Probability of winning two games in a row with infinite recursion

Question

I think this is different than the traditional gambler's ruin question answered here: Geometric Expectation problem?

In that question, Calvin and Hobbes play a deuces match, in which Calvin wins an individual point with $p$ probability, independently. To win the match, Calvin must win by two points.

Let's change the formulation just a tiny bit. Now, instead of a win by two rule, the winner is the first person to win two points in a row. I'm getting stuck in the recursions here, because now, if Calvin wins one point and loses the next, he's no longer back at the start. Hobbes has one point under his belt. He only needs one more to win the match, not two. Similarly, if Calvin loses the first point but wins the second, he's not back at zero. He now only needs to win once more to win the match.

I can simulate this fairly easily, but is there a closed-form solution? What is Calvin's overall probability of winning the match?

Answer 1

Let $P$/$Q$ be the probability C wins if he has respectively won/lost the previous game.

Then $P=p+(1-p)Q$ and $Q=pP$.

Then $P=\frac{p}{p^2-p+1},Q=\frac{p^2}{p^2-p+1}$.

Probability P wins is $$pP+(1-p)Q=\frac{2p^2-p^3}{p^2-p+1}.$$

Answer 2

Let's denote the probability that Calvin wins if he currently leads $k$ points as $P_k$. See how this doesn't depend on the number of turn they are in.

Then: $$\begin{aligned} P_2 &=1 \\ P_1 &= pP_2 + (1-p)P_0,\\ P_0 &= pP_1+(1-p)P_{-1},\\ P_{-1}&=pP_0+(1-p)P_{-2},\\ P_{-2} &=0 \end{aligned} $$

To solve for $P_0$ one can put $P_{-1}$ and $P_1$ into equation for $P_0$: $$ P_0 = p(p + (1-p)P_0) + (1-p)pP_0,\\ (1-2p(1-p))P_0=p^2,\\ P_0=\frac{p^2}{1-2p(1-p)} $$

Answer 3

There are really only two states to be concerned with, Calving has won the last point or Hobbes has won the last point. Let $C$ be the probability that Calvin wins the game, given that he won the last point, and let $H$ be the probability that Calvin wins the game, given that Hobbes won the last point.

We have $$\begin{align}C&=p+(1-p)H\\ H&=pC\end{align}$$

which you can easily solve for $C$ and $H$.

The probability that Calvin wins is $$pC+(1-p)H$$ by considering what happens on the first roll.