# Prob question on games

I am playing in a racquetball tournament, and I am up against a player I have watched but never played before. I consider three possibilities for my prior model: we are equally talented, and each of us is equally likely to win each game; I am slightly better, and therefore I win each game independently with probability 0.6; or he is slightly better, and thus he wins each game independently with a probability 0.6. Before we play, I think that each of the three possibilities is equally likely. In our match we play until one player wins three games. I win the second game, but he wins the first, third, and fourth. After this match, in my posterior model, with what probability should I believe that my opponent is slightly better than I am?

I am getting an answer of 17/30. Can someone verify if it is correct? Thanks a lot

• How did you get 17/30? – JLee Nov 16 '14 at 22:23
• It is long answer which I did it in EXCEL. I just need someone to let me know if they solved the problem and what their answer is. If you want I can email my EXCEL sheet. – Satish Ramanathan Nov 16 '14 at 22:25
• Steps to follow: What is the probability of $3:1$ wins for a better, an equally talented, a worse player? Then use Bayes – Hagen von Eitzen Nov 16 '14 at 22:26
• @HagenvonEitzen, I found the prior probability in a long way. They are as you mentioned, $\frac{2}{3}, \frac{1}{2},\frac{1}{3}$. Is there a short trick to finding it out. And you know as well,that I complicate things. Wondering if you could suggest something? – Satish Ramanathan Nov 16 '14 at 22:32
• @HagenvonEitzen, is the answer provided by the responder correct?. I would appreciate if you could vet the answer. I believe he is correct but would want someone like you to say it is correct. – Satish Ramanathan Nov 17 '14 at 17:53

Let $p$ denote your a priori probability of winning a game .The pmf of $p$ is $$\mathbb{P}(p=\alpha) = 1/3$$ for $\alpha \in \lbrace 0.4, 0.5, 0.6 \rbrace$. Then by Bayes' \begin{align} \mathbb{P}(p=\alpha \mid \text{LWLL}) &= \frac{\mathbb{P}(\text{LWLL} \mid p = \alpha)\mathbb P (p =\alpha)}{\mathbb P (\text{LWLL})}\\ &= \frac{(1-\alpha)^3\alpha \frac{1}{3} }{\sum_\alpha \mathbb P (\text{LWLL} \mid p = \alpha ) \mathbb P(p=\alpha)}\\ &= \frac{(1-\alpha)^3\alpha}{0.1873} \end{align} For $\alpha = 0.4$ this becomes approximately $0.46$.

• how did you get the pmf $P(p=\alpha) = \frac{1}{3}$? – Satish Ramanathan Nov 17 '14 at 0:07
• @satishramanathan since each case is equally likely and there are three cases – Slug Pue Nov 17 '14 at 0:09
• Where are you finding the prori probability of one winning on all three cases when your opponent is better, equally talented, and worse,. In other words, where are you using the fact that the probability of your opponent winning the game is 0.6 other than the time when you calculate the P(LWLL)? – Satish Ramanathan Nov 17 '14 at 0:18
• well, that is denoted by $\mathbb P (\text{LWLL} \mid p = \alpha)$. $\alpha$ is your probability of winning, his is $(1-\alpha)$. So the case you are interested in is substituted as $\alpha = 0.4$ at the end. – Slug Pue Nov 17 '14 at 0:27
• Priori probability of you winning the tournament should not have any information of the event of winning the match (LWLL). Don't you think, it is more complicated of finding one winning in each possibility (of course each of one of them are equally likely of a scenario, not winning though). I thought you should calculate the priori probability of winning the tournament under the condition that each player needs to win three games first to be a winner. I have a habit of overcomplicating things. Just clarify this, please – Satish Ramanathan Nov 17 '14 at 0:36

Draw a tree diagram.

Event you've witnessed is LWLL. What is the probability of this under each case?

Work your way backwards through tree diagram to calculate posterior probabilities.

Now calculate the expected probability of your friend winning a random game under posterior model.