Problem 30.1 Easy Free 1 variation

Red and Black Pairs

A standard 52-card deck is shuffled and turned over two cards at a time, forming 26 pairs. A pair of two black cards goes to the dealer’s pile; a pair of two red cards goes to your pile; a mixed pair is set aside and discarded. After all 26 pairs have been processed, you win $10 if your pile has more cards than the dealer’s, and otherwise (including a tie) you win nothing. What is the fair price to pay to play this game?

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. A mixed pair removes exactly one red and one black card, so the red and black cards that survive are always equal in number.

Fix any order of the deck. Let $m$ be the number of mixed pairs. Let $r$ be the number of cards in your pile, all red, and $b$ the number in the dealer’s pile, all black. Every red card ends up either in your pile or in a mixed pair, and likewise for black. Counting each colour gives

26 = r + m, \qquad 26 = b + m .

Hence $r=b$ for every shuffle. The game always ends in a tie, so you never win the $10. The expected payoff is $0$ , and that is the fair price.

\boxed{\$0}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 30.1a

A standard 52-card deck is shuffled and dealt two cards at a time. Alice keeps any pair of two black cards, you keep any pair of two red cards, and a mixed pair is discarded. Once the deck is exhausted, you and Alice pool whatever cards you have kept, reshuffle them, and repeat the same process, continuing round after round until every card of one colour has been eliminated. If red cards are the ones still remaining at that point, you win $100; otherwise you win nothing. What is the fair price to pay to play this game?

Work it out, then compare with the answer.

Solution

Suppose a round starts with $k$ red and $k$ black cards. The counting argument of the main problem applies within the round. If $m$ mixed pairs are discarded, then $k-m$ reds and $k-m$ blacks are kept. The pooled deck therefore again has equal numbers of each colour, starting from $k=26$ .

By induction the red and black counts are equal after every round. A colour is eliminated only when its count is $0$ , and at that moment the other count is $0$ as well.

The game ends with probability $1$ . Each round with $k\ge1$ has a positive probability of containing a mixed pair, and for odd $k$ it is certain. So the count strictly decreases infinitely often until it reaches $0$ .

When the game ends, no red cards remain. You never win, and the fair price is $0$ .

\boxed{\$0}

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