# Given three six-sided dice, what is the probability that the value of the third will be greater than the sum of those of the first two?

A die is rolled three times, or three dice are rolled. What is the probability that the third die values greater than the sum of the first two?

(assuming six-sided dice, but I would be interested in a generalization to N-sided dice, or perhaps a scenario where the dice each have different numbers of sides)

## 1 Answer

The total number of possibilities is $6^3=216$. Now we will look separately at the different favourable events possible.

Certainly the last die cannot be $1$ or $2$, because the sum of numbers of the faces of the first two dice is at least $2$.

Suppose $3$ falls on the last dice. Then only $1$ and $1$ could have fallen on the first two dice,so that is one possibility.

Suppose $4$ falls on the last dice. Then there are two possibilities, namely, $(1,2)$ and $(2,1)$ (here my notation $(a,b)$ is where $a$ comes on the first dice and $b$ on the second dice), but then even the three possibility, namely $(1,1)$ is possible, since $1+1<4$. So there are $3$ possibilities in total.

Suppose $5$ falls on the last dice. All the possibilities for $4$ and $3$ come again, which are $3$ in total. The others are $(1,3)$, $(2,2)$ and $(3,1)$, so that gives $6$ possibilities.

Finally, suppose $6$ falls on the last dice. All the possibilities for $5$ and $4$ and $3$ come again, which are $6$ in total. The others are $(1,4)$, $(2,3)$,$(3,2)$ and $(4,1)$, so that gives $10$ possibilities.

In total, we have $1+3+6+10=20$, and the probability would then be $\frac{20}{216}=\frac{5}{54}$.

Being a little paranoid about getting answers right, I am going to enumerate all the possibilities again and count them here, so you can see I have not left anything out: $(1,1,3)$,$(1,1,4)$,$(1,2,4)$,$(2,1,4)$,$(1,1,5)$,$(1,2,5)$,$(2,1,5)$,$(1,3,5)$,$(2,2,5)$,$(3,1,5)$,$(1,1,6)$,$(1,2,6)$,$(2,1,6)$,$(1,3,6)$,$(2,2,6)$,$(3,1,6)$,$(1,4,6)$,$(2,3,6)$,$(3,2,6)$,$(4,1,6)$. That is $20$ possibilities.

For general $N$ sided dice, you can start with $3$ and work upwards like I have done (you would have surely seen a pattern in $1,3,6,10$ which are the possibilities at each stage) to get your answer.