The Question

Which is more likely: rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?

My Work

First we have to determine the probability of two die summing to $9$. I began by calculating the sample space. There are $6*6 = 36$ different ways the two die could sum. Now to figure out how many of those ways sum to $9$. This is essentially the same question as how many integer solutions to the equality $x_1 + x_2 = 9$ where $1 \leq x\le6$ ($x$ represents the two dice in this case). The number of ways to do this is $\binom{8}{7} - 2$ (where $2$ represents the case where $x\geq6$. So the probability of rolling two dice which sum to $9$ is $\frac{6}{36}$

Figuring out how many ways $3$ dice sum to $9$ is a similar process. Find all solutions to $x_1 + x_2 + x_3 =9$ where $1\le x\le6$ which is $\binom{8}{6} - 3 = 25$ ways to do this (3 represents all the solutions where $x_i$ is greater than $6$). There are $6^3$ ways $3$ dice can be rolled therefore the probability $\frac{25}{6^3}$.

Computing the probabilities I found $\frac{6}{36} > \frac{25}{6^3}$ Therefore, it is more likely with 2 dice.

The Problem

My book says it is more likely with 3 dice. Can anyone point out where I went wrong in my approach, and how to correct it?


In your computation of how many ways for two dice to sum to 9, you subtracted 2 for the two cases in which $x_1$ is 7 or 8, but forgot to subtract 2 for the two cases in which $x_2$ is 7 or 8.

Using a formula is nice, but especially when what's going on is as simple as two numbers from 1 to 6 summing to 9, double-check it by writing them out!

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  • $\begingroup$ $x_1 + x_2 = 9$ $y_1 + y_2 = 7$ (Give each $x$ a 1) $y_1 + z_1 = 1$ (for the case when one integer equals 7 or more) then we multiply this case by 2 and subtract it from the total for the correct answer, right? $\endgroup$ – Dunka Nov 21 '14 at 6:25
  • $\begingroup$ Sure, that works, though it seems overly complicated for the two dice case. (Also, I could figure out what you meant, but it took me a moment because you didn't define anything.) Can you visualize the answer for the two dice case? $\endgroup$ – aes Nov 21 '14 at 6:33

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