This analysis models the probability of achieving a 25-game winning streak in the High-Low card game. The rank order is **Ace High** (Rank 13) and **Two Low** (Rank 1). The Base Card ($C_1$) is removed before the Reveal Card ($C_2$). The expected value of winning a single round is constant due to the deck's symmetry.
The total sample space ($\Omega$) for one round is the set of all ordered pairs $(C_1, C_2)$ where $C_1$ is drawn from 52 cards and $C_2$ is drawn from the remaining 51.
$$\left|\Omega\right| = 52 \times 51 = 2652$$The total number of Stalemate outcomes ($S_{Total}$):
$$S_{Total} = 13 \times 4 \times 3 = 156$$The total Non-Stalemate outcomes ($NS_{Total}$):
$$NS_{Total} = 2652 - 156 = 2496$$Let $R$ be the rank of $C_1$ ($R \in \{1, \dots, 13\}$, where $R=1$ is a Two and $R=13$ is an Ace). The optimal strategy is to guess Higher for $R \le 6$ and Lower for $R \ge 7$.
The total number of favourable winning pairs ($W_{Total}$) is calculated using the formula where $N_{\text{fav}}(R) = 4 \cdot \max(13-R, R-1)$ is the number of favourable $C_2$ cards for rank $R$:
$$W_{Total} = \sum_{R=1}^{13} 4 \cdot N_{\text{fav}}(R)$$Due to the symmetry of the ranks (the set of favourable cards is preserved when the ranks are reversed):
$$W_{Total} = 4 \cdot \left[ \sum_{R=1}^{6} 4(13-R) + 4(6) + \sum_{R=8}^{13} 4(R-1) \right] = 1920$$ $$P_W = \frac{W_{Total}}{\left|\Omega\right|} = \frac{1920}{2652}$$Let $P(\text{Win}')$ be the conditional probability of a win given $\text{Not Stalemate}$:
$$P(\text{Win}') = \frac{W_{Total}}{NS_{Total}} = \frac{1920}{2496} = \frac{10}{13}$$The probability of achieving a streak of $N=25$ wins is the product of the probabilities of the independent trials:
$$\text{P}(\text{Streak of } 25) = \prod_{i=1}^{25} P(\text{Win}'_i) = \left(\frac{10}{13}\right)^{25}$$The final numerical result is:
$$\text{P}(\text{Streak of } 25) \approx 0.00165$$The odds against winning the streak are $1 / P(\text{Streak of } 25)$, or $\mathbf{1 \text{ in } 606}$.