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A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, and two marked 4. Tickets are drawn at random from the box without replacement until a number appears that has appeared before. Let X be the number of draws that are made.

so the probability when X = 1 is: $4*\frac{2}{8}*\frac{1}{7}$ which is $\frac{1}{7}$. when X = 2, in my understanding, the probability should be $4*\frac{2}{8}*\frac{6}{7}*\frac{1}{6}$, which is $\frac{1}{7}$ again. I doubt my answer since I don't think the two probabilities will be the same, but I couldn't spot where my reasoning went wrong. Any help will be appreciated. Thank you.

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2 Answers 2

up vote 2 down vote accepted

The question asks what is the expected number of draws before a match appears. There is no way to get a match on the first draw. On the second draw, the probability of a match is 1/7. On the third draw, (assuming no match on the first) the probability is 2/6. On the fourth (if we get that far) it is 3/5. On the fifth, it is 1. So the average is 2(1/7)+3(6/7)(2/6)+4(6/7)(4/6)(3/5)+5(6/7)(4/6)(2/5).

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thank you! I think I got it now! –  Gigi Nov 7 '11 at 5:25

You are right, the probabilities are not the same, although I am a little puzzled by your case "$X=1$." The smallest possible value of the number of draws until we get our first match is $2$. But perhaps you are not counting the last draw.

So let $Y$ be the number of draws until the first match, including the draw that got us that match. It is easy to see that $P(Y=2)=\frac{1}{7}$. It doesn't matter what we first drew. The probability that the next draw matches it is $\frac{1}{7}$.

For $Y=3$, there are two possible patterns, which I will call $abb$ and $aba$ (where $a\ne b$). We don't care what the first draw is, call it $a$. The probability of getting something different ($b$) on the next draw is $\frac{6}{7}$. And given we got $a$ then $b$ on the first two draws, the probability of drawing $b$ on the third is $\frac{1}{6}$. So the probability of a pattern of type $abb$ is $\frac{6}{7}\cdot\frac{1}{6}$.

Similarly, given that we got the pattern $ab$ on the first two draws, the probability of getting $a$ on the third is $\frac{1}{6}$. Thus $$P(Y=3)=\frac{6}{7}\cdot\frac{1}{7}+\frac{6}{7}\cdot\frac{1}{6}=\frac{2}{7}.$$

The calculation can be collapsed into one step. Given that we got something different from $a$ on the second draw, there are $2$ (out of $6$) tickets that will give us a match, so the probability is $\frac{6}{7}\cdot\frac{2}{6}$. The mistake in the calculation is in the last term of your product. You had $\frac{1}{6}$ there instead of the correct $\frac{2}{6}$. The $4\cdot\frac{2}{8}$ part was unnecessary but harmless.

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