# What's the chance that exactly 2 2s will have appeared before the 2nd 8?

A standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.

What is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?

I think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.

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Hint: one approach is to recognize that all the other cards are just fluff. There are only ${8 \choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count. You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability. The first are failures, the second are successes if you have two 2's already. For example, one is that you start with three 2's. This is $\frac 12 \cdot \frac 37 \cdot \frac 26=\frac 1{14}$. Another is 2828. This is $\frac 12 \cdot \frac 47 \cdot \frac 12 \cdot \frac 35=\frac 3{35}$

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Oh! I think I got it from that hint! The only way for this to hold is for the string of combos be 8228[some ordering of 2 2s and 2 8s], 2828[some ordering of 2 2s and 2 8s], or 2288[some ordering of 2 2s and 2 8s]. Hence, the solution is 3 * (4 choose 2) / (8 choose 4), which is approximately 0.257! :D ... right? – David Faux Sep 9 '12 at 4:03
@DavidFaux: Good thought, but I wouldn't be sure that all the cases are equally probable. 8228 is $\frac 48 \frac 47 \frac 36 \frac 35=\frac 3{35}$ while 2288 is $\frac 48 \frac 37 \frac 46 \frac 35$, which is the same, so I think it works. I would report the result as $\frac 9{35}$, which is exact. – Ross Millikan Sep 9 '12 at 4:20
Hmm, I think the combinations are equal because non-2 and non-8 cards are all fluff. If we make the 8s non-unique, and 2s non-unique, this basically becomes a problem involving strings of four 8s and four 2s, right? – David Faux Sep 9 '12 at 4:34
@DavidFaux: That is right. The more I think, I agree with your calculation. I was worried that 22228888 would be less probable than 82282882, but now I don't think so. It is easy to think things are equally probable when they are not. – Ross Millikan Sep 9 '12 at 4:39

As was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.

We count the number of "favourable" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$.

The pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is $$\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$ If we want to simplify a bit, we get $\dfrac{9}{35}$.

Remark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.

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