Probability Theory – Problem 6: Find the expectation and variance of

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

Each kill has three mutually exclusive and exhaustive outcomes: equipment 1 drops (probability $p=0.2$), equipment 2 drops (probability $q=0.1$), or no equipment drops (probability $r=0.7$), and all kills are mutually independent. Define $X$ as the number of kills needed until the first drop of either equipment 1 or equipment 2, and $Y$ as the number of additional kills needed after obtaining one piece of equipment to obtain the other. Clearly $\tau = X + Y$, and $X$ and $Y$ are independent, so $E[\tau] = E[X] + E[Y]$ and $\text{Var}(\tau) = \text{Var}(X) + \text{Var}(Y)$. Step 1. Analysis of the distribution of $X$: $X$ is the number of kills until the first "success," where "success" means dropping equipment 1 or equipment 2, with success probability $s = p + q = 0.2 + 0.1 = 0.3$. Thus $X$ follows a geometric distribution with parameter $s=0.3$ (defined by $P(X=k) = (1-s)^{k-1}s$, $k=1,2,\dots$), whose expectation is $E[X] = \frac{1}{s}$ and variance is $\text{Var}(X) = \frac{1-s}{s^2}$. Substituting $s=0.3$ gives $E[X] = \frac{1}{0.3} = \frac{10}{3}$ and $\text{Var}(X) = \frac{1-0.3}{0.3^2} = \frac{0.7}{0.09} = \frac{70}{9}$. Step 2. Analysis of $E[Y]$ via the law of total expectation: Let $Z$ denote the type of equipment obtained first ($Z=1$ for equipment 1, $Z=2$ for equipment 2). Then $P(Z=1) = \frac{p}{s} = \frac{0.2}{0.3} = \frac{2}{3}$ and $P(Z=2) = \frac{q}{s} = \frac{0.1}{0.3} = \frac{1}{3}$. When $Z=1$, we need the first drop of equipment 2, with per-trial success probability $q=0.1$, so the geometric expectation is $\frac{1}{q}$; when $Z=2$, we need the first drop of equipment 1, with per-trial success probability $p=0.2$, so the geometric expectation is $\frac{1}{p}$. Thus $E[Y|Z=1] = \frac{1}{0.1} = 10$ and $E[Y|Z=2] = \frac{1}{0.2} = 5$. By the law of total expectation, $E[Y] = 10 \times \frac{2}{3} + 5 \times \frac{1}{3} = \frac{20 + 5}{3} = \frac{25}{3}$. Step 3. Analysis of $\text{Var}(Y)$ via the law of total variance $\text{Var}(Y) = E[\text{Var}(Y|Z)] + \text{Var}(E[Y|Z])$. First, compute the expected conditional variance: the variance of a geometric distribution is $\frac{1 - p_{\mathrm{success}}}{p_{\mathrm{success}}^2}$, so $\text{Var}(Y|Z=1) = \frac{1-q}{q^2} = \frac{1-0.1}{0.1^2} = 90$ and $\text{Var}(Y|Z=2) = \frac{1-p}{p^2} = \frac{1-0.2}{0.2^2} = 20$. Thus $E[\text{Var}(Y|Z)] = 90 \times \frac{2}{3} + 20 \times \frac{1}{3} = \frac{180 + 20}{3} = \frac{200}{3}$. Next, compute the variance of $E[Y|Z]$: the possible values of $E[Y|Z]$ are 10 and 5, with probabilities $\frac{2}{3}$ and $\frac{1}{3}$ respectively. Thus $E[(E[Y|Z])^2] = 10^2 \times \frac{2}{3} + 5^2 \times \frac{1}{3} = \frac{200 + 25}{3} = 75$, and $(E[Y])^2 = \left(\frac{25}{3}\right)^2 = \frac{625}{9}$, so $\text{Var}(E[Y|Z]) = 75 - \frac{625}{9} = \frac{675 - 625}{9} = \frac{50}{9}$. Therefore $\text{Var}(Y) = \frac{200}{3} + \frac{50}{9} = \frac{600 + 50}{9} = \frac{650}{9}$. Step 4. $E[\tau] = \frac{10}{3} + \frac{25}{3} = \frac{35}{3}$, and $\text{Var}(\tau) = \frac{70}{9} + \frac{650}{9} = \frac{720}{9} = 80$.

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