I have a standard $52$ deck of cards, shuffled randomly.
If I flip cards off the top until a Ace is shown, what is the average number of cards expected to be flipped over, including the Ace?
Further to this, what would be the average numbers of cards expected to be flipped for the second Ace, then the third Ace, and then finally the fourth?
My personal thoughts were that it would be $4/52$ chance, so take the reciprocal of this and it's $13$ cards on average.
Take the number of cards left, and so, for example, let's say it was $13$ cards the first time. $52-13=39$. $1/(3/39)$ is the average number next time, which is $13$ cards, so this seems to fit the pattern.
However when I actually do this, the number on average has been much lower. So I'm assuming there is something I'm missing?
Best Answer
Re: Replacement
The expected number of draws with replacement until getting the first ace would be $13$. However, without replacement, which is what applies here, the expected number of draws is smaller. This makes sense because the non-aces are not being replaced to be drawn a second time.
Without Replacement
The probability that the $n^\text{th}$ ace is at position $k$ is the number of ways for $n-1$ aces to be in the first $k-1$ positions and $4-n$ aces to be in the last $52-k$ positions. That is, $$ \frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}} $$ The mean position of the $n^\text{th}$ ace is $$ \begin{align} \sum_{k=1}^{52}\frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}}k &=\sum_{k=1}^{52}\frac{\binom{k}{n}\binom{52-k}{4-n}}{\binom{52}{4}}n\\ &=\frac{\binom{53}{5}}{\binom{52}{4}}n\\[3pt] &=\frac{53n}{5} \end{align} $$