We roll $4$ fair, $6$-sided dice. What is the expected value of the product of the two smallest scores?
For example, if we rolled, say, $1,6,4,5$ then the product of the two smallest scores is $1\cdot 4=4$, or if we rolled $1,1,3,6$ then it would be $1\cdot 1=1$.
Of course we could calculate every outcome and manually compute the expectation but I'm wondering if there's a nicer or quicker way to get it, or just as useful to me would be a decent approximation.
My thoughts: Let $X$ denote the minimum of the $4$ rolls. Then $$\mathbb{E}(X)=\sum^{6}_{n=1}\mathbb{P}(X\ge n)=\sum^{6}_{n=1}\Big(\frac{n}6\Big)^4$$ I was then hoping the find the second smallest similarly by finding the expected value of the minimum of the remaining $3$, but these are dependent on $X$ and so I am unable to compute this.
A very rough approximation can be found by considering that by symmetry we expect the two smallest to lie in $\{1,2,3\}$, so the second lowest score (which we'll now denote by $Y$) should lie in $[X,3]$, we may then take $\mathbb{E}(Y)$ as the midpoint of this interval and take the expected value of the product of $X$ and $Y$ as $$\mathbb{E}(X)\cdot\mathbb{E}(Y)=\mathbb{E}(X)\cdot\mathbb{E}(\frac{X+3}2)=\frac12\big(\mathbb{E}(X^2)+3\mathbb{E}(X)\big)$$ but $X$ and $Y$ are not independent so I don't think this is at all a good approximation.
Any insight is appreciated. Thank you.
Going through the $6^4=1296$ cases one obtains the expected value $$E={7469\over1296}\doteq5.76312\ .$$ Your argument is very shaky. Note that the product of the two smallest numbers can be as large as $36$. Generally speaking: The result of the throw is a multiset of cardinality $4$ on $[6]$. Going through all these multisets and their probabilities might be as cumbersome as counting the $1296$ possible cases individually.