# Question from a test concerning dice probability

There are 4 dice in the following colors: blue, yellow, black and red. They're rolled at the same time. Consider $X_1$ to be the quantity of dice in which the face value scored was one. Let $X_2, ... , X_4$ be defined in the same manner. Therefore, for example, $X_1$ can be equal to 0,1, 2 , 3 or 4 at maximum, because you can get at maximum 4 times the number 1. That would be if you rolled 1, 1 ,1 and 1. $X_2$ = 3 would mean that you got three times the number two and the remaining die can be any number. The 4 dice are rolled SIMULTANEOUSLY. Let Y = $X_1X_2X_3X_4$. What values Y can get? Get $E(Y)$

• i think you need to clear up your questiln. why do you get 3 for rolling 1 on the 2nd dice roll? – Lost1 Dec 17 '13 at 15:17
• I understand it's confusing but the problem itself is confusing, i've tried to be clear as i can. 'The 4 dice are rolled SIMULTEANOUSLY' so there's no 2nd dice roll. X4 = 4 means you got 4 times the number 4 SIMULTANEOUSLY. – matt_zarro Dec 17 '13 at 15:45

You can only have $Y>0$ if $X_{i}>0$ for each $i\in\left\{ 1,2,3,4\right\}$.

This combined with $X_{1}+X_{2}+X_{3}+X_{4}=4$ leads to $X_{1}=X_{2}=X_{3}=X_{4}=1$ hence $Y=1$. This shows that $Y$ can only take the values $0$ and $1$.

Then:

$\mathbb{E}Y=P\left\{ X_{1}=X_{2}=X_{3}=X_{4}=1\right\}$.

There are $4!=24$ ways to this event, each having probability $6^{-4}$.

So:

$\mathbb{E}Y=24\times6^{-4}$