We roll a die until we get a six and denote the number of rolls by $X.$ Then we take a fair coin and we repeatedly flip it until we get $X$ heads. We denote the number of coin flips needed by $Y.$

Find the conditional probability mass function $Y$ given $X=x$.

Given $$ f_{Y|X}(y|x)= \frac{f_{Y,X}(x,y)}{f_{X}(x)}= \frac{\text{Something}}{(5/6)^{x-1}(1/6)} $$ Obviously I cant figure out the something, I believe it is binomial, but am unsure of the proper parameters.


The probability that $Y=y$ given that $X=x$ is the probability we get $x-1$ heads in the first $y-1$ tosses, and then get a head on the $y$-th toss.

Let $y\ge x\ge 1$. The probability of $x-1$ heads in the first $y-1$ tosses is $\binom{y-1}{x-1}(1/2)^{x-1}(1/2)^{y-x}$. Multiply by $1/2$ for the head on the $y$-th toss. Our probability simplifies to $$\binom{y-1}{x-1}\left(\frac{1}{2}\right)^y.$$ We have used the negative binomial distribution.

Remark: The solution assumed that the coin is fair. Minor modification solves the problem for a coin that has probability $p\gt 0$ of head. The conditional probability is then $\binom{y-1}{x-1}p^x(1-p)^{y-x}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.