A and B play until one scores 2 points in a row.

Question

A and B play until one scores 2 points in a row, who will win. Probability of A or B scoring a point is $a$ or $b$, respectively. What is the probability that A wins?

I know the correct answer is $\frac{a^2}{1-2ab}$, based on this.

I was wondering what is wrong with the following approach. We form the following tree. The green/red rectangles are where A/B has won.

Tree showing possible sequences leading to the winning of A.

We can see the following pattern for the top and bottom rows (green boxes where A has won):

Top: $a^2, a^3b, a^4b^2, \ldots$

Bottom: $a^2b, a^3b^2, a^4b^3, \ldots$

When we add the probabilities, both top and bottom rows are geometric series with a factor of $ab$. So,

Sum of the top row: $\frac{a^2}{1-ab}$

Sum of the bottom row: $\frac{a^2b}{1-ab}$

Total probability: $\frac{a^2(1+b)}{1-ab}$

Questions: This not equal to $\frac{a^2}{1-2ab}$. What am I missing?

Answer 1

Your game is not the same as the game in the linked question.

I did not look at your infinite tree. Anyway: The result is correct.

Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations $$p|_A=a\cdot 1+ b\cdot p|_B\>,\qquad p|_B= a \cdot p|_A+ b\cdot 0\ ,$$ resulting in $$p|_A={a\over 1-ab},\qquad p|_B={a^2\over 1-ab}\ .$$ It follows that $$p|_O=a\cdot p|_A+b\cdot p|_B={a^2(1+b)\over 1-ab}\ .$$

Answer 2

Let $P(A)=\text{Probability of A winning, }P(B)=\text{Probability of B winning.}$

We are actually supposed to find $\dfrac{\text{Success Probability}}{\text{Success Probability + Failure Probability}},$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.

This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $\frac{\text{P}(A)}{\text{P}(A\cup B)}=\frac{\text{P(A)}}{\text{P(A)}+\text{P(B)}}$ since both are mutually exclusive ($A$ and $B$ cannot both win).

Answer 3

The question you linked is different from the question you have asked.

In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.

This is not the same game--for example, consider the sequence $abb$.