# The "last of aces"

You have a throughly shuffled deck of 52 cards. Each time you choose one card from the deck. The drawn card is put back in the deck and all 52 cards are again throughly shuffled. You continue this procedure until you have seen all four different aces. What are the expected value and the standard deviation of the number of times you have to draw a card until you have seen all four different aces?

Clearly the problem is that I can draw the same ace over and over again, so I should count the number of cards that I draw between the $$k-1$$th success and the $$k$$th success (ie to say, for example, the drawn cards between the third and the fourth ace, each different from the other). But the fact the aces must be different linked to the fact that once drawn an ace this ace is put back in the deck implies that once I draw the third ace the "game" starts from scratch, so I have to calculate the number of trials (drawn cards) necessary to have the first success (ie to say the fourth and last ace not seen yet).

So if $$X_i=[$$# drawn cards before the $$i$$th ace$$]$$, we have that $$X_1,X_2,X_3,X_4\sim\operatorname{Geo}(p)$$.

How can I use this to solve the problem? Thanks in advance for any help.

Let $$X$$ denote the random variable denoting the number of trials to get $$4$$ aces. We can divide $$X$$ into $$4$$ separate random variables, which we will call $$X_1$$ (the number of trials to pull the first ace), $$X_2$$ (the number of trials to pull the second ace after pulling the first one), $$X_3$$ (the number of trials to pull the third ace after pulling the second one) and $$X_4$$ (the number of trials to pull the fourth ace after pulling the third one). The variables' probability distribution is the geometric distribution. For a geometric distribution, we have $$E[Y]=\frac1p$$ and $$\text{Var}(Y)=\frac{1-p}{p^2}$$.

The chance of pulling our first ace in a trial is $$\frac4{52}=\frac1{13}$$ so $$E[X_1]=13$$ and $$\text{Var}(X_1)=156$$. The chance of pulling our third ace in a trial is $$\frac3{52}$$ so $$E[X_2]=\frac{52}3$$ and $$\text{Var}(X_2)=\frac{2548}9$$. The chance of pulling our third ace in a trial is $$\frac2{52}=\frac1{26}$$ so $$E[X_3]=26$$ and $$\text{Var}(X_3)=650$$. The chance of pulling our third ace in a trial is $$\frac1{52}$$ so $$E[X_3]=52$$ and $$\text{Var}(X_4)=2652$$.

Because $$X_1$$, $$X_2$$, $$X_3$$ and $$X_4$$ are independent, we can directly add the expected values and variances to get the expected value and variance of $$X$$. We have

$$E[X]=E[X_1]+E[X_2]+E[X_3]+E[X_4]=13+\frac{52}3+26+52=108\frac13$$

and

$$\text{Var}(X)=\text{Var}(X_1)+\text{Var}(X_2)+\text{Var}(X_3)+\text{Var}(X_4)=156+\frac{2548}9+650+2652=\frac{33670}9$$

To find the standard deviation we take the square root of the variance, that is:

$$\sigma=\sqrt{\text{Var}(X)}=\sqrt{\frac{33670}9}=\frac{\sqrt{33670}}3\approx61.164$$

• The correct result of the standard deviation is $61,16$. Sep 13, 2020 at 13:30
• The discrepancy is probably because your textbook uses less significant digits.
– Kyky
Sep 13, 2020 at 13:53
• I think you have a typo in your solution. $64.164$ should be $61.164$. Sep 13, 2020 at 14:07
• Just realized my mistake. Sorry about that.
– Kyky
Sep 13, 2020 at 14:15