# Probability - pmf for alternating probabilities

Two people, Amanda and Bella, will have consecutive arm wrestling matches until one of them wins two matches in a row and is declared the winner. Amanda wins a given match with probability $\frac{3}{5}$. Assuming that the matches are independent, give the probability mass function of the number of matches required to declare a winner.

I've been thinking about the pmf in terms of what is required for an even number of matches versus an odd number of matches, but my pmf is getting complicated and I'm wondering if there's an easier way.

The number of matches $W$ required to declare a winner is $W=1+A$ with probability $u=\frac35$ and $W=1+B$ with probability $v=1-u$, where $A$ is the number of matches required to declare a winner after a match has been won by Amanda and $B$ is the number of matches required to declare a winner after a match has been won by Bella. In turn, $A=1$ with probability $u$ and $A=1+B'$ with probability $v$, where $B'$ is distributed like $B$. Likewise, $B=1$ with probability $v$ and $B=1+A'$ with probability $u$, where $A'$ is distributed like $A$.
Now, we translate all these stochastic identities in terms of the generating functions $w(s)=E(s^W)$, $a(s)=E(s^A)$ and $b(s)=E(s^B)$. Each one yields an identity between generating functions, namely, $$w(s)=usa(s)+vsb(s),\quad a(s)=us+vsb(s),\quad b(s)=vs+usa(s).$$ Solving for $w(s)$ and using $u+v=1$ yields $$w(s)=\frac{u^2+v^2+uvs}{1-uvs^2}s^2=(u^2+v^2+uvs)s^2\sum_{n\geqslant0}(uv)^ns^{2n}.$$ Thus, for every $n\geqslant1$, $$P(W=2n)=(u^2+v^2)(uv)^{n-1},\qquad P(W=2n+1)=(uv)^n.$$
• The players cannot win if $n=1$ because there must be at least two matches for one to win. – pocketlizard Sep 22 '14 at 16:19
• Indeed. And surely you noted that the first non zero term in $w(s)$ is the $s^2$ term. – Did Sep 22 '14 at 18:47