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This is a problem from a brazilian probability book called "Probabilidade e Variáveis Aleatórias" (Probability and Random Variables) [MAGALHAES, M. N. 2015]:

"Cards are drawn without replacement from a deck of 52 cards until all four aces are chosen. Find the expectation of the number of draws using conditioning."

I've tried some approaches explained in other related questions/answers (Hypergeometric Distribution, Negative Hypergeometric) but I didn't manage to do that using conditioning. The correct answer is supposed to be 52 according the book. I don't know whether it makes sense, considering the results shown in similar questions.

I'm thinking about the property $E(X) = E(E(X|Y))$ but I'm not making any progress.

I would appreciate any help.

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    $\begingroup$ How could the expected value of the variable $X$ (number of draws w/o replacement until all four aces drawn) be $52$ when, if we draw all the cards then necessarily the fourth ace must have turned up at draw $52$ or before? That is, always $4 \le X \le 52$ so expected value must be something between $4$ and $52.$ $\endgroup$ – coffeemath Apr 8 '17 at 5:18
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    $\begingroup$ I think $52$ makes sense with replacement. $\endgroup$ – samerivertwice Apr 8 '17 at 8:53
  • $\begingroup$ @RobertFrost Yes, with replacement changes the question and may make the answer come out $52,$ though may be harder to set up. Without replacement I got $204/5=40.8.$ $\endgroup$ – coffeemath Apr 8 '17 at 21:34
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    $\begingroup$ @coffeemath actually thinking about it; with replacement, you expect to draw any $4$ aces by the $52^{nd}$ draw - by simple expected value of the binomial. You would still not expect to have drawn all $4$. $\endgroup$ – samerivertwice Apr 9 '17 at 6:59
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What they're looking for is; probability of drawing the first ace, times probability of drawing the 2nd ace given that you drew the first, times probability of drawing the third given that you drew the other two, times probability that you draw the fourth given that you drew the other three.

You will need to either;

A) calculate on the basis of drawing some arbitrary ace at each step, or

B) calculate on the basis of drawing any given ace and then multiply by the number of orders in which the aces can be drawn $(4!)$.

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  • $\begingroup$ Ok, thanks! I will try that and see what I get. $\endgroup$ – allanvc Apr 9 '17 at 13:34

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