- Get link
- Google+
- Other Apps

Assuming without loss of generality that team \(1\) is the winner in sudden-death overtime. As we have two independent Poisson processes, the probability of this occurring is \(\frac{\lambda_1}{\lambda_1 + \lambda_2}\). Thus, the overall likelihood we'd like to maximize is \[L = e^{-\lambda_1} \frac{{\lambda_1}^n}{n!} e^{-\lambda_2} \frac{{\lambda_2}^n}{n!} \frac{\lambda_1}{\lambda_1 + \lambda_2}.\] Letting \(l = \log{L}\) we get \[l = -{\lambda_1} + n \log{\lambda_1} - {\lambda_2} + n \log{\lambda_2} - 2 \log{n!} …