# Expected Value Probability choosing same color balls in one hand

Urn 1 has one black ball and one white ball Urn 2 has 2 black 2 white Urn 3 has 2 black 3 white If in each round, you randomly draw three balls, one from each urn, and continue until you draw three balls of the same color in one round (the balls are put back after each round, so there is replacment), and X is the number of rounds it takes to get three balls of the same color, what is the expectation of X?

Could someone write out/ explain the formula I would Use to calculate this?

• Since there are equal numbers of black and white balls in Urns 1 and 2, it's easy to see that the probability of "success" on any given round is $1//4$: Whatever color you pick from Urn 3, it'll be matched with probability $1/2$ from Urn 2 and $1/2$ from Urn 1. – Barry Cipra Oct 9 '17 at 14:30

In each round $r$ with probability of $$p = P(\text{three balls are black}) + P(\text{three balls are white}) = (\frac{1}{2}\times \frac{2}{4} \times \frac{2}{5}) + (\frac{1}{2} \times \frac{2}{4} \times \frac{3}{5})$$ we have 3 balls with same color.
Let $X$ be the number of rounds it takes to get three balls of the same color. $X$ has geometric distribution ($X\sim G(p)$).
We know if $Y\sim G(q)$, then $E(Y) = \frac{1}{q}$. Thus, $$\bbox[5px,border:2px solid #C0A000]{E(X) = \frac{1}{p} = \frac{1}{\frac{1}{2}\times \frac{2}{4} \times \frac{2}{5} + \frac{1}{2} \times \frac{2}{4} \times \frac{3}{5}}}$$