You turn over the cards 2 at a time, if they are both red, you keep the cards, if they are both black I keep the cards. If one is red and the other is black then neither you nor I get a card. If you have more cards then I you win $100 otherwise you win nothing. What is the maximum amount you are willing to pay to play this game?

  • $\begingroup$ How big is the deck? $\endgroup$ – Gerry Myerson Feb 15 '14 at 5:47
  • $\begingroup$ 52 cards - a regular deck $\endgroup$ – Random Feb 15 '14 at 5:49
  • $\begingroup$ OK, maybe you should edit that information into the question. But maybe one way to approach the problem is to solve it first for a 2-card deck, then 4-card, 6-card, looking for a pattern or a recursion. $\endgroup$ – Gerry Myerson Feb 15 '14 at 5:50
  • $\begingroup$ okk. Thanks! Can you give me an example? My math skills are rather poor.. $\endgroup$ – Random Feb 15 '14 at 5:51
  • 2
    $\begingroup$ Actually, the question I should have asked was, how many red cards in the deck, and how many black? If the numbers are the same, then for each time there are 2 red cards turned over, there must be a time when two black cards are turned over, so you never win, and shouldn't pay anything. $\endgroup$ – Gerry Myerson Feb 15 '14 at 5:55

As indicated in my comment, I wouldn't pay anything to play this game, since I can't win.

Assuming that there are exactly as many red as black cards in the deck (for example, if it's an ordinary 52-card deck), then at the end of the game, I will have turned over two red cards exactly as often as I will have turned over two black cards. Thus, I will have exactly as many cards as you, and I won't win any money. So, I shouldn't pay any money to play.

  • $\begingroup$ Okay, so Gerry Myerson won't pay anything. Who will? >:) (I guess that isn't a math question anymore though.) $\endgroup$ – PyRulez Mar 11 '16 at 1:22

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