# Expected number of cards drawn before drawing a $4$ or $5$

I'm working on the following problem:

Compute the number of expected cards drawn from a standard 52 card deck (without replacement) until a $4$ or $5$ is drawn.

I tried to model it using a geometric distribution, but am running into problems since the probability of drawing a $4$ or $5$ increases with each successive card drawn. Could this problem be approached using Markov Chains?

• It is hardly going to be Markov Chains, since there is a finite number of cards. It will be pure combinatorics, I guess. – String Apr 22 '15 at 19:22
• If $P(n)$ denotes the probability of drawing $4,5$ after exactly $n$ cards, we have $P(1)=8/52,P(2)=(1-P(1))\cdot 8/51=44\cdot 8/(52\cdot 51)$ for a start. – String Apr 22 '15 at 19:24
• See: math.stackexchange.com/questions/245354/expected-value-of-sums Just replace "ace" with "$4$ or $5$". – ploosu2 Apr 22 '15 at 19:32

Let's call the 4s and 5s "special" cards. Add a joker to the deck and pretend it's an additional special card, so that there are now $9$ special cards in a deck of $53$ cards. Now shuffle all the cards up and then deal them out, face down, in one big circle. If you think about it, the average distance between consecutive special cards is $53/9$. Now locate the joker and think of it as identifying the "top" of the deck. The average distance to the next special card (which is now either a 4 or a 5) is still $53/9$.

• Yes, indeed! This is incredible. – String Apr 22 '15 at 21:15

The number of ways you can draw $n$ cards from $52$ is $P(n,52)=\frac{52!}{(52-n)!}$. Note that order matters here. The number of ways all of those are not $4,5$ will be $P(n,44)=\frac{44!}{(44-n)!}$. Here $P$ stands for permutations.

Now let $P(n)$ (not the same meaning of $P$) denote the probability that we fail the first $n-1$ draws and succeed in the $n$-th. Then \begin{align} P(n)&=\frac{P(n-1,44)}{P(n-1,52)}\cdot\frac{8}{(52-(n-1))}\\ &=\frac{(52-(n-1))!44!}{52!(44-(n-1))!}\cdot\frac{8}{(52-(n-1))}\\ &=\frac{8(52-n)!44!}{52!(44-(n-1))!} \end{align} and we can "simply" sum $$E=\sum_{n=1}^{45}n\cdot P(n)=\frac{53}{9}=5.\overline 8$$ to get the expected number of cards to draw.

• I believe you should only be summing until $n=44$ as that is the last n in the probability space. As well, anything bigger than 44 for $n$ will result in a negative number for your factorial. – Paddling Ghost Apr 22 '15 at 19:39
• @PaddlingGhost: There is a lot of confusion here about indices ... Sorry, I will try to clean it up. – String Apr 22 '15 at 19:44
• i think it looks good now. – Paddling Ghost Apr 22 '15 at 19:51
• So our expected values are not matching. Would you mind taking a look at my answer to see if you can find any errors? – Paddling Ghost Apr 22 '15 at 19:53

For this problem, we can develop a formula for $P(x=k)$. Where the variable x represents the draw on which we obtain a success. $$P(x=1)=2/13$$ $$P(x=2) = 44/52*8/51$$ $$P(x=3)=(44/52)*(43/51)*(8/50)$$continuing in this fashion, we obtain the formula: $$P(x=k)=8*(44!/52!) \frac{(52-k)!}{(45-k)!}$$ Thus, we can use the definition of expected value: $E(x)=\sum_{k=1}^{45}k\cdot P(x=k)$ Unfortunately, I know no method to solve this sum. So of course, I resorted to wolfram which spit out $53/9$. Thus, $$E(x) = 53/9 = 5.888...$$

• A couple of things. First $52-8=44\neq 42$. And for $P(x=3)$ we have $44\cdot 43$ in the numerator whereas it is $52\cdot 51\cdot 50$, so not equally many factors. Thus it should be $(52-k)!$ to have $k$ factors, but $(44-k+1)!$ to only have $k-1$ factors for that product. Finally, we may not succeed until drawing the $45$-th card. Not only the $43$-th. But the last bit there hardly makes a difference - they are both so unlikely. – String Apr 22 '15 at 20:00
• Made some silly mistakes there. Thanks for the help in cleaning it up. – Paddling Ghost Apr 22 '15 at 20:10
• No problem. Looks good now! (+1) – String Apr 22 '15 at 20:12