100%"? $\endgroup$
    – Dave
    May 19 at 14:28
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    $\begingroup$ The answer must depend on how many "consequent" games there are, who participates in them, and exactly what you mean by "the same exact person has survived," because that phrase is ambiguous. $\endgroup$
    – whuber
    May 19 at 15:36
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    $\begingroup$ It depends on when you're asking / what led to the question. If you identify the person before the first attempt and ask "what are the chances this specific person will make it twice?", you get a very different answer than looking at say 300 people after the fact and noticing that at the end that 'surviving twice' happened to at least one of them and asking "hey that seems weird, what are the chances of doing that???" ... that second case (in effect, choosing a person after the fact to ask a question about) has a far higher probability of two successes. ... ctd $\endgroup$
    – Glen_b
    May 20 at 1:52
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    $\begingroup$ ...ctd If instead you notice a particular person survived one trial and you then ask about the chances of another success, it's different again. Almost all such questions are purely specifying the event post-hoc for which the answers are difficult to figure out (e.g. you need to know what would trigger the question) $\endgroup$
    – Glen_b
    May 20 at 1:53

1 Answer 1


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.

  • $\begingroup$ I do not follow what you mean about clairvoyance. $\endgroup$
    – Dave
    May 19 at 15:18
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    $\begingroup$ In the first game, someone will survive 100% of the time, but to correctly predict exactly who will survive in the first game would only occur 1% of the time, and perhaps indicate clairvoyance, i.e., rejecting the null hypothesis of non-clairvoyance. This is exactly the motivating argument behind the titular character of The Lady Tasting Tea, a famous book about statistics. $\endgroup$
    – Noah
    May 19 at 15:30
  • $\begingroup$ Thanks Noah! I'm happy with your answer but something else came to my mind. Let's say this was not a survival game but a win/loss game (where the chance of winning = 0.01 for each player per game). Now if we allow each player to participate multiple times (N times) doesn't that increase the chance of any player to win 2 consequent games by chance? so I wonder if we also need to correct the calculations you made with the number of games (N) to be able to accept or reject non-clairvoyance? or N doesn't matter; any 2 consequent wins will be enough to reject the null hypothesis at 0.5 level? $\endgroup$
    – Hunar
    May 20 at 12:23
  • $\begingroup$ @Hunar that's a far more complicated question, and one I am not qualified to answer. $\endgroup$
    – Noah
    May 20 at 14:26

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