Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a probability problem with cards and the expected value of a card rank.

I have a deck of 52 cards. I draw cards without replacement. While drawing cards from the deck, a first ace is drawn at rank $k$ (that is the $k^{th}$ card drawn is an ace, all previous were not). We want to find the expected number of additional draws until we get an ace.

My idea is to follow this route:

If I call $X$ the random variable of the rank of the second ace, $N=52$ the total number of cards, $p = N -k$ the remaining number of cards after the first ace is drawn, the idea is to plug the expectation value of $X$ $$E[X] = \sum_{i=1}^{p-3}i P(X=i)$$

But the formula doesn't seem to simplify. What would be your take at this?

share|improve this question
1  
I would think that it would have to see with all the options: once the first ace $a_1$ appears at the k-th step, then the 2nd can appear in the (k+1)st, (k+2)nd,...52nd step, each time with probability $\frac {3}{52-k-t}$, where t is the number of cards removed after the k-th card. –  gary Aug 3 '11 at 15:52
    
Actually, if you have a specific formula, maybe we can help you simplify it; the one I gave, I think, is along what you have in mind. –  gary Aug 3 '11 at 15:53
    
Correction: the 2nd ace must have appeared at the (by the ) 50th step. –  gary Aug 3 '11 at 16:13
add comment

2 Answers

up vote 3 down vote accepted

We have $3$ Aces left, and $52-k$ cards left. We will find the expected number of additional draws until we get an Ace.

We change the notation, and solve a more general problem. There are $b$ boys and $g$ girls. People are chosen, without replacement, until we get a boy. Let $X$ be the number of draws. We want to find $E(X)$. In our card problem, we have $b=3$ and $g=52-k-3$.

Label the girls $G_1, G_2, \dots, G_g$. For $i=1$ to $g$, define the random variable $Y_i$ by

$$Y_i=1 \quad\text{if $G_i$ is drawn before a boy},\quad \text{$Y_i=0$ otherwise}.$$

Then

$$X=1 +(Y_1+Y_2+\cdots +Y_g),$$ since $\sum E(Y_i)$ is the mean number of girls before the first boy.

We will be finished if we can find the $E(Y_i)$. These are all the same, so we find $E(Y_1)$.

But note that $$E(Y_1)=P(Y_1=1)=\frac{1}{b+1}.$$

This is because in the group of $b+1$ people consisting of $G_1$ and the $b$ boys, $G_1$ must be chosen first, and all possible orders among these $b+1$ people are equally likely. To see this, it is necessary to think of the selections going on until everybody is chosen.

Thus $$E(X)=1+\frac{g}{b+1}=\frac{b+g+1}{b+1}.$$

Comment: There are less pleasant, but more combinatorial ways to solve the problem. We can without much difficulty find an expression for the probability that the first boy is chosen on the $i$-th choice, and use the ordinary expression for expectation. It is a bit of a mess, and simplifying it is tricky. But it is easier now that we know, from the simpler calculation above, what the answer should be!

share|improve this answer
    
+1. I agree on pleasant. –  Did Aug 3 '11 at 17:48
    
Great André, thank you, that is a very neat explanation. Looking at your track record on this site one can tell you're really good, but how did the idea of writing X as a sum of Y come out? Is it a common pattern? –  alex_C Aug 3 '11 at 17:55
2  
I have seen (and taught) the standard indicator random variable way of dealing with the geometric distribution. Had to think a bit about how to adapt to this situation, since it has been a while. By the way, one can without much trouble adapt to find mean waiting time until the $p$-th boy. –  André Nicolas Aug 3 '11 at 18:02
add comment

The five strings of non-aces can be permuted amongst each other, conserving probability. Note that some of the strings may be empty and where there are multiple empty strings, the 120 possible "abstract" permutations will produce fewer than 120 physical permutations of the cards.

The symmetry implies that the expected length each string of non-aces is the same, or 48/5 for a standard deck of cards.

The expected waiting time to the first ace, or from the first to the second ace, or ace $k$ to ace $k+1$, is the preceding number $+1$. That is the expected length of a full interval of non-aces followed by an ace. So 53/5 is the answer for a standard deck.

The general answer is (cards+1)/(aces+1), by the same argument.

share|improve this answer
    
Thanks this is a great way to solve the problem. –  alex_C Aug 4 '11 at 8:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.