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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?

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    No, you’re not. Suppose that you roll the dice blindfolded. Someone then tells you that you rolled either $3$ or $5$, but he won’t tell you which. Given that information, what’s the probability that you rolled $3$? **That’s** the question that Pr(you rolled $3$|you rolled $3$ or $5$) answers.2011-09-26

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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 $3$.

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    Thanks for the clarification. I thought the conditional probability was interpreted "Given you have rolled a 3 or a 5 in your previous roll, what is the probability that you have rolled a 3 on this roll". I misinterpreted it as two events happening at different times.2011-09-26
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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.

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    Do you mean ``the probability of getting a 6 **or** a head is 1/2 + 1/6``?2013-01-15
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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?"

$P(X=k)=(2/36)(34/36)^{k-1}$ (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(Xk)\\ &=\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 $1/3$.

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    @Dilip: But the question as stated here is different. From the way the OP stated this question, I could pretty much tell what the previous question was (given the answer from that problem).2011-09-27