One Coin, One Die

Let $C$ be the outcome of a coin toss, and let $D$ be the outcome of a die roll. Then $X \; = \; C_1 + C_2 + \cdots + C_D$ , where $C_1,C_2,...,C_D$ are independent, and all distributed as $C$ . Thus $E[t^X|D=k] \; = \; E[t^{C_1+C_2+\cdots+C_k}] \; = \; E[t^C]^k \; = \; G_C(t)^k$ where $G_C$ is the probability generating function of $C$ . Thus $E[t^X|D] \; = \; G_C(t)^D$ and hence $G_X(t) \; = \; E[t^X] \; = \; E[E[t^X|D]] \; = \; G_D(G_C(t))$ Thus $\begin{aligned} G_X'(t) & = \; G_D'(G_C(t))G_C'(t) \\ G_X''(t) & = \; G_D''(G_C(t))G_C(t)^2 + G_D'(G_C(t)) G_C''(t) \end{aligned}$ and since we know that $\mu_X \; = \; G_X'(1) \hspace{2cm} \sigma_X^2 \; = \; G_X''(1) + G_X'(1) - G_X'(1)^2$ is it a simple bit of algebra to show that $\mu_X \; = \; \mu_C \mu_D \hspace{2cm} \sigma_X^2 \; = \; \mu_C^2 \sigma_D^2 + \mu_D \sigma_C^2$ Similarly we can show that $G_Y(t) = G_C(G_D(t))$ and hence that $\mu_Y \; = \; \mu_C\mu_D \hspace{2cm} \sigma_Y^2 \; = \; \mu_D^2\sigma_C^2 + \mu_C \sigma_D^2$ Since we have $\mu_C = \tfrac32$ , $\sigma_C^2 = \tfrac14$ , $\mu_D = \tfrac72$ , $\sigma_D^2 = \tfrac{35}{12}$ , we easily calculate that $\mu_X \; = \; \mu_Y \; = \; \tfrac{21}{4} \hspace{2cm} \sigma_X^2 \; = \; \sigma_Y^2 \; = \; \tfrac{119}{16}$

---

If the die were $n$ -sided, we would have $\mu_D = \tfrac{n+1}{2}$ and $\sigma_D^2 = \tfrac{n^2-1}{12}$ , and then $\mu_X \; = \; \mu_Y \; = \; \tfrac{3(n+1)}{4} \hspace{2cm} \sigma_X^2 \; = \; \sigma_Y^2 \; = \; \tfrac{(n+1)(3n-1)}{16}$ While $\mu_X$ and $\mu_Y$ are always equal, and while $\sigma_X^2$ and $\sigma_Y^2$ are the same for any fair $n$ -sided die, this will no longer be true if the die stops being fair. It is easy to see that the outcome $\sigma_X^2 = \sigma_Y^2$ is equivalent to the identity $\sigma_D^2 \; = \; \tfrac{1}{3}\mu_D(\mu_D-1)$ which holds for any fair die, but not for a biased die.

Mean of X = mean of die times mean of coin = $3.5*1.5 = 5.25$

Mean of Y = mean of coin times mean of die = $1.5*3.5 = 5.25$

Both standard deviations are also the same and can be shown to equal $\frac{\sqrt{119}}{4}$

So $\mu_{X}=\mu_{Y}$ and $\sigma_{X}=\sigma_{Y}$

It is not intuitively obvious that the standard deviations should be the same. Their histograms do not resemble one another but I worked them out exactly a few years ago. My guess is this would work with dice with different numbers of sides as well, but I cannot fathom a formula.