Rolling n m-sided dice - number of outcomes that are less than the first roll

Question

Say I roll 2 fair 6-sided dice and sum the outcomes. If I roll these dice again , in how many ways can I get a new sum that is less than the previous one? I know that if one die is rolled instead, there are $(m^2-m)/2$ outcomes that will be less than the first roll. Can this be extended to rolling n dice?

Answer 1

Let $X_1$ be a random variable equal to the sum of the first roll of the $n$ dice with $m$ sides, and let $X_2$ be the sum of the second roll. By symmetry, $P(X_1>X_2)=P(X_1<X_2)$, which together with $P(X_1<X_2)+P(X_1=X_2)+P(X_1>X_2)=1$ implies that $$ P(X_2<X_1)=\frac12\Big(1-P(X_1=X_2)\Big) $$ There is a clever trick we can use to simplify calculating $P(X_1=X_2)$; it turns out that $$ P(X_1=X_2)=P\big(X_1+X_2=n(m+1)\big) $$ In other words, the event that the two rolls are the same has the same probability that the two rolls have "complementary" sums. To see this, given an outcome where $X_1=X_2$, consider what happens when you replace each value $i$ in the second roll with $m+1-i$. The resulting rolls will now be complementary. Since this is a bijection of outcomes, the probabilities of the events are the same.

Note that $P(X_1+X_2=n(m+1))$ is the probability that the sum of $2n$ dice with $m$ sides is equal to $n(m+1)$, which is the middle value. This can be calculated explicitly as follows: $$ P(\text{$2n$ dice with $m$ sides sum to $k$})=\sum_{k=0}^{\lfloor (k-1)/(m+1)\rfloor} (-1)^j\binom{2n}j\binom{k-1-j(m+1)}{2n-1} $$ Asymptotically, using this answer, the probability is about $\frac1m\sqrt{\frac{3}{\pi n}}$.

Answer 2

If $(2,1)$ is considered same as $(1,2)$, here is how it would look for $2$ dice for example -

Number of ways to get a sum of $n$ with two dice is say, $p$.

Then, $p = \lfloor \frac{n}{2} \rfloor$ where $\lfloor \, \rfloor$ is floor function.

Say, number of ways to get a sum lower than $n$ in the second roll of two dice is $q$.

Then, $q = \sum \limits_{i = 2}^{n-1} \lfloor \frac{i}{2} \rfloor \, (3 \le n \le 12)$. If $n = 2$, number of ways to get lower sum is obviously zero.

So it turns out to -

a) If $n$ is even then $q = \frac{n(n-2)}{4}$

b) If $n$ is odd then $q = \frac{(n-1)^2}{4}$