Proguessing

Let $C_j$ be the event that Harshil guesses the $j$ th card correctly. For any $0 \le a,b \le 26$ , let $X_{a,b}$ be the event that the first $a+b$ cards comprise $a$ Red and $b$ Black cards. Then $P[X_{a,b}] \; = \; \frac{\binom{26}{a}\binom{26}{b}}{\binom{52}{a+b}} \hspace{2cm} 0 \le a,b \le 26$ Conditioning on what has happened in the first $j-1$ card turns, we see that $P[C_j] \; = \; \sum_{a+b=j-1} P[C_j|X_{a,b}]P[X_{a,b}] \hspace{2cm} 1 \le j \le 52$ Harshil's optimal strategy is to guess that the next card to be turned up will be the colour of which fewer has turned up so far (and so is more likely to be the next colour chosen). Thus $P[C_{a+b+1}|X_{a,b}] \; = \; \frac{\mathrm{max}(26-a,26-b)}{52-a-b} \hspace{2cm} 0 \le a,b \le 26\,,\,a+b < 52$ and hence $P[C_j] \; = \; \sum_{a+b=j-1} \frac{\mathrm{max}(26-a,26-b)}{52-a-b} \times \frac{\binom{26}{a}\binom{26}{b}}{\binom{52}{a+b}} \hspace{2cm} 1 \le j \le 52$ Let $Z_j$ be the random variable equal to $1$ if the $j$ th card is guessed correctly, and $0$ otherwise. Then $Z = \sum_{j=1}^{52}Z_j$ is the total number of correct choices, and $E[Z_j] = P[Z_j=1] = P[C_j]$ , and so the expected number of correct guesses made is $\sum_{j=1}^{52}P[C_j] \; = \; \sum_{a+b\le51} \frac{\mathrm{max}(26-a,26-b)}{52-a-b} \times \frac{\binom{26}{a}\binom{26}{b}}{\binom{52}{a+b}} \; = \; \frac{3724430600965475}{123979633237026}$ making the answer $\boxed{3848410234202501}$ .