Suppose we roll 2 n-sided dice, but we keep re-rolling the second while it is equal to the first (we guarantee different face values). What is the EV of the product?
Now clearly if they could be the same it’s the EV of a single n-sided die squared, but the distinct component makes this EV less.
I was tempted to say we can find the EV conditioned on the first being (n+1)/2 and take the average of the product of the second due since that is closed-form-able since it only relies upon knowing the closed form of the sum of the first n numbers and the sum of the first n squares. However, this can’t work when n%2 is 1 and I’m not even sure this is an appropriate condition to make to begin with for n%2=0 dice.
This can obviously be brute forced, but I wanted to know if there’s a closed form here.
Model the tosses with random variables $X_1,X_2$ which are iid uniform on $\{1,\ldots, n\}$, note that $$P(X_1=i, X_2=j|X_1\neq X_2) = \begin{cases} 0 &\text{if } i=j \\ \frac{1}{n^2-n} &\text{if } 1\leq i\neq j \leq n. \end{cases}$$ Compute $$\begin{aligned}E[X_1X_2 | X_1\neq X_2] &=\sum_{1\leq i,j\leq n} ij P(X_1=i, X_2=j|X_1\neq X_2) \\&= \frac{1}{n^2-n}\sum_{1\leq i\neq j\leq n} ij \\ &= \frac{1}{n^2-n}\Big(\sum_{1\leq i,j\leq n} ij - \sum_{1\leq i\leq n} i^2\Big) \\&= \frac{1}{n^2-n}\Big(\big[\sum_{1\leq i\leq n} i\big ]^2 - \sum_{1\leq i\leq n} i^2\Big) \\&= \frac{(n+1)(3n+2)}{12}. \end{aligned}$$