# Conditional Probability that sum of dice is even

If two dice are rolled and let $X and Y$ be the two random variables. What is the conditional probability that $X+Y$ is even when $X$ is odd. And when $X$ is odd? What is the total Probability of $X+Y$ to be even?

I am using the Bayes theorem but the answer is coming out 1. I am confused. • With minimal computation, in particular without Bayes' Theorem, we can see that each conditional probability is $1/2$. But the probability that $X+Y$ is even is not obtained by adding. – André Nicolas Jul 9 '16 at 3:12

Let $A$ be the event $X+Y$ is even, let $B$ be the event $X$ is odd, and let $C$ be the event $X$ is even. By the Law of Total Probability we have $$\Pr(A)=\Pr(A\mid B)\Pr(B)+\Pr(A\mid C)\Pr(C).\tag{1}$$ It looks as if you calculated $\Pr(A\mid B)$ and $\Pr(A\mid C)$ correctly. They are both $1/2$. But $\Pr(B)=\Pr(C)=1/2$. Substituting in (1) we get $\Pr(A)=(1/2)(1/2)+(1/2)(1/2)=1/2$.
There are many other ways to compute the probability that $X+Y$ is even. We can do it the long way, by counting the number of ordered pairs $(x,y)$ where $x+y$ is even, and $1\le x\le 6$, $1\le y\le 6$, and dividing by $36$. A much simpler way related to the answer above is that whatever the first roll is, the probability the second roll results in an even sum is $1/2$.
• I am slightly confused since $$\Pr(A\mid B)=\frac{\Pr(B\mid A)\cdot \Pr(A)}{\Pr(B)}$$ is used to compute $\Pr(A)$ but has $\Pr(A)$ in itself? It seems I do not completely understand how you apply Bayes to substitute in $(1)$. – Christian Ivicevic Jul 27 '16 at 13:55
• @ChristianIvicevic: I could not read what OP wrote well, just saw that $\Pr(A\mid B)$ looked right. But I did not compute $\Pr(A\mid B)$ using Bayes. I used a direct calculation.Given that $B$ holds, that is, $X$ is odd, then with probability $1/2$, $Y$ will be even and therefore $X+Y$ odd, and with probability $1/2$ $Y$ will be odd and thefore $X+Y$ will be even. So $\Pr(A\mid B)=1/2$. – André Nicolas Jul 27 '16 at 15:36