You repeatedly toss two fair coins together until both coins show heads. Let $X$ and $Y$ denote the number of heads resulting from the tosses of the first and second coin, respectively. What is the joint probability mass function of $X$ and $Y$? What are the marginal distributions of $X$ and $Y$? What is $P(X=Y)$?

suppose have to throw $n$ times in order to get $(H,H)$. X=x is the number of heads in the first coin and so $Y=n-x$. Therefore,

$$ P(X=x,Y=y) = P(X=x, Y=n-x) $$

In other words, this is geometric with p=1/4 and $x+y-1$ trials before we observe $HH$. Thus,

$$ P(X=x, Y=y) = (3/4)^{x+y-1} 1/4 $$


$$ P(X=x) = \sum_{y=1}^{n-x} (3/4)^{x+y-1} 1/4 $$


$$ P(Y=y) = \sum_{x=1}^{n-y} (3/4)^{x+y-1} 1/4 $$

now, I get stuck in $P(X=Y)$. Do we need to sum over the diagonal?


1 Answer 1


I don't think we have $X+Y=n$.

We repeated throw two coins together until both show heads simultaneously, we don't stop tossing one of them at any point of time.

In each toss, we can get $(H,H), (H,T), (T,H), (T,T)$.

If $X=x$ and $Y=y$, it means we have $x-1$ times of $(H,T)$ and $y-1$ times of $(T,H)$, exactly one $(H,H)$ and it is possible to have as many $(T,T)$.

\begin{align} Pr(X=x, Y=y) &=(0.25)^{x-1}(0.25)^{y-1}(0.25) \sum_{i=0}^\infty(0.25)^i \\ &=(0.25)^{x+y-1}\cdot \frac{1}{1-0.25} \end{align}

From there, we can compute the marginal distributions of $X$ and $Y$ as well.

\begin{align} Pr(X=Y) &= \frac{1}{1-0.25}\sum_{x=1}^\infty (0.25)^{2x-1} \end{align}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .