Probability with card deck flips 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!
 A: 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.
A: Mjqxxx has shown by a recursion that the probability that the card following the first club in the deck is also a club is $13/52$.  Here is another proof of that fact.
We start by asking a related question: What is the probability that the card following the first club in the deck is the ace of clubs?  There are $52!$ possible orderings of the deck, all of which we assume are equally likely.  We ask how many of those orderings have the ace of clubs immediately following the first club.  Suppose we remove the ace of spades from the deck and shuffle the remaining cards.  There are $51!$ possible orderings of the remaining cards, and in each case there is only one place to insert the ace of clubs so it is the next card after the first club; so there are $51!$ orderings in which the ace of clubs is the next card after the first club.  So the probability that the ace of clubs is the next card is
$$\frac{51!}{52!} = \frac{1}{52}$$
We can say the same about any particular club: the probability that it is the next card after the first club is $1/52$.  So the probability that the next card after the first club is a club is
$$13 \cdot \frac{1}{52} = \frac{13}{52}$$
