How can 20 balls, 10 white and 10 black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it?

Intuitively i know the right answer: put 1 white, 0 blacks in one urn and 9 white and 10 blacks in the other, but i just want to arrive at it with more mathematical arguments.

Here's my attempt:

Define the events:

W:= the ball taken is white

B:=the ball taken is black

$U_1$:=the ball taken is from urn 1

$U_2$:= the ball taken is from urn 2.

We are interested in W event, which can be written as the following disjoint union:

$W= WU_1\cup WU_2 $.

Hence, $P(W)=P(W|U_1)P(U_1)+P(W|U_2)P(U_2)$.

Supposing that we put $u$ balls in the first urn, and that from these $w$ are white, we have the following distribuition:

$U_1:$ $u$ balls, $w$ white and $u-w$ black.

$U_2:$ $20-u$ balls, $10-w$ white and $20-10+w$ black.

Therefore, $P(W)=\frac{w}{2u}+\frac{10-w}{2(20-u)}$

I'm actually having trouble in order to maximize it. Can someone help?

  • 1
    $\begingroup$ You could hold $u$ constant and find optimal $m$. Then find optimal $u$ $\endgroup$
    – Henry
    Apr 9 '17 at 12:42

We have for $1 \leq u \leq 19$ and $\max\{0, u - 10\} \leq w \leq \min\{u, 10\}$, $$ P(W) = \frac{w}{2u} + \frac{10 - w}{2(20 - u)} = \frac{10}{2(20 - u)} + w\left(\frac{1}{2u} - \frac{1}{2(20-u)}\right) $$ If $u \geq 10$, then $\frac{1}{2u} \leq \frac{1}{2(20 - u)}$, thus $w$ should be equal to $u - 10$ to maximize $P(W)$. In this case, $$ P(W) = \frac{u - 10}{2u} + \frac{1}{2} $$ and it is maxmized when $u = 19$. The corresponding $P(W)$ is $\frac{14}{19}$.

If $u < 10$, then $\frac{1}{2u} > \frac{1}{2(20 - u)}$, thus $w$ should set as $u$ to maximize $P(W)$. When $w = u$, we have $$ P(W) = \frac{10 - u}{2(20 - u)} + \frac{1}{2} $$ It is maximized when $u = 1$. The corresponding $P(W) = \frac{14}{19}$.

Now consider the special case when one urn contains all ball, which is not captured by the formula of $P(W)$. $P(W)$ in this case is $\frac{1}{4} < \frac{14}{19}$.

By above, we conclude that the optimal strategy is to put $10$ black balls and $9$ white balls in the first urn and $1$ white ball in the second urn.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.