# Average number of cards drawn without replacement when first, second, third, and fourth aces appear

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?

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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}

• Is there something wrong here? Is the downvote because of something missing or wrong in the answer? I think the question provides sufficient context. – robjohn Apr 30 '18 at 16:09
• Thanks robjohn. Only if you feel inclined, my actual real world example has 59 cards in the deck (it's a different card game). So what would the equation look like? Same thing 4 cards the same, without replacement. – Grumpycat May 2 '18 at 3:26
• @Grumpycat: Hmm... $59$ instead of $52$. What might we change in the reasoning above that would change $52$ to $59$? Ponder the answer above a bit and give it a try. – robjohn May 2 '18 at 16:21