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This is kind of an extension question to this: Expected value of sums. Suppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get the first ace and the first king (including the ace/king depending on which comes out first)?

My solution:

  1. Let the first ace be $A$ and the first king be $K$. By symmetry, half the time we would expect $A$ to come before $K$.
  2. Now we have a situation like this $----A----K----$ i.e. $3$ spaces. Let's call them $X_1, X_2,$ and $X_3$ going from left to right.
  3. Now all non aces and non kings can go in $X_1, X_2 and X_3$ with equal probability. There are $52 - 8 = 44$ such cards so at the moment, we have $E(X_1) = \frac{44}{3}, E(X_2) = \frac{44}{3}, E(X_3) = \frac{44}{3}$
  4. For A to be the first ace, the other 3 aces can only go in spaces $X_2 or X_3$ and each of these aces goes in with probability $\frac{1}{2}$ giving $3 * \frac{1}{2} = \frac{3}{2}$.
  5. For the remaining 3 kings, they must all go in $X_3$ with probability 1.
  6. So $E(X_1) = \frac{44}{3}, E(X_2) = \frac{44}{3} + \frac{3}{2} = \frac{97}{6} and E(X_3) = \frac{44}{3} + \frac{3}{2} + 3 = \frac{115}{6}$
  7. Going back to this $----A----K----$, the expected number of cards drawn before we get an ace and a king is $E(X_1) + 1 + E(X_2) + 1 = \frac{197}{6}$

Is this correct? I don't have the answer because I made it up myself.

amWhy
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Apex345
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  • The lengths of $X_2$ and $X_3$ seem to be dependent on how many aces and kings they contain, in a non-trivial way. I've given a solution that has clean separation between the aces-kings group and the types other types cards. – user3257842 Aug 01 '24 at 15:20
  • Thanks for your reply. Why would inlcuding the aces and kings in $X_2$ and $X_3$ be a problem? – Apex345 Aug 01 '24 at 16:49

2 Answers2

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Each sequence of cards contains a subsequence of $8$ elements made entirely out of aces and kings. They split the remaining $44$ cards into $9$ intervals. The average expected length of such an interval is $44/9$, which becomes the expected average "distance" in the deck between $2$ consecutive cards in the $8$ card subsequence. So, the average expected position of the $n$'th element of the $8$ element subdeck along the whole deck becomes $(44/9 +1) * n$

Now we need to figure out how long along the isolated $8$ card subdeck we must go before having both an ace and a king. By symmetry, we can assume the first element is an ace (it's either ace or king). In the remaining $7$ cards, the $4$ kings split the $3$ aces with an average distance between kings of $3/(4+1)$ aces . So (assuming we drew ace first) the first king occurs at average position $3/5 + 1$ in the remainder $7$ card deck, so $3/5 + 2$ at the $8$ deck. At this position, we will have both an ace and a king. Plugging in $n = 3/5 + 2$ in the previous formula, we get $(44/9 +1) * (3/5 + 2) = \frac{689}{45}$

user3257842
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  • I get the first paragraph of your answer. For the second paragraph, what do you mean by: 'the $4$ kings split the $3$ aces with an average distance between kings of $3/(4+1)$ aces'. – Apex345 Aug 01 '24 at 17:09
  • There are $8$ cards: $4$ aces and $4$ kings. The first card is an ace, so we focus on the remaining $7$ cards: $3$ aces and $4$ kings. In those $7$ cards, we're interested in how many aces occur before the first king occurs. Just as our original $8$ cards split the remaining $52- 8 = 44$ into $8+1 = 9$ intervals, so do the $4$ kings split the $7-4 = 3$ cards (which are only aces) into $4+1 = 5$ intervals. So an average of $3/5$ aces occur (within those $7$ cards) before the first king occurs. – user3257842 Aug 01 '24 at 17:26
  • Got it now thanks! How did you come up with this solution so quickly? What gave you this intuition? Also do you know where I am going wrong with my solution? Thanks – Apex345 Aug 01 '24 at 21:16
  • In your solution, you treat the probability of an $A$ not appearing in $X_2$ to be equal to the probability of it appearing. Ie. you take the aces "out of the deck", then put them back in. That is not correct, as the probability-number actually depends on how many combinations there are of $X_3$, which grows when $X_3$ contains more aces. – user3257842 Aug 02 '24 at 09:31
  • Take a simpler problem for intuition on where the approach is wrong: Consider a deck made of $4$ Kings and $3$ aces. We're interested only in how many aces appear on average before the first $K$. A naive approach would take the aces away from the deck, then consider the problem $___ K ___$ , where $X_1$ (first interval) is empty, $X_2$ is just $3$ kings. So $E(X_1) = 0$ and $E(X_2) = 3$ . Then we add back the $3$ aces, obtaining (incorrectly) that E(aces before first K) = $3/2$. – user3257842 Aug 02 '24 at 09:36
  • But consider the cases individually: $AAAKKKK$ ($3$ aces before first $K$) is just one, whereas $0$ aces before first $K$ is many: $KAAAKKK$, $KAKAKAK$, $KAAKAKK$ ... etc. The aces have way more room to permute after the fist $K$ than they do before it, so the average nr. of $A$s before first $K$ gets much smaller than $3/2$. So the error would be in removing things from the deck then adding them back naively, rather than considering the whole deck and exploiting symmetries between similar portions. The correct average nr. of aces before first $K$ here is $3/5$. – user3257842 Aug 02 '24 at 09:38
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In general if a deck contains $n$ cards among which are $k\leq n$ specific ones then by symmetry it can be proved that the expectation of the number of draws (without replacement) needed to arrive at the first specific card equals:$$\frac{n-k}{k+1}+1=\frac{n+1}{k+1}\tag1$$

If $X$ denotes the number of draws needed to arrive at the first card that can be qualified as ace or king then on base of $(1)$ we find: $$\mathbb EX=\frac{52+1}{8+1}=\frac{53}9$$

WLOG we assume that this drawn card happened to be an ace.

After this drawing we go on by drawing from a deck that contains $52-X$ cards among which there are $4$ kings.

Now if $K$ denotes the number of extra draws needed for arriving at a first king then again using $(1)$ we find the following expression:$$\mathbb E(K|X)=\frac{(52-X)+1}{4+1}=\frac{53-X}5$$

Then we find:$$\mathbb EK=\mathbb E[\mathbb E[K|X]]=\mathbb E\left[\frac{53-X}5\right]=\frac{53-\mathbb EX}{5}=\frac{53-\frac{53}9}5=\frac{424}{45}$$

As final answer for the expectation of number of draws needed for arriving at a first ace and a first king we find:$$\mathbb E(X+K)=\mathbb EX+\mathbb EK=\frac{53}9+\frac{424}{45}=\frac{689}{45}$$

drhab
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  • Thanks for this answer. I got it now. However, I don't understand what I was doing wrong in my method? – Apex345 Aug 02 '24 at 08:16
  • If $A$ and $K$ denote the first ace and the first king in point 2 then space $X_3$ contains on average more non- aces and non-kings than the other $2$ spaces. – drhab Aug 02 '24 at 09:15
  • Could you please post a reference, possibly online, for the derivation of (1)? – gboffi Aug 02 '24 at 09:23
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    @gboffi Add a special card so that now there are $n+1$ cards among which $k+1$ are special. Randomly place the cards in a circle. Evidently the $k+1$ 'strings' of non-special cards that arise have equal average length. So this average length must be $\frac{n-k}{k+1}$. This does not change if we turn the circle into a row by removing one of the special cards. – drhab Aug 02 '24 at 10:02