A well known problem: https://en.wikipedia.org/wiki/Problem_of_points
Suppose players A and B decided to play till 3 wins and the game stopped after 2:1 to A. There are 4 possible scenarios to game ending, while in 3 of them A wins. In two scenarios that A wins in round 4 the round 5 is not relevant anymore. I wondered, about the intuition why these two scenarios: (1) A wins in round 4, A wins in round 5; (2) A wins in round 4, B wins in round 5; are counted as two different scenarios?
Thanks!
Assume that the game is fair, and that each player has a 50% chance of winning in any given round. Rather than counting, one could compute probabilities as follows:
So one arrives at the correct answer this way, but notice that the probabilities of the outcomes are different. This is fine for such a small example, but suppose we were playing until 75 wins and the current score was 46:43...this method would be tedious to calculate.
But we can also consider the game differently. Suppose the current score is $r$:$s$, and the number of required wins is $n$. Instead of playing until one player or the other wins $n$ times, simply play the game $2n-r-s-1$ times, after which a winner is assured. Then, go back through each outcome and determine the winner.
In this case, each individual outcome of the additional $2n-r-s-1$ games has the same probability of occurring, namely $\frac{1}{2^{2n-r-s-1}}$. So calculating the probability of a player winning reduces to counting the number of outcomes of this extended game where the player wins. This is a much easier problem to both calculate and derive a general solution for.