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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?

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    $\begingroup$ It is hardly going to be Markov Chains, since there is a finite number of cards. It will be pure combinatorics, I guess. $\endgroup$ – String Apr 22 '15 at 19:22
  • $\begingroup$ 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. $\endgroup$ – String Apr 22 '15 at 19:24
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    $\begingroup$ See: math.stackexchange.com/questions/245354/expected-value-of-sums Just replace "ace" with "$4$ or $5$". $\endgroup$ – ploosu2 Apr 22 '15 at 19:32
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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 $53/9$.

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

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    $\begingroup$ 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. $\endgroup$ – Paddling Ghost Apr 22 '15 at 19:39
  • $\begingroup$ @PaddlingGhost: There is a lot of confusion here about indices ... Sorry, I will try to clean it up. $\endgroup$ – String Apr 22 '15 at 19:44
  • $\begingroup$ i think it looks good now. $\endgroup$ – Paddling Ghost Apr 22 '15 at 19:51
  • $\begingroup$ So our expected values are not matching. Would you mind taking a look at my answer to see if you can find any errors? $\endgroup$ – Paddling Ghost Apr 22 '15 at 19:53
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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, $$E(x) = 53/9 = 5.888...$$

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  • $\begingroup$ 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. $\endgroup$ – String Apr 22 '15 at 20:00
  • $\begingroup$ Made some silly mistakes there. Thanks for the help in cleaning it up. $\endgroup$ – Paddling Ghost Apr 22 '15 at 20:10
  • $\begingroup$ No problem. Looks good now! (+1) $\endgroup$ – String Apr 22 '15 at 20:12

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