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You pull cards out of a shuffled deck until you get a jack and a queen. Given see the jack first, what is the expected number of cards between the jack and the queen?

I believe this is a conditional expectation problem. Our goal is to subtract the expected position of the jack from the expected position of the queen. However, I am lost on how to find this other than writing out all possible combinations which isn't feasible. I know how to find the expected position of either card without the condition. Do I use this and compute variance in some way? Kind of lost, any guidance is appreciated.

RobPratt
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shrizzy
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  • Since you are given that the Jack appeared first, you can (instead) pretend that the problem is asking for the expected number of cards between the first Jack, and the first Queen, regardless of which comes first. For $~k \in {1,2,3,\cdots,45},~$ let $~f(k)~$ denote the probability that the first [Jack or Queen] appears on card $~k.~$ Then, under the assumption that the first Jack or Queen occurred on card $~k,~$ let $~g(k)~$ denote the expected number of cards between this [Jack or Queen] and the very next [Queen or Jack]. ...see next comment – user2661923 Sep 16 '23 at 04:18
  • Then, the desired computation is $$\sum_{k=1}^{45} \left[f(k) \times g(k)\right].$$ So, the problem reduces to computing $~f(k)~$ and $~g(k).~$ If (for example), this problem was assigned to you as part of a Probability class, then the class should have given you prior training in the computation of $~f(k)~$ and $~g(k).$ – user2661923 Sep 16 '23 at 04:20
  • For what it's worth, the approach taken in the previous comments, though simple to envision, is fairly primitive. I suspect that there is a much more elegant way of envisioning the problem, that would expedite the calculations. However, the creativity required for the more elegant vision is such that I question what the problem composer's intended solution is. – user2661923 Sep 16 '23 at 04:25
  • With respect to the start of my very first comment, notice that the problem is symmetric with respect to the occurrences of a Jack and Queen (i.e. there are 4 of each). So, you would expect the same answer if you were given that the Queen appeared first. Then, notice that there are only two possibilities, each of which lead to the same answer: either the Queen appeared first or the Jack appeared first. – user2661923 Sep 16 '23 at 04:29

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Suppose that the first J occurs in position $i\in[45]$. Then, there are $52-i$ cards remaining, $4$ of which are queens. A classical result shows that the expected number of cards to flip the first queen must then be $(48-i)/5$. Now the probability that the first J occurs in position $i\in[45]$ is given by $\binom{52-i}{7}/\binom{52}{8}$, so the expectation is $$ \sum_{i=1}^{45}\frac{48-i}{5}\frac{\binom{52-i}{7}}{\binom{52}{8}} = \frac{379}{45} \approx 8.42 $$ (evaluated using WolframAlpha).

Tai
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    As a sanity check, it's actually fairly simple to see that it should definitely be at least $44/9\approx 4.88$, since the expected number of cards between any of the 8 jacks or queens is $(52 - 8) / (1 + 8) = 44/9$. – Tai Sep 17 '23 at 01:26
  • Can you explain how the probability that the first J occurs in position i arises? – Octot Sep 18 '23 at 18:31
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    It's the same as the probability that the first of 8 cards falls in the $i$-th position. There are $\binom{52}{8}$ ways to choose the positions of the $8$ cards. Of these, the number of ways that the first one appears in position $i$ is $\binom{52-i}{7}$, since the one of them is in position $i$, and the rest of the 7 come from the last $52-i$ positions. – Tai Sep 18 '23 at 19:10
  • To my eye it looks like we only consider the number of ways that exactly 1 out of the 4 jacks appears in position i. However, if I multiplied everything by 4 to account that any of the jacks could appear in position i, the answer would be clearly wrong. I just don't see how this includes the possibilities that every single one of the jacks could appear in position eye. – Octot Sep 18 '23 at 21:09
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    If we count the permutations of the 8 cards (4 jacks and 4 queens), they would also appear when permuting the $\binom{52}{8}$ quantity in the denominator, and then they would cancel out. – Tai Sep 18 '23 at 21:14