# Expectation of the number of cards drawn without replacement until get all four aces (using conditioning)

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.

• 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 $52.$ – coffeemath Apr 8 '17 at 5:18
• I think $52$ makes sense with replacement. – samerivertwice Apr 8 '17 at 8:53
• @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 $204/5=40.8.$ – coffeemath Apr 8 '17 at 21:34
• @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 $4$. – samerivertwice Apr 9 '17 at 6:59

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 $(4!)$.