Casino Card Game

A casino deals a standard 52-card deck two cards at a time. When both cards are black they go to the dealer's pile, when both are red they go to your pile, and a mixed pair is discarded. After all $26$ pairs are dealt you win $100 if your pile is strictly larger than the dealer's, and otherwise you get nothing. The casino lets you name your price. How much should you pay to play? (Hint, count how many red and black cards each outcome removes.)

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#Counting the cards

The deck holds $26$ red and $26$ black cards, dealt as $26$ pairs. Let $p$ be the number of red-red pairs that reach your pile and let $k$ be the number of mixed pairs. Each mixed pair carries off exactly one red and one black card.

Every red card sits in a red-red pair (two reds) or a mixed pair (one red), so

2p + k = 26. \tag{1}

Let $d$ be the number of black-black pairs that reach the dealer. Every black card sits in a black-black pair (two blacks) or a mixed pair (one black), so

2d + k = 26. \tag{2}

Subtracting the two equations cancels the shared $k$ ,

2p + k = 2d + k \implies p = d. \tag{3}

Each colour starts with 26 cards. Every mixed pair removes one red and one black, so the reds split as 2p paired plus k discarded and the blacks split as 2d paired plus k discarded. Both totals equal 26, which forces p equal to d. Your pile and the dealer's pile therefore hold the same number of cards on every deal.

#Why you can never win

Your pile holds $2p$ cards and the dealer's holds $2d = 2p$ cards, equal on every possible deal. The mixed pairs strip away one red and one black together, so they can never tilt the balance. Your pile is therefore never strictly larger, and

\PP(\text{your pile larger}) = 0. \tag{4}

Every game is a tie, and a tie pays nothing, so in effect the dealer always wins.

#The fair price

A fair price equals the expected payout, which here is

\mathbb{E}[\text{payout}] = \PP(\text{win}) \cdot \$100 = 0. \tag{5}

You should pay nothing. The fair price is $0.