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.
Best Answer
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).