91Ó°ÊÓ

In the context of this exercise, a Poisson process is a mathematical way to describe random events that occur over time. Here, the points scored by Teams 1 and 2 follow independent Poisson processes with rates \(\lambda_1\) and \(\lambda_2\), respectively. This means the average number of points scored per unit of time is determined by these rates.
One key property of a Poisson process is that events occur independently of the last event. In this match scenario, it implies that the scoring of points by one team does not affect the likelihood of the other team scoring.

• The probability of a certain number of events (points) in a given time interval follows a Poisson distribution.
• It's memoryless – past events don't impact future events.
• The process is characterized by a single parameter, the rate, \(\lambda\).

For our problem, understanding the Poisson process helps determine the probabilities of each team scoring at every moment during the game. This serves as a foundation for analyzing who is likely to reach the winning condition first.

In this problem, the recurrence relation is essential to express the probability of Team 1 winning given the current score difference. It is a mathematical expression that uses known values of a function to determine other values. For Team 1 to win, we analyze how their probability of winning changes from one point difference to another.

The recurrence relation derived here is:\[ W_i = pW_{i+1} + (1-p)W_{i-1} \]This equation expresses the probability \(W_i\) of Team 1 winning when they are \(i\) points ahead.

• \(p\) is the probability that Team 1 scores the next point.
• \(1-p\) is the probability that Team 2 scores the next point.

The strategy is to use known boundary conditions to solve this relation:

• \(W_k = 1\): Team 1 wins if they are \(k\) points ahead.
• \(W_{-k} = 0\): They lose if Team 2 is \(k\) points ahead.

By iteratively using the recurrence relation, we zero in on \(W_0\), which gives the probability of Team 1 winning when the scores are tied.

In the gambler's ruin approach to this problem, the winning probability for Team 1 determines whether they win the match by achieving the required score difference first. Given that both teams play under Poisson processes, the exact probability of winning is influenced by their respective scoring rates.

The "goal" conditions for the match are:

• Team 1 needs to be \(k\) points ahead to win (\(W_k = 1\)).
• Team 2 wins if they are \(k\) points ahead (\(W_{-k} = 0\)).

With Poisson processes, the probability \(p\) that Team 1 scores the next point is \(\frac{\lambda_1}{\lambda_1 + \lambda_2}\). By solving the recurrence relations, we obtain the winning probability formula for Team 1 as:\[W_0 = \frac{1-\left(\frac{\lambda_1}{\lambda_2}\right)^k}{1-\left(\frac{\lambda_1}{\lambda_2}\right)^{2k}}\]This formula gives the likelihood that Team 1 will win, starting from a tied score. It's a beautiful way of connecting the behaviors under Poisson processes with classic gambler's ruin concepts, showing how strategic insights can be calculated mathematically.