Three Urns

The first urn holds $1$ white ball and $2$ black; the second holds $2$ white and $1$ black; the third holds $2$ white and $2$ black. A blindfolded man moves one ball from the first urn to the second, then one ball from the second to the third, then draws a ball from the third. Is it an even chance that the ball he draws is white?

Solution

It is not quite even. The chance of white is $\tfrac{31}{60}$, a shade above one half.

Follow the two transfers. The ball leaving the first urn is white with probability $\tfrac13$, black with probability $\tfrac23$. That changes the second urn before the next draw:

• With probability $\tfrac13$ the second urn becomes $3$ white, $1$ black, so the ball passed on to the third is white with probability $\tfrac34$.

• With probability $\tfrac23$ it becomes $2$ white, $2$ black, so the ball passed on is white with probability $\tfrac12$.

So the ball arriving at the third urn is white with probability $$\frac13 \cdot \frac34 + \frac23 \cdot \frac12 = \frac14 + \frac13 = \frac{7}{12}.$$ The third urn, originally $2$ white and $2$ black, gains this ball and then has five balls. Its white count is $2$ plus the newcomer, so the chance the final draw is white is $$\frac{1}{5}\left(2 + \frac{7}{12}\right) = \frac{1}{5} \cdot \frac{31}{12} = \frac{31}{60} \approx 0.517.$$ The blindfold tilts the odds very slightly towards white.