What is the probability that some event happens before another event? In the previous question I asked the $\mathbb{P}(X < Y )$  where $X$ is the number of rolls(two dice) it takes to roll a sum of 3 and $Y$ is the the number of rolls it takes to roll a sum of 5. The probability was $\frac{\mathbb{P}(X)}{\mathbb{P}(X)+\mathbb{P}(Y)} = \frac{(2/36)}{(4/36+2/36)} = \frac{1}{3}$. I am not sure why you divide though by $\mathbb{P}(X)$ or $\mathbb{P}(Y)$. Can some one explain this to me intuitively?
 A: Let $\mathrm P(n)$ denote the probability that a single die roll gives the result $n$.
Consider the first time a $3$ or a $5$ is rolled.  The probability that the result of that roll is $3$, given that we know it's either $3$ or $5$, is $\mathrm P(3)$ out of $\mathrm P(3) + \mathrm P(5)$ — that is to say, $\frac{\mathrm P(3)}{\mathrm P(3) + \mathrm P(5)}$.

There's some notational confusion involved in your question, perhaps prompted by a slight abuse of notation in Thijs Laarhoven's answer to the earlier question.  By convention, $\mathrm P(Q)$ (sometimes written as $\mathrm P[Q]$ or $\mathrm{Pr}[Q]$ or in various similar ways by different authors) denotes that probability that the event $Q$ occurs.  An event is something that either happens or not, such as "$X < Y$" (where $X$ and $Y$ are random variables) or "the result of the die roll is $3$".  An event is not the same as a random variable; in particular, if $X$ is a random variable, the expression $\mathrm P(X)$ is meaningless.
The slight abuse of notation I alluded to is that Thijs was, in effect, using "$3$" as a shorthand for the event "the result of the die roll is $3$" in the expression $\mathrm P(3)$.  (I used the same notation in the first part of my answer above.)
A more correct way to express the answer might be to let $N$ be a random variable denoting the outcome of a single die roll.  Then the conditional probability that $N = 3$, given that $N \in \{3,5\}$, is
$$\mathrm P(N = 3 \,|\, N \in \{3,5\}) = \frac{\mathrm P(N = 3)}{\mathrm P(N \in \{3,5\})} = \frac{\mathrm P(N = 3)}{\mathrm P(N = 3) + \mathrm P(N = 5)}.$$
In particular, this equals the probability that the first die roll which is either a $3$ or a $5$ is, in fact, a $3$.
A: Imagine you throw a coin on one hand and a dice on the other hand simultaneously. You want to know what is going to happen first: Will you get a head (on the coin) first or will you get a 6 first  (on the dice). Intuitively you would say that it is probably easier to get a head first than a 6, because the probability of getting a head is a half but getting a 6 is a sixth. The probability of getting a 6 and a head is 1/2 + 1/6. The probability that you get a head before the 6 is a fraction of that previous probability of getting both, namely (1/2)/[1/2 + 1/6]  and similarly the probability of getting a 6 before a head is a fraction of the previous quantity, namely (1/6)/[1/2 + 1/6]. When you add both together, you get 1, because either you get a head first or a 6 first. Hope this is clear.
A: I believe that the question you asked is a bit different from the question answered. The question you asked was "If $X$ is the number of rolls to get a $3$ and $Y$ is the number of rolls to get a $5$, what is the probability that $X<Y$?" 
$P(X=k)=(2/36)(34/36)^{k-1}$ (don't roll a $3$ on the first $k-1$ rolls, then roll a $3$) and $P(Y>k)=(32/36)^k$ (don't roll a $5$ on the first $k$ rolls), so
$$
\begin{align}
P(X<Y)
&=\sum_{k=1}^\infty P(X=k)P(Y>k)\\
&=\sum_{k=1}^\infty(2/36)(34/36)^{k-1}(32/36)^k\\
&=(2/36)(32/36)\sum_{k=0}^\infty(34/36)^k(32/36)^k\\
&=64/1296\frac{1}{1-(34/36)(32/36)}\\
&=4/13
\end{align}
$$
The question which was answered was "what is the probability that a $3$ is rolled before a $5$?" That is, "ignoring all other rolls, what is the probability of rolling a $3$ instead of a $5$?". Since the probability of rolling a $3$ is $2/36$ and the probability of rolling a $5$ is $4/36$, the probability of rolling one or the other is $6/36$, and of that event, $2/36$ is rolling the $3$. Thus, given that we rolled either a $3$ or a $5$, the probability that we rolled a $3$ would be $(2/36)/(6/36)=1/3$.
The first answer $(4/13)$ is smaller than $1/3$ since there is a chance that $X=Y$. Performing a similar calculation, we get $P(X=Y)=1/26$. Thus, $P(X\le Y)=9/26$, a bit above $1/3$.
