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Three $6$-sided fair dice are rolled. In $10$ independent throws, Let $A$ be the number of times all the sides are the same and let $B$ be the number of times only two sides are the same. Find $E(6AB)$:

Here is how I approached this:

I know that $E(6AB) = 6E(AB)$. I know that $E(AB)$ for a multinomial distribution is:

$$n(n-1)\cdot\text{probability of }A\text{ occurring}\cdot\text{probability of }B\text{ occurring}$$

Probability of $A$ occurring is $6\left(\dfrac16\right)^3$. Probability of $B$ occurring is $6\left(\dfrac16\right)^2 \left(\dfrac56\right)$

Therefore $$E(AB) = 10 \cdot 9 \cdot 6 \left(\frac16\right)^3 \cdot 6 \left(\frac16\right)^2 \cdot \frac56$$ and $E(6AB)$ is just $6$ times the quantity to the left.

I would like to know if my approach above is correct since I don't get the correct solution of $25/4$ Thanks for the help.

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It's unclear (to me) what $$n(n-1)\cdot\text{probability of }A\text{ occurring}\cdot\text{probability of }B\text{ occurring}$$ refers to. $A$ and $B$ were introduced as numbers, not as events. If I assume that this is an abuse of notation and you're using these letters to refer both to events and to the number of occurrences of those events, the formula seems to be wrong: If $A$ and $B$ are both the certain event, the expectation value of the product of the numbers of their occurrences should be $n^2$, not $n(n-1)$. –  joriki Sep 25 '11 at 19:54
    
A and B are the events of rolling 3 identical faces and two identical faces respectively. I think my error is in calculating the probability of B. The n(n-1) is correct because that is part of the explicit formula for calculating E(XY) of a multinomial distribution. –  icobes Sep 25 '11 at 19:56
    
Just ignoring my comments isn't going to improve anything. You addressed neither the inconsistency that I pointed out (the fact that $A$ and $B$ are introduced as numbers, not events), nor the counterexample that I gave for your formula. If you think the counterexample is invalid, please point out why. –  joriki Sep 25 '11 at 20:03
    
From Sasha's answer, it seems that a meaningful reply to my comment would have been "This formula holds only if $A$ and $B$ refer to two different possible outcomes for the multinomial distribution". –  joriki Sep 25 '11 at 21:11
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1 Answer 1

up vote 2 down vote accepted

Let $C$ be the number of throws when all dice have different outcomes. Then $A+B+C=10$, and the triple $\{A,B,C\}$ follows multinomial distribution with parameters $n=10$ and probabilities $p_1 = \frac{6}{6^3} = \frac{1}{36}$, $p_2 = \frac{3 \cdot 6 \cdot 5}{6^3} = \frac{5}{12}$ and $p_3 = \frac{6 \cdot 5 \cdot 4}{6^3} = \frac{5}{9}$. It is easy to see that $p_1 + p_2 + p_3 = 1$.

The moment-generating function for multinomial distribution is $\phi(t_1,t_2,t_3) = \left( p_1 \mathrm{e}^{t_1} + p_2 \mathrm{e}^{t_2} + p_3 \mathrm{e}^{t_3} \right)^{10}$.

Then $\mathrm{E}(A B) = \left. \partial_{t_1} \partial_{t_2} \phi(t_1,t_2, t_3)\right\vert_{t_i=0} = 10 \cdot 9 \cdot p_1 p_2 \cdot \mathrm{e}^{t_1+t_2} \cdot \left( p_1 \mathrm{e}^{t_1} + p_2 \mathrm{e}^{t_2} + p_3 \mathrm{e}^{t_3} \right)^7\vert_{t_i=0} = 90 p_1 p_2 $.

Therefore $\mathrm{E}(6 A B) = 6 \mathrm{E}(A B) = 6 \cdot 90 \cdot p_1 p_2 = \frac{25}{4}$.

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