Question

Teams 1 and 2 are playing a match. The teams score points according to independent Poisson processes with respective rates \(\lambda_{1}\) and \(\lambda_{2} .\) If the match ends when one of the teams has scored \(k\) more points than the other, find the probability that team 1 wins. Hint: Relate this to the gambler's ruin problem.

Short answer

The probability that team 1 wins the match when they score \(k\) more points than team 2 can be found using the gambler's ruin problem formula. Given independent Poisson processes with rates \(\lambda_1\) and \(\lambda_2\), the probability of team 1 winning is: $$P_k = \frac{1 - \left(\frac{\lambda_2}{\lambda_1}\right)^k}{1 - \left(\frac{\lambda_2}{\lambda_1}\right)^{2k}}$$

Step-by-step solution

Defining the Variables

Let's define the following variables: - \(n_i\): the total number of points scored by both teams after \(i\) seconds. - \(z_i\): the cumulative number of points scored by team 1 after \(i\) seconds. The total number of points scored by team 2 after \(i\) seconds is given as \((n_i - z_i)\).

Probability of Team 1 Winning

Let \(P_i\) denote the probability that team 1 wins given that \(z_i = i\), i.e., team 1 has scored \(i\) points more than team 2. We can write the difference in points after \(i\) seconds as: $$z_{i+1} - z_i = \begin{cases} 1 & \text{if both teams score a point in the next second} \\ -1 & \text{if neither team scores a point in the next second} \end{cases}$$ Now, we can express the probability \(P_i\) in terms of probabilities at the next step, as follows: $$P_i = P(\text{Team 1 wins } | z_i = i) = P(z_{i+1} = i+1) P_i + P(z_{i+1} = i-1) P_{i-2}$$ Here, we are expressing \(P_i\) as the sum of the probabilities of team 1 winning given that its lead increases by one point, and of team 1 winning given that its lead decreases by one point.

Finding the Probability of Team 1 Scoring and Team 2 Scoring in the Next Second

The probability of team 1 scoring and team 2 not scoring in the next second is given as: $$P(z_{i+1} = i+1) = P(\text{Team 1 scores})P(\text{Team 2 does not score}) = \lambda_1 e^{-\lambda_1} e^{-\lambda_2} = \lambda_1 e^{-(\lambda_1+\lambda_2)}$$ Similarly, the probability of team 1 not scoring and team 2 scoring in the next second is: $$P(z_{i+1} = i-1) = P(\text{Team 1 does not score})P(\text{Team 2 scores}) = \lambda_2 e^{-\lambda_1} e^{-\lambda_2} = \lambda_2 e^{-(\lambda_1+\lambda_2)}$$

Substituting the Probabilities in the Equation

Substituting the probabilities found in Step 3 into the equation in Step 2, we get: $$P_i = \lambda_1 e^{-(\lambda_1+\lambda_2)} P_i + \lambda_2 e^{-(\lambda_1+\lambda_2)} P_{i-2}$$ Rearranging the equation, we find the probability \(P_i\): $$P_i = \frac{\lambda_2}{\lambda_1}P_{i-2}$$

Applying the Gambler's Ruin Formula

Applying the formula for the gambler's ruin problem, the probability of team 1 winning (i.e., reaching a lead of \(k\) points) is given by: $$P_k = \frac{1 - \left(\frac{\lambda_2}{\lambda_1}\right)^k}{1 - \left(\frac{\lambda_2}{\lambda_1}\right)^{2k}}$$ So, the probability that team 1 wins the match is \(P_k\).