Expected number of tosses so that 1 out of 3 bins has 2 balls in it You have 3 bins.  You can toss balls one at a time to the bins until any one of them has 2 balls in it, and then you stop.
The tosses are independent, and each bin is equally likely to be hit.  It's impossible to not make it in a bin on any given toss.
What is the expected number of tosses so that 1 bin has 2 balls in it?
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Not sure if it's important, but by the pigeon hole principle we know that the maximum number of balls tosses required to do this would be 4.
 A: It can't happen in 1 toss.  In 2 tosses, the first toss doesn't matter (probability 1), but the second must match the first (probability $\frac{1}{3}$), so it happens in 2 tosses with probability $1\cdot\frac{1}{3}=\frac{1}{3}$.  In 3 tosses, the first toss doesn't matter, the second toss must not match the first toss ($\frac{2}{3}$), and the third toss must match one of the first two ($\frac{2}{3}$), so it happens in exactly 3 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$.  In 4 tosses, the first toss doesn't matter, the second toss must not match the first toss, the third toss must not match either of the first two tosses ($\frac{1}{3}$), and the fourth toss will match one of the first three no matter what (1), so it happens in exactly 4 tosses with probability $1\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot 1=\frac{2}{9}$.  Note that the sum of these three probabilities is 1, which confirms that there is no need to look past 4 tosses.  The expected number of tosses is $2\cdot\frac{1}{3}+3\cdot\frac{4}{9}+4\cdot\frac{2}{9}=\frac{26}{9}$.
A: Hint:  So what is the chance that you get success in two throws?  Call it x.  In exactly three throws?  Call it y.  And that tells you the chance that you will need four is 1-x-y.  Then what is the expected number of throws?
