# Card game - Insidious casino

Here's the question: A casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer's pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties) you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game? Here's the solution: This surely is an insidious casino. No matter how the cards are arranged, you and the dealer will always have the same number of cards in your piles. Why? Because each pair of discarded cards have one black card and one red card, so equal number of red and black cards are discarded. As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same. The dealer always wins! So we should not pay anything to play the game. What I don't understand about the solution is this part: "As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same". I know that for each pair of discarded cards, one black and one red cards are discarded. But how is this statement enough to conclude that 'the number of red cards left for you and the number of black cards left for the dealer are always the same'? Isn't this completely ignoring obtaining a pair of black cards or a pair of red cards??? ## 3 Answers So there are 26 red cards and 26 black cards. If when you are done, you have$p$pairs of mixed cards (1 red and 1 black) then then the total number of red and black left over (ie not in the mixed pairs) is$26-p$. This is the case for black and red. Number of pairs of reds and number of pairs of blacks is therefore$(26-p)/2$. It also allows you to deduce that$p$is going to be even. So counter offer to the casino. I'll pay if the number of mixed pairs is odd and you pay if it's even! If it helps to visualize, try playing the game with only 2 red and 2 black, then 3 red and 3 black and so on, and you can see that the piles will always be the same. Either 0 piles or 1 pile each with 4 cards total, either 0, 1, or 2 piles each with 6 cards total. So if you gain an advantage by making a pile of your own before your opponent, you have x cards of your color left in the deck, but the opponent has x + 2. Since a match requires one card from each of your color your opponent will always have +2 cards until they make a match of their own. However, the same logic applies to them, so they can never gain an advantage either. But still win the money when the piles end up the same. While the other answers have already explained that you literally have zero percent chance to win given the current rules, I figured it was worth checking how the odds would change if you were also given the option to discard a pair of cards before revealing them. In other words, when choosing to discard there is a chance that you will discard a pair of red or black. As a function of the (randomly) discarded pairs of cards, I simulated the chance to win (average of 1e7 runs per data point). If you were to always discard the first pair of cards and no others, then the odds would jump from 0 to just under 25% chance to win. The best odds (~39% chance to win) are achieved when approximately half of the deck is discarded. So interesting alternatives (while maintaining the house's advantage) are to offer \$25 and then discard the first pair independent of what it is or to split the deck and offer \\$40 on either of the decks.

• What a difference a discard makes! Exploring a puzzle around its edges, as here, adds extra sophistication (sometimes also art, as in this case) to Puzzling. – humn Sep 4 '17 at 14:02