Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.
I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.
Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.
PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.
Best Answer
Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$ Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$
Given any particular value of $X,$ the probability that the twenty-sided die will roll higher is $$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$
The overall probability that the twenty-sided die will roll higher than the total on the other three dice is \begin{align} P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\ &= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n). \end{align}
The last line of that set of equations is just the expected value of $1 - \frac1{20}X.$ That is, \begin{align} P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\ &= 1 - \frac1{20} \mathbb E[X] \\ &= 1 - \frac1{20} \left(\frac{21}{2}\right) \\ &= \frac{19}{40}. \end{align}
If the question is actually the one posed in the original question body rather than in the original title, namely the probability that $X > Y,$ then we simply observe that for any given value of $X,$ $$ P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$
The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$ We find that \begin{align} P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\ &= \frac1{20} \mathbb E[X] - \frac1{20}\\ &= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\ &= \frac{19}{40}. \end{align}
This should not be surprising, because it also follows from $P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$