I'm not good with math which is why I'm turning to you here :) So, I'm preparing for a tabletop role playing game session and I'm not happy with the rules when it comes to character creation. The rules say to roll 4d6 (Four six-sided dices) for each stat and remove the the dice with the lowest number. The result of the remaining three dices will be the value for that stat. This is too random so I would like to calculate a pott that the players can then divide into the different stats themselves. There are 6 different stats. So what I want to calculate is the... median? result of six 4d6 rolls where the lowest dice of each throw is removed.
Super grateful for any help as my brain is hurting!
You seem to be content with the mean, but since I’d already worked out how to obtain the median, I might as well post it.
We’ll need the number of ways for $3$ integers in $[k,l]$ to sum to $s$. This can be treated as the problem of distributing $s-3k$ balls into $3$ bins with limited capacity $l-k$. Ignoring the limited capacity would yield the count $\binom{s-3k+2}2$, but we have to subtract the cases that violate the constraints. Since the distribution is symmetric with respect to reflection at the central sum $\frac32(k+l)$, we only need the cases up to that point, and then at most one bin can exceed its capacity. There are $3$ choices for that bin, and $s-2k-l-1$ balls remain to be distributed after $l-k+1$ have been placed to violate the constraint, so the corrected count is
$$ \binom{s-3k+2}2-3\binom{s-2k-l+1}2\;, $$
where, contrary to the usual convention, the binomial coefficient should be taken to be zero if the upper index is negative.
Now we can distinguish $24$ cases according to the lowest value $v$ rolled and the number $n$ of dice that show that value. For $n=1$, we can apply the count derived above to the other three dice that show values in $[v+1,6]$, yielding
$$ \binom{s-3v-1}2-3\binom{s-2v-7}2\;. $$
For $n=2$, no constraints need to be considered in the lower half of the distribution, so the number of ways to form the sum $s$ by distributing $s-v$ balls to the two remaining dice in $[v+1,6]$ is $s-3v-1$.
For $n=3$, there is just one way each to form the sums from $3v+1$ to $2v+6$, and for $n=4$ the only possible sum is $3v$.
We also need to include a factor $\binom4n$ to take into account the different ways to choose the $n$ dice with the lowest value.
This yields the following table:
$$ \begin{array}{c|ccccc|ccccc|ccccc|cccccc|r} n&1&&&&&2&&&&&3&&&&&4&&&&&&\Sigma \\\hline v&1&2&3&4&5&1&2&3&4&5&1&2&3&4&5&1&2&3&4&5&6 \\\hline 3&&&&&&&&&&&&&&&&1&&&&&&1\\ 4&&&&&&&&&&&4&&&&&&&&&&&4\\ 5&&&&&&6&&&&&4&&&&&&&&&&&10\\ 6&4&&&&&12&&&&&4&&&&&&1&&&&&21\\ 7&12&&&&&18&&&&&4&4&&&&&&&&&&38\\ 8&24&&&&&24&6&&&&4&4&&&&&&&&&&62\\ 9&40&4&&&&30&12&&&&&4&&&&&&1&&&&91\\ 10&60&12&&&&24&18&&&&&4&4&&&&&&&&&122\\ 11&72&24&&&&18&24&6&&&&&4&&&&&&&&&148\\ 12&76&40&4&&&12&18&12&&&&&4&&&&&&1&&&167\\ 13&72&48&12&&&6&12&18&&&&&&4&&&&&&&&172\\ 14&60&48&24&&&&6&12&6&&&&&4&&&&&&&&160\\ 15&40&40&28&4&&&&6&12&&&&&&&&&&&1&&131\\ 16&24&24&24&12&&&&&6&&&&&&4&&&&&&&94\\ 17&12&12&12&12&&&&&&6&&&&&&&&&&&&54\\ 18&4&4&4&4&4&&&&&&&&&&&&&&&&1&21\end{array} $$
The table confirms Travis Willse’s result in the comments that the median is $12$ and the mode is $13$, with the mean of $\frac{15869}{6^4}\approx12.2446$ in between.