Card Game: In-Between with a Standard Deck

Game Theory · Easy · Free problem

In the card game "In-Between," played with a standard 52-card deck (ranks 2 through 10, J, Q, K, A with four suits each), two cards are drawn uniformly at random without replacement and revealed. You then choose a stake $x \in [0, W]$, where $W$ is your current bankroll. A third card is drawn from the remaining 50 cards:

  • If the third card's rank is strictly between the two revealed ranks, you win $+x$.
  • Otherwise (including ties with either boundary card), you lose $-x$.

Suppose the revealed ranks are 5 and Q (Queen).

  1. Compute the exact probability of winning, accounting for the two cards already removed.
  2. Express the expected value of one round as a function of $x$.
  3. State the risk-neutral optimal stake $x^{*}$ under the constraint $x \in [0, W]$.

Hints

  1. List the ranks strictly between 5 and Q. How many distinct ranks is that, and how many cards of each rank remain after the two revealed cards are removed?
  2. The third card is drawn uniformly from the 50 remaining cards. The win probability is (number of cards with winning ranks) / 50.
  3. Compute $E[\text{payoff}] = x \cdot P(\text{win}) - x \cdot P(\text{lose})$. If this is negative for all $x > 0$, the optimal stake is zero.

Worked Solution

How to Think About It: This is a conditional probability problem with a card-counting twist. You see a 5 and a Q, so the winning ranks are 6, 7, 8, 9, 10, J -- six distinct ranks. The question is how many of those cards remain in the 50-card deck after removing the two revealed cards. The ranks of the removed cards are 5 and Q, so none of the winning-rank cards have been removed. Once you have the probability, the EV and optimal stake follow immediately. Gut check: 6 winning ranks times 4 suits = 24 winning cards out of 50 remaining, so the win probability should be close to

4/50 = 0.48$. Since that is below 0.5, the game has negative EV per dollar staked.

Quick Estimate: Win probability is roughly 4/50 = 0.48$. EV per dollar staked is (0.48) - 1 = -0.04$, so slightly negative. The optimal risk-neutral stake should be $x = 0$.

Approach: Count winning cards in the remaining deck, compute exact probability, then derive EV.

Formal Solution:

Part 1: Win probability

The rank ordering is: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. The revealed ranks are 5 and Q. The ranks strictly between 5 and Q are: 6, 7, 8, 9, 10, J -- that is 6 ranks.

Each of these 6 ranks has 4 cards in the deck. Since neither revealed card shares a rank with any winning rank (one is a 5, the other is a Q), all $6 \times 4 = 24$ winning cards remain among the 50 cards left in the deck.

The third card is drawn uniformly from these 50 cards, so: $P(\text{win}) = \frac{24}{50} = \frac{12}{25}$

Part 2: Expected value

The payoff is $+x$ with probability

2/25$ and $-x$ with probability
3/25$. Therefore: $E[\text{payoff}] = x \cdot \frac{12}{25} - x \cdot \frac{13}{25} = x \left(\frac{12 - 13}{25}\right) = -\frac{x}{25}$

Part 3: Optimal stake

Since $E[\text{payoff}] = -x/25 \leq 0$ for all $x \geq 0$, with equality only at $x = 0$, the risk-neutral optimal strategy is: $x^{*} = 0$

You should not bet. The game has negative expected value for this particular spread because the gap between 5 and Q is not wide enough to give you a majority of winning cards.

Answer: The win probability is

2/25 = 0.48$. The expected payoff is $-x/25$. The optimal risk-neutral stake is $x^{*} = 0$ (do not bet).

Intuition

In-Between is a great example of how card counting drives betting decisions. The key variable is the width of the gap between the two revealed cards relative to the full rank spectrum. With 13 ranks total and a gap of 6 (from 5 to Q exclusive), you might naively expect roughly 6/13 of the deck to be winners. But 6/13 is about 0.46, which is below 0.50, so the bet is unfavorable. The exact calculation (24/50 = 0.48) confirms this.

The practical lesson is that in any binary bet, you need $P(\text{win}) > 0.50$ for the risk-neutral player to want positive exposure. In-Between only becomes favorable when the gap is at least 7 ranks wide (e.g., 3 and J, giving 7 winning ranks and $P = 28/50 = 0.56$). This same edge-calculation discipline -- compute the exact probability, check whether the EV is positive, and size accordingly -- is the foundation of any trading or betting strategy.

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