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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?

husnucuk
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    These problems are nearly the same, since the last ace is the first ace if you started from the bottom of the deck. – lulu Aug 02 '21 at 17:58
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    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. – Henry Aug 02 '21 at 18:03
  • @Henry That's what I was starting to worry about, but can you elaborate on how to distinguish between the two? – husnucuk Aug 02 '21 at 18:07
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    It is the distinction between mode and mean of a non-symmetric distribution – Henry Aug 02 '21 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. – husnucuk Aug 02 '21 at 18:25
  • @lulu Sure, it's easy to beat the problem that way. But isn't drawing from the bottom of the deck cheating? (joking). – David Aug 02 '21 at 18:40
  • To emphasize the observation pointed out by lulu, there is a very clear bijection between arrangements of the deck where the first ace is in spot $k$ and arrangements of the deck where the last ace is in spot $53-k$ seen by simply reversing the order in which cards appear in the deck. The existence and properties of such a bijection should be obvious if one is practiced in these types of problems, but if not then it is worth taking a moment to look at it in greater detail. Now... thanks to that bijection, we can then relate the two problems closely and use the solution of one for the other. – JMoravitz Aug 02 '21 at 18:47
  • So... the expected value of the position of the first ace is position $10.6$... that makes the expected position of the last ace $53-10.6 = 42.4$ – JMoravitz Aug 02 '21 at 18:50
  • With your edit... "I'm looking for the difference between the expected position and the most probable position..." I do not understand why you are asking this. What do you mean by "what is the difference"? They are completely different questions and can have and do have completely different answers and methods of obtaining the answer. The one is the average result... if we were to repeat the experiment many times and average the results... The other is just whichever outcome was most likely. Why should you have thought any different? – JMoravitz Aug 03 '21 at 16:20
  • "Why shouldn't the answer to the first problem be $1$ as well?" The answer to the question of what the most probable position the first ace would be in is $1$. The answer to the question of what the expected value of the position of the first ace is $10.6$... These two facts do not contradict one another. They are each correct answers to their own distinctly different questions. – JMoravitz Aug 03 '21 at 16:23
  • Perhaps an analogy would help... Suppose I have a deck of cards... Eight of these cards have the number $1$ on it. Two of these cards have the number $10000$ on it. A person can play the game by drawing one of these cards and winning the number of points that the card shows. What is the average amount of points a player would have if they played the game? $\frac{2\cdot 10000 + 8\cdot 1}{10}$. What is the most likely outcome? $1$. These are.... again... and I can't emphasize this enough.... different questions. – JMoravitz Aug 03 '21 at 16:25
  • @JMoravitz I'm learning about probability and statistics newly, thank you so much for the explanation. The question that is making it harder for me to understand is the example I gave: "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 (52 card) for all rounds?" Since the mode is the $52nd$ card, I should always select the last card, right? Once again, thank you so much for explaining patiently. – husnucuk Aug 04 '21 at 03:42
  • @husnucuk: The answer to the question in your last comment is yes, you should select the mode, which is the last card, for the reason that you correctly provided in the question. – joriki Mar 07 '24 at 22:43

2 Answers2

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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.

Bertrand Einstein IV
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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 $42.4$.

paw88789
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