Consider the following game: five numbers are chosen randomly in the interval [0..1] with uniform distribution. The player is shown each number in turn and asked if it is the highest. The game proceeds until the player either answers incorrectly (false positive and false negative are both losing outcomes) or correctly guesses that the present number is the highest (correctly answering that the current number is the highest is an instant win, since the player could presumably answer correctly that one of the others is).
If the player's goal is to maximize the probability of answering correctly whether each given number is the highest, the player should answer "yes" when shown a number which is larger than any seen thus far, and which is larger than the 1-1/2^(1/n), where n is the number of unseen numbers remaining (e.g. if the third number is sqrt(2), and the first two numbers were smaller than that, there's a 50% probability that both of the remaining numbers are less than that, so either "yes" and "no" would have a 50% probability of being correct; if the value is higher, then "yes" will more likely be correct; if it's lower, then "no" will be most likely.
If the player's goal is to maximize the probability of winning, however, the calculations would seem to get more complicated. When asked whether the fourth number is highest, the player should answer "Yes" if it's greater than 50%, and greater than all numbers seen thus far. If the player answers correctly, the player wins (a player who correctly states that the fourth number is not the biggest will have no trouble answering that the fifth is). When answering whether the "third" number is the biggest, however, the reward for correctly identifying that it is the biggest (i.e. an instant win) is larger than the reward for identifying that it is not (since the player will still risk a loss on the next number).
Is there any nice way to determine the threshold for each number where the player should announce that it's the biggest (assuming it's larger than any values seen thus far)? Attempting to partition a 3-dimensional space to decide what to do on the third guess seems awkward but not impossible, but going beyond that to figuring out the optimal strategy on the second guess would seem unworkable absent some simplification I'm not seeing.