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Here is an interesting brainteaser I came across that I have not been able to solve:

You are playing a game in which you and an opponent select cards (without replacement) from a 100-card deck, which are numbered 1-100. You each flip over a card initially. If your card value is higher than theirs, you win 1\$. If their card is higher than yours, you have two options: flip over a second card (in which case you win 1\$ if it is higher than their card and lose 2\$ if both your cards are smaller) or not flip over a second card and lose 1\$. What is the fair value of the game

I am stuck on when you should flip a second card and when you should not.

Bepop
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1 Answers1

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Suppose your initial card has value $x$ and the opponent's card has initial value $y$, with $y>x$. What is the probability that a card selected from the remaining 98 cards is $y+1$ or higher? Once you have worked that out, then you can work out the expected gain from flipping over the second card.

smcc
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  • I understand that P(Y = i|Y>X) = 2i/N(N+1). I'm not sure how to use this to find the probability that another card selected is greater than it. – Bepop Sep 02 '23 at 16:16
  • To get the probability that a randomly drawn card has a value higher than $y$, you just need to work out how many cards are greater than $y$ and then divide by 98 (the number of cards left in the pack after the initial draw of two cards). – smcc Sep 02 '23 at 16:20
  • So the probability a randomly drawn card has a value higher than y is (100-y)/98, so we can use \sum (100-i)/98 *P(Y=i|Y > X) – Bepop Sep 02 '23 at 16:23