# 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?

-- 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.

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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 $2\cdot\frac{1}{3}+3\cdot\frac{4}{9}+4\cdot\frac{2}{9}=\frac{26}{9}$.