# Dice and probability

Two fair six-sided dice are rolled. If the sum of the numbers obtained is $$4$$, find the probability that the numbers obtained on both dice are even.

I'm not sure if it's $$\frac{1}{36}$$ or $$\frac{1}{3}$$.

• Number of states, sum of two even is four : $$\#\{(2,2)\}=1$$ Number of states, sum of two dice is four : $$\#\{(1,3),(2,2),(3,1)\}=3$$ – Fardad Pouran Jun 15 '14 at 5:03

$4$ can obtained by $1+3,2+2,3+1$

So, the probability $\displaystyle=\frac{\text{No. of favorable cases}}{\text{No. of possible cases}}=\frac13$

• That is because the sum has already been given and the size of our sample space reduces to 3? – Pratik Jun 15 '14 at 5:04
• @Pratik, Exactly – lab bhattacharjee Jun 15 '14 at 5:04

The important point is that you are first given that the sum is $4$.

This can only happen rolling $(1,3), (2,2)$ or $(3,1)$.

So given that the sum is $4$, we now have a sample size of $3$, out of which only $1$ entry, i.e., $(2,2)$, has both dice showing an even number.

Therefore, the answer is $1/3$.

If, however, you were just asked: What is the chance of rolling $(2,2)$?

Then your sample size would be all the possible rolls with two dice, and you would see that only $1$ of the $36$ possibilities is $(2,2)$. In this case, different from yours, the answer would be $1/36$.

• Thank you. That was very helpful – Pratik Jun 15 '14 at 5:11
• @Pratik You're welcome! Hopefully the problem is clearer now... – Benjamin Dickman Jun 23 '14 at 7:23

This problem is related to conditional probability or Bayes theorem. The probability P(A) of getting the sum 4 is 3/36. Probability of getting even numbers on each dice and sum 4 is $$P(A \cap B)=$$1/36. So conditional probability is

$$P(A \cap B)/P(A) =\frac{1/36}{3/36}=1/3$$