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This is conditional probability mass problem.

Thoughts: I get the basic pieces. I know the probability of taking y rolls to get a 4 is $ (\frac{5}{6})^{y-1}(\frac{1}{6}) $, and that the probability of getting x 3s in those y rolls is $ {{y}\choose{x}}(\frac{1}{6})^{x}(\frac{5}{6})^{y-x} $.

Problem: I'm not sure what to do with information to get $ p_{X|Y}(X=x | Y=y) $. I feel like I'm missing something about counting for the 3s, like I should exclude them from the $ p_Y(y) $ part.

I know $ p_{X|Y}(x | y) = \frac{p_{X,Y}(x,y)}{p_Y(y)}$, but I'm struggling to find the joint PMF. This is why I don't think my "just multiply them" approach works; If I just multiply $p(X=x)$ and $p(Y=y)$, then I'm just dividing that product by $p(Y=y)$ and end up with $p(X=x)$ again.

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    Why isn't it just the probability of getting x 3s in y-1 rolls of a 5-sided die? – 2017-02-08
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    I think it is, if I'm reading the other answer right. Thanks! – 2017-02-09

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Given that a single roll isn't a 4, the probability that it is a 3 is $\mathbb P(3\text{ and not }4)/\mathbb P(\text{not }4)=\mathbb P(3)/\mathbb P(\text{not }4)=(1/6)/(5/6)=1/5$. This is intuitively obvious, since if it is not a 4 it is equally likely to be any of the other 5 numbers.

Given that the first $Y-1$ rolls are known not to be 4s, then, the number of those which are 3s is given by $\mathbb P(X=x|Y=y)=\binom{y-1}{x}(1/5)^x(4/5)^{y-1-x}$.

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    Thanks for the clear explanation! This makes sense to me, so I'll go ahead and accept this. – 2017-02-09