Calvin and Hobbes play a match consisting of a series of games, where Calvin has
probability p of winning each game (independently). They play with a “win by two”
rule: the first player to win two games more than his opponent wins the match. Find
the probability that Calvin wins the match (in terms of p) by conditioning.
$W$ = "Calvin wins by 2 by moving from $k$ to $k+2$",
$"G_1 = C"$ is "Calvin won game 1"
Anyways, we have
$P(W) = P(W|G_1 = C)P(G_1=C) + P(W|G_1=H)P(G_1=H) = p^2 + qP(W|G_1=H) = p^2+q(p*P(W) + P(W) * p)$
When we solve for $P(W)$ we get $P(W) = p^2/1-2qp$
This is right, but I wonder if $P(W|G_1=H) = p*P(W) + P(W)*p$ is a weird way to think of the problem. Basically $P(W)$ is the probability of moving to the right by 2, and so if we're at $i-1$ and want to get to $i + 2$, we can either move to right by 1 then 2 ($p*P(W)$), or 2 then 1 ($P(W)*p$). Is this a weird/wrong way of thinking of it?
By the way, I am aware of the method of conditioning on the outcomes of the first 2 games. But I wouldn't have thought of this way of conditioning without looking at the solution. Are there any tips regarding what I should condition on?
Best Answer
Let $l$ be the state where $C$ is level with $H$,
$a$ be the state where $C$ is $1$ ahead of $H$
and $b$ be the state where $C$ is $1$ behind $H$
At start, we are in state $l$, and we want the $Pr$ that $C$ ultimately wins from start, so we get the following equations:
$l=p*a +(1-p)*b\;\;$[From level, $C$ gets $1$ ahead or $1$ behind]
$a =p +(1-p)*l\;\;$[From $1$ ahead, $C$ wins or drops down to level]
$b=p*l\;\;$[From $1$ behind, $C$ must climb back to level for win]
Solving, we get P($C$ wins) = $\dfrac{p^2}{2p^2-2p+1}$
Alternate method
With $C's$ win/lose probabilities of $p$ and $q$ for any game,$\;(q=1-p)$
Suppose P(C ultimately wins) $= w$, then
either C wins in two games, with $Pr=p^2$,
or returns to square one (restart) with $Pr=pq+qp = 2pq$
Thus $w = p^2 +2pq\cdot w$, which yields
$w = \dfrac{p^2}{1-2pq} = \dfrac{p^2}{2p^2-2p+1}$