Probability Theory – Problem 56: Find the expectation and variance of

In a certain game, killing a particular monster has a fixed "drop rate" for producing items $1,2$. Suppose that after killing the monster, item $1$ drops with probability $0.2$, item $2$ drops with probability $0.1$, and no item drops with probability $0.7$. Let $\tau$ denote the minimum number of kills required to collect both items $1$ and $2$. Find the expectation and variance of $\tau$.

Each kill of the monster has three mutually exclusive and exhaustive outcomes: drop of item 1 (probability $p=0.2$), drop of item 2 (probability $q=0.1$), and no drop (probability $r=0.7$), with each kill being independent. Define $X$ as the number of kills until the first drop of either item 1 or item 2, and $Y$ as the number of additional kills needed after obtaining one item to obtain the other. Clearly $\tau = X + Y$, and $X$ and $Y$ are independent, so $E[\tau] = E[X] + E[Y]$ and $\text{Var}(\tau) = \text{Var}(X) + \text{Var}(Y)$. Analysis of the distribution of $X$: $X$ is the number of kills until the first "success," where "success" means dropping item 1 or item 2, with success probability $s = p + q = 0.2 + 0.1 = 0.3$. $X$ follows a geometric distribution with parameter $s=0.3$ (defined as $P(X=k) = (1-s)^{k-1}s$, $k=1,2,\dots$), with expectation $E[X] = \frac{1}{s}$ and variance $\text{Var}(X) = \frac{1-s}{s^2}$. Substituting $s=0.3$: $E[X] = \frac{1}{0.3} = \frac{10}{3}$, $\text{Var}(X) = \frac{1-0.3}{0.3^2} = \frac{0.7}{0.09} = \frac{70}{9}$. Analysis of the expectation of $Y$: We use the law of total expectation. Let $Z$ denote the type of the first item obtained ($Z=1$ for item 1, $Z=2$ for item 2). Then $P(Z=1) = \frac{p}{s} = \frac{0.2}{0.3} = \frac{2}{3}$, $P(Z=2) = \frac{q}{s} = \frac{0.1}{0.3} = \frac{1}{3}$. When $Z=1$, we need to obtain item 2 for the first time, with success probability $q=0.1$ per trial, giving geometric expectation $\frac{1}{q}$; when $Z=2$, we need to obtain item 1 for the first time, with success probability $p=0.2$ per trial, giving geometric expectation $\frac{1}{p}$. Thus $E[Y|Z=1] = \frac{1}{0.1} = 10$, $E[Y|Z=2] = \frac{1}{0.2} = 5$. By the law of total expectation: $E[Y] = 10 \times \frac{2}{3} + 5 \times \frac{1}{3} = \frac{20 + 5}{3} = \frac{25}{3}$. Analysis of the variance of $Y$: We use the law of total variance $\text{Var}(Y) = E[\text{Var}(Y|Z)] + \text{Var}(E[Y|Z])$. First, the expectation of the conditional variance: the variance of a geometric distribution is $\frac{1-\text{success prob.}}{\text{success prob.}^2}$, so $\text{Var}(Y|Z=1) = \frac{1-q}{q^2} = \frac{1-0.1}{0.1^2} = 90$, $\text{Var}(Y|Z=2) = \frac{1-p}{p^2} = \frac{1-0.2}{0.2^2} = 20$. Substituting: $E[\text{Var}(Y|Z)] = 90 \times \frac{2}{3} + 20 \times \frac{1}{3} = \frac{180 + 20}{3} = \frac{200}{3}$. Next, the variance of $E[Y|Z]$: $E[Y|Z]$ takes values 10 and 5 with probabilities $\frac{2}{3}$ and $\frac{1}{3}$, respectively. So $E[(E[Y|Z])^2] = 10^2 \times \frac{2}{3} + 5^2 \times \frac{1}{3} = \frac{200 + 25}{3} = 75$, and $(E[Y])^2 = \left(\frac{25}{3}\right)^2 = \frac{625}{9}$. Hence $\text{Var}(E[Y|Z]) = 75 - \frac{625}{9} = \frac{675 - 625}{9} = \frac{50}{9}$. Therefore $\text{Var}(Y) = \frac{200}{3} + \frac{50}{9} = \frac{600 + 50}{9} = \frac{650}{9}$. $E[\tau] = \frac{10}{3} + \frac{25}{3} = \frac{35}{3}$, $\text{Var}(\tau) = \frac{70}{9} + \frac{650}{9} = \frac{720}{9} = 80$.

Expectation: $\dfrac{35}{3}$, Variance: $80$