# Expected number of draws until the first/last ace in a standard deck

## Update

Thanks for both of the answers (+1), but I'm looking for an explanation for the difference between the expected number of draws until I get the first/last ace and the most probable location of the first/fourth ace. For example, if I were playing a game with someone, where we were guessing the location of the fourth ace round after round for a new shuffled deck, would it make the most sense to pick the last spot ($$52nd$$ card)?

Both problems are regarding a standard deck of cards.

What is the expected number of draws until I draw the first ace?

From here, I can clearly deduce that the value is $$10.6$$. However, I get confused when I try to solve the next problem.

What is the expected number of draws until I draw the fourth ace?

Mathematically, I want to say $$52$$ ($$52nd$$ card). The reason is if I were say the $$51st$$ card, we would have fewer possibilities to place the remaining cards as we can't place an ace on the last spot. So, isn't it most probable that the ace lies on the last card if we keep playing this game again and again?

I'm essentially very confused because if the answer is the last card (to the second problem), then shouldn't the answer to the first problem be the first card (1) as well?

• These problems are nearly the same, since the last ace is the first ace if you started from the bottom of the deck.
– lulu
Commented Aug 2, 2021 at 17:58
• The most probable position of the first Ace is the first card and of the last Ace is the last card, but that is not their expected/average positions. Commented Aug 2, 2021 at 18:03
• @Henry That's what I was starting to worry about, but can you elaborate on how to distinguish between the two? Commented Aug 2, 2021 at 18:07
• It is the distinction between mode and mean of a non-symmetric distribution Commented Aug 2, 2021 at 18:17
• @Henry Thank you for explaining. So, if I were playing a game with someone, where we were guessing the location of the fourth ace, it would make the most sense to pick the last spot, right? I'm happy to accept your answer if you write one even if you repeat your last comment. Commented Aug 2, 2021 at 18:25

The solution to this problem is almost identical to the post you linked:

Consider the case where we only have 4 cards and all 4 are aces, how many would we expect to draw before getting all aces? Trivially 4.

For $$\color{red}4$$ cards with $$4$$ aces: $$E(N)=\sum_{k=1}^4k\cdot P(k)=4\cdot \frac{1}{1}=4=\frac{\color{red}{20}}5.$$

Now considering the case where we have 5 cards with 4 aces, there is only 1 ordering that all 4 aces will be pulled out immediately and 4 orderings that will require you to draw all 5 cards. The total number of orderings is of course given by $${5 \choose 1}=5$$

For $$\color{red}5$$ cards with $$4$$ aces: $$E(N)=\sum_{k=1}^5k\cdot P(k)=4\cdot \frac{1}{5} + 5\cdot \frac{4}{5}=\frac{\color{red}{24}}5.$$

And now we may do the case for 6 cards, which isn't too different:

For $$\color{red}6$$ cards with $$4$$ aces: $$E(N)=\sum_{k=1}^6k\cdot P(k)=4\cdot \frac{1}{15} + 5\cdot \frac{4}{15} + 6\cdot \frac{10}{15}=\frac{\color{red}{28}}5.$$

At this point the pattern should be clear, and infact you can prove it by induction pretty easily, but if we were to define $$E_i(N)$$ as the expected number of draws to get 4 aces out of $$i$$ cards with a total of 4 aces, we see the recursion relation:

$$E_{i+1}(N)=E_i(N) \frac{i \choose i-4 }{i+1 \choose i+1-4} + (i+1)\frac{i \choose i-3}{i+1 \choose i+1-4} \text{ }\text{ , } E_4=4$$

This can be solved explicitly as $$E_i(N)=\frac{4 (i+1)}{5}$$, ofcourse as mentioned previously you could just look at the pattern, assume the answer and prove it by induction, this is much easier than solving the recurrence relation.

From here we find that for 52 cards and 4 aces, $$i=52$$ and hence $$E_{52}=\frac{4}{5}(52+1)=\frac{212}{5}=42.4$$

Therefore on average we will draw $$42.4$$ cards before drawing all $$4$$ aces out of a deck of $$52$$ cards.

The $$48$$ non-ace cards go into five (some possibly empty) runs of non-ace cards (before the first ace; between the first and second aces; ...; after the fourth ace).

So the expected length of each of these non-ace runs is $$\frac{48}{5}=9.6$$. So the expected value of the first ace is at position $$10.6$$. Similarly, the expected position of the last ace is at position $$42.4$$.