Probability with card deck flips

Question

Here is my problem : we flip cards from a 52-card standard well-shuffled deck until the first club appears. I am looking to calculate the probability that the next card at the $k+1$th flip is also a club given that the $k$th flip is a club. Let $T$ be the flip on which we encounter the first club. Thanks to this answer I get
$$\mathbb{E}[T]=\frac{53}{14} \approx 3.7857$$
Now let $Y_n=1$ if we flip a club on the $n$th flip and $Y_n=0$ if we flip another suit. The number of clubs flipped amongst the first $n$ flips would be
$$C_n=\sum_{k=1}^n Y_k$$
with $C_T=1$. After the $n$th flip, we have $\tilde{X}_n$ clubs remaining in the deck with proportion $X_n$:
$$X_n =\frac{\tilde{X}_n}{52 – n}, \ \tilde{X}_n = 13 – C_n$$
with $\tilde{X}_T=12$. So
$$X_T = \frac{13-C_T}{52 – T} = \frac{12}{52 – T}$$
We get
$$\mathbb{E}[X_T] = \frac{12}{52 – 3.7857} \approx 0.2489$$
the probability that the next card is a club. Can I use $\mathbb{E}[T]$ in the denominator like this? Thanks!

Answer 1

Fix the number of clubs ($k$). Let $Q_n$ be the desired probability for a deck of $n \ge k$ cards with exactly $k$ clubs. Either the first card is a club (with probability $k/n$), in which case the second card is a club with probability $(k-1)/(n-1)$; or else the first card is not a club (with probability $1-k/n$), and we've reduced to the $n-1$ case. That is: $$ Q_n=\frac{k(k-1)}{n(n-1)}+\frac{n-k}{n}Q_{n-1}, $$ with the boundary condition that $Q_k=1$.

Playing with this recursion soon leads to the conjecture that $Q_n=k/n$, which may be proved by induction: $$ Q_{n-1}=k/(n-1)\implies \\ Q_n=\frac{k(k-1)}{n(n-1)}+\frac{n-k}{n}\cdot\frac{k}{n-1}=\frac{k(k-1)+(n-k)k}{n(n-1)}=\frac{-k+nk}{n(n-1)}=\frac{k}{n}. $$ In other words, the card after the first club is exactly as likely to be a club as any other card in the deck.