Probability that the median roll of 21 rolls is 6 Let's say I roll a fair 6-sided dice 21 times. Assume  tracks the median of the 21 numbers rolled. What is ( = 6)?
So far, I believe that this implies at least 11 of the rolls have to be 6, and the other 10 can be any number. I would simply take 1 - (5/6)^11. However, I'm not sure if that's the right way to solve this problem.
 A: You are correct that you need at least $11$ rolls to be $6$ for the median to be $6$.  You should explain the reasoning behind your expression.  It seems you are choosing a specific set of $11$ rolls, calculating the chance that all of them are not $6$ with the $(\frac 56)^{11}$ and subtracting from $1$.  This gets the chance that at least one of those $11$ rolls is a $6$, which is not what you want.
You need to look up the binomial distribution, where each roll has two possibilities, a $6$ or not a $6$.  You then compute the chance there are $11\ 6$s, $12\ 6$s, etc. and add them up.  It would be easiest to do this in a spreadsheet where you can make a column from $11$ to $21$ for the number of $6$s and copy down to get each probability, then add the column.
A: If we have the median is 6, this means that 11 of the rolls must be 6 and the other 10 are less than or equal to 6.
As such, probability that there are up to 10 rolls that are not 6. Define another random variable $X$, which is the number of rolls out of the 21 that are not 6. We can calculate the probability of the number of non-six rolls as $P(X=x) = {21 \choose x} \left(\frac{5}{6}\right)^x \left(\frac{1}{6}\right)^{21-x}$. This is a binomial distribution that we can calculate the probability of the median as its cumulative distribution.
$$P(M=6) = P(X\leq10) = \sum_{x=0}^{10} {21 \choose x} \left(\frac{5}{6}\right)^x \left(\frac{1}{6}\right)^{21-x} = 0.00018730$$
