High-Low Winning Streak Probability Analysis (Ace High)

Summary of Card Depletion Analysis and Probability Adjustment

The core mechanism is that the Base Card ($C_1$) is **removed** before the Reveal Card ($C_2$) is drawn. This reduces the total possible outcomes for any round from $52 \times 52$ to $52 \times 51 = 2652$. Due to the symmetry of the ranks (the set of favourable cards is preserved even with Ace High), the **expected probability of winning any single round remains constant** at the value derived below.

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Part 1: Single Round Probabilities (Expected Value)

A round consists of two draws, $C_1$ and $C_2$. Total possible ordered pairs: $\mathbf{2652}$.

1. Probability of Stalemate ($P_S$)

A stalemate occurs if $\text{Rank}(C_1) = \text{Rank}(C_2)$. Since $C_1$ is removed, there are 3 cards remaining of that rank.

$$\text{Total Stalemate Pairs} = 13 \times 4 \times 3 = 156$$ $$P_S = \frac{156}{2652} = \frac{1}{17} \approx 5.88\%$$

2. Probability of a Win ($P_W$) - Optimal Guess

The player always makes the **optimal guess** (Higher for ranks 1-6; Lower for ranks 7-13). The total number of **favourable** winning pairs ($W_{\text{Total}}$) is the sum of the optimal outcome counts for all 52 possible Base Cards. Let $R$ be the rank of $C_1$ ($1 \le R \le 13$):

$$W_{\text{Total}} = 4 \cdot \left[ \sum_{R=1}^{6} 4(13-R) + 4(6) + \sum_{R=8}^{13} 4(R-1) \right] = 4 \times 480 = 1920$$

The overall probability of a Win is:

$$P_W = \frac{1920}{2652} = \frac{480}{663} \approx 72.40\%$$

3. Probability of a Loss ($P_L$)

The probability of a loss is the remaining probability ($1 - P_S - P_W$):

$$\text{Total Loss Pairs} = 2652 - 1920 - 156 = 576$$ $$P_L = \frac{576}{2652} = \frac{144}{663} \approx 21.72\%$$

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Part 2: Winning Streak of 25

The challenge requires 25 wins without an intervening loss. We use the probability of a Win in a **simplified round** ($P(\text{Win}')$) by excluding the stalemate outcome:

$$P(\text{Win}') = \frac{P_W}{1 - P_S} = \frac{1920}{2496} = \frac{10}{13}$$ $$P(\text{Win}') \approx 0.76923$$

The probability of achieving a streak of 25 wins is the product of 25 independent simplified rounds:

$$\text{P}(\text{Streak of } 25) = \left(\frac{10}{13}\right)^{25}$$ $$\text{P}(\text{Streak of } 25) \approx (0.76923)^{25} \approx 0.00165$$

The probability of achieving a winning streak of 25 is approximately $\mathbf{0.165\%}$.