Solving distribution using conditional probability

Question

A dice is tossed untill 3 even numbers apear. Let Y be the number of times "1" showed up. Find the distribution of Y+3.

My idea was: Let L be the number of tosses, i managed to get that L is distributed negetive binominially, and Y|L is distributed binominially with L-3 as the number of trials and p=1/2. Adding all p(Y=k-3|L=n)*p(L=n) should get me p(Y+3=k), but im having a hard time solving this sum. Any ideas?

Answer 1

A die is tossed until the third even result occurs and we count the rolls of 1 among them.   We know the count for rolls, $L$, is negative binomial $(3, 1/2)$ and the count for 1 among those rolls, $Y$, is conditionally binomial $(L-3, 1/3)$ for a given $L$, but let us not focus on that distraction.

Instead let us count only rolls of 1, 2, 4, or 6 and ask: What is the distribution for that count until the third even result ?

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$Y$ is the count of 1 before the third even roll.   We aren't interested in results of 3 or 5 so... let us ignore those sides whenever they turn up and only count the rolls of interest. On the rolls we don't ignore, the results from 1,2,4,6 each still occur without bias, so we are effectively rolling a fair four-sided die. $Y+3$ is the count of rolls of interest until the third even result.

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$\begin{align}\mathsf P(Y=y) &= \sum_{x=y+3}^\infty\mathsf P(Y=y,X=x) \\ &=\sum_{x=y+3}^\infty \dfrac{(x-1)!~1^y3^32^{x-3-y}}{y!~2!~(x-3-y)!~6^x} \\ &= \text{can be done but let's do it the easy way} \\[3ex] \mathsf P(Y+3=k) &= {\text{effectively the probability for rolling }\\k-3\text{ ones before the third not-one on a four sided die.}} \end{align}$