Giving randomly 5 red balls and 5 blue balls to 5 kids, each kid get 2 balls. What is the expected value of number of kids got 2 different balls?

I solve it by:

Let $X_i$ be random variable that holds whether kid $i$ got different balls (i.e. red and blue), or same balls (i.e. red and red or blue and blue).

$X_i = 1 \iff$ got red and blue. $X_i = 0 \iff$ got blue and blue or red and red.

Get probabilities: $P(X_i = 0) = red\cdot red + blue\cdot blue = \frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}2{} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.$

From here, $P(X_i = 1) = 1 - P(X_i = 0) = \frac{1}{2}$.

Find $E(X_i = 1) =\sum_1^5 X_i\cdot P(X_i = 1) = 5\cdot \frac{1}{2} = \frac{5}{2}$.

It's not the answer (got 0/9 on the exam sheet).

Why? What is the correct answer? Thanks in advance.

  • $\begingroup$ Note that given your initial conditions it is impossible to get RR BB for all of the kids. At least one kid will end up with two different colored balls. $\endgroup$ – adam May 13 '15 at 16:56

Your approach with indicator variables using linearity of expectation went exactly in the right direction. You just used the wrong probabilities. Give the first child a ball; no matter which colour it is, $5$ of the remaining $9$ balls are of the other colour, so the probability that the second ball you give them is of the other colour is $\frac59$. Thus the expected number of kids with different colours is $5\cdot\frac59=\frac{25}9$.


I solved it with some brute force but I found it to be the easiest way. If there were a higher number of balls and kids It would be better to use some algebra.

You can have $5$, $3$ or $1$ kid(s) with different colored balls. You only need to count in how many ways each can happen.

For {{R,B},{R,B},{R,B},{R,B},{R,B}} there's only one possibility.

For {{R,B},{R,B},{R,B},{R,R},{B,B}} there's $5\times 4=20$ options (5 possible kids who have the red balls, then 4 possible kids who have the blue balls).

For {{R,B},{R,R},{R,R},{B,B},{B,B}} there's $5\times 6=30$ options (5 possible kids who have different balls and $4\choose 2$ to order the other 4).

All together you have $51$ possibilities. Now let's calculate the mean of kids with different boys:

$$\mu = 5\times\frac{1}{51}+3\times\frac{20}{51}+1\times\frac{30}{51}= \frac{5+3\times 20+30}{51}=\frac{95}{51}\\ \mu \approx 1.863 $$

Hope it helped you.

  • $\begingroup$ These $51$ outcomes are not equiprobable, so you can't calculate a probability as the fraction of favourable outcomes. There are $10!$ elementary outcomes (the permutations of the $10$ balls), so the denominator of the expected value of an integer can't contain a factor of $17$. $\endgroup$ – joriki Jun 19 '16 at 11:21

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