Obtain a sum of $4$ before a sum of $2$

Question

I am trying to solve the following problem.

Suppose you a roll two fair $6$-sided dice repeatedly. What is the probability that you obtain a sum of $4$ before a sum of $2$.

It is clear to me that the probability you obtain a sum of $4$ or $2$ is $3/36$ or $1/36$ respectively. However, I do not understand how to determine the probability of obtaining a sum of $4$ before a sum of $2$. Does the solution involve an infinite sum? I do not understand.

Answer 1

As you have correctly observed, the probability is $\frac{3}{36}$ and $\frac{1}{36}$ to get a sum of $4$ and a sum of $2$ (respectively).

Now, it follows that there are $3$-times chances to get a sum of $4$ over a sum of $2$ (that implies a pair of $1$ for both the dice), since $\frac{\frac{3}{36}}{\frac{1}{36}}=3$. Thus, we have finally ${\frac{3}{3+1}}=\frac{3}{4}$ (number of wins over number of total cases, ratio).

Answer 2

One way to approach these sorts of questions is as follows:

(First, convince yourself that there is zero chance of this process continuing into eternity.)

Then, look at the conditional probabilities with respect to the first throw:

• You have a $\frac{3}{36}$ chance to roll a four, at which point your chance of rolling a four first is certainty.
• You have a $\frac{1}{36}$ chance to roll a two, after which rolling a four first is impossible.
• You have a $\frac{32}{36}$ chance of rolling neither, at which point you are back where you started and have the same probability you started with.

Plug this in, and if $P$ is the probability you are looking for then we get the equation $P = \frac{3}{36} (1) + \frac{1}{36} (0) + \frac{32}{36}(P)= \frac{1}{12} + \frac{8}{9}P$