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

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?

• Deos the order of the outcomes count? If you roll, say, 1,2,4 is it a different "way" from 4,1,2? If you provide this information I will be able to help you.
– user799688
Commented Oct 13, 2020 at 13:38
• @Vlad the order doesn't count. Commented Oct 15, 2020 at 17:07

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, $$P(X_1>X_2)=P(X_1, which together with $$P(X_1X_2)=1$$ implies that $$P(X_2 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 $$\frac1m\sqrt{\frac{3}{\pi n}}$$.

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 $$q = \frac{(n-1)^2}{4}$$