Question

Fair diced, which is unbiased. Each throw is independent.

Step 1. You roll a six-sided die. Let X be the (random) number that you obtain. Step 2. You roll X six-sided dice. Let Y be t

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\small \text{A six-sided fair die is rolled and the outcome is noted in a random variable } X. \text{ So, } X \text{ takes values from } \\ \text{1 to 6 with equal probability of 1/6.} \\ \\ \text{Now, given } X = x, x \text{ other six-sided fair dice are rolled and their sum is denoted by } Y. \text{ If } Y_1, \dots, Y_x \\ \text{denote the outcome of } x \text{ different rolls, then each } Y_i \text{ has the same distribution as } X \text{ and also } \\ Y = Y_1 + \dots + Y_x. \\ \text{We know that, } E(X) = \frac{1}{6}(1+2+3+4+5+6) = 3.5. \text{ And from above we can say, } \\ \begin{align*} E(Y | X = x) &= E(Y_1 + \dots + Y_X | X = x) \\ &= E(Y_1 + \dots + Y_x) \\ &= E(Y_1) + \dots + E(Y_x) \\ &= x\cdot E(Y_1) & (\text{since } Y_i \text{'s are identical in distribution}) \\ &= 3.5x & (Y_1 \text{ has same distribution as } X \text{ and } E(X) = 3.5) \end{align*} \\ \text{Then using \textit{Law of Total Expectation}, we get }\\ E(Y) = E(E(Y | X = x)) = E(3.5X) = 3.5E(X) = 3.5*3.5 = 12.25.

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