There are three cards in front of you, all face down. Each card has a real number written on it. You know that all three numbers are different, but you do not know what the numbers are.
You are allowed to choose a card and turn it over. At this point, you can either keep the card, or discard the card, and turn over a second card. Once you discard a card, you cannot return to it.
You have same option with the second card: You can either keep the second card, or discard it, and turn over the third card. If you discard the second card, your only option left is to turn over the third card and keep it.
At the end, if you have the card with the highest number on it, then you win a valuable prize. Otherwise, you leave empty-handed. If you follow the optimal strategy, then what is the probability that you win the prize?
I have no idea how to do this problem, any help is appreciated.
You can improve on a 1/3 chance as follows.
Automatically look at, but skip the 1st card.
If the 2nd card is lower than the 1st card, automatically skip it, and take the 3rd card. Otherwise, automatically take the 2nd card.
Denoting the 1st card as A, the second as B, and the third as C, and letting the order of the cards listed below represent highest to lowest on each line, the possibilities are
L : ABC
L : ACB
W : BCA
W : BAC
W : CAB
L : CBA
Thus, you have a 1/2 chance of winning.