A fair coin is rolled until at least two heads and at least two tails are obtained. Let
the random variable X denote the roll in which the second head is obtained and let the
random variable Y denote the roll in which the second tail is obtained.
How do I find the joint probability mass function of X and Y?
I don't know how to start this question.
Any help is much appreciated.
Best Answer
If the first two outcome is $HH$, then we have $X=2$, the smallest value that $Y$ can take is $4$.
Similarly, if the first two outcome is $TT$, then we have $Y=2$, the smallest value that $X$ can take is $4$.
If the first two outcome is $HT$ or $TH$, then $\min(X,Y)=3$.
That is we have $\min(X,Y) \in \{2, 3\}$ and $\max(X,Y) \ge 4$.
Let $n \ge 4$.
To compute $P(X=2, Y=n)$, the first two tosses $H$, the last outcome is $T$, from the $3$rd to $(n-1)$-th outcome, there is exactly one $T$.
That is consider $$HHW\ldots WT$$
where we have $n-3$ $W$'s, we have to choose $1$ of them to be $T$. We have $\binom{n-3}{1} = n-3$. The probability to get a particular sequence is $\frac1{2^n}$, hence the corresponding probability is
$$P(X=2, Y=n) = \frac{n-3}{2^n}$$
We also have $P(X=n, Y=2) = P(X=2, Y=n)$ by symetry.
Now, let's compute $P(X=3, Y=n)$, in this case, the first two tosses is either $HT$ or $TH$, the third position must be $H$, the last position is $T$. The remaining position must be $H$.
Hence we have
$$P(X=3, Y=n)=\frac{2}{2^n}$$
We have $P(X=n, Y=3) = P(X=3, Y=n)$ by symmetry.