If a team finishes a 4-game season with a 3-1 record (3 wins, 1 loss), they have a winning percentage of 75%. But what is (was) the probability that they would win any one game? There are 4 ways that a team can win 3 and lose 1: WWWL, WWLW, WLWW, & LWWW. If P=probability of winning one game, then the probability of a 3-1 season is: 4 x (P^3) x (1-P) [For some reason, I can't get LaTeX to work] If P=0.75, that comes out to 0.421875. Of course, with a 75% probability of winning one game, there is also a 31.64 chance of finishing 4-0, a 21.1% chance of finishing 2-2, a 4.69% chance of finishing 1-3, and a 0.39% chance of finishing 0-4. Can I reverse engineer that formula and set it = 0.75: 4 x (P^3) x (1-P) = 0.75 Then P^3 - P^4 = 0.75/4 I don't know how to solve that for P or even if that would solve the problem.

\(P^{3} - P^{4}\) has a global maximum of \(27 / 256 \approx 0.1055\) (obtained with \(P = 3 / 4\)). That's less than \(3 / 16 = 0.1875\), so you won't be able to solve \(P^{3} - P^{4} = 3 / 16\) with a real value of \(P\). If it did have a solution, \(P^{4} - P^{3} + Q = 0\) is the problem of finding the roots of a quartic polynomial. It's possible to solve these analytically (see here for instance), but it usually isn't worth the trouble. As for TeX, the same tags still seem to work (this is the first time I'm using them since the software was updated), but I had to refresh the browser window and they didn't show properly when I was previewing the post.

After doing a little more research, it looks like this is a statistics problem. As I mentioned above, as long as the probability of winning one game is >0 and <1, there are 5 possible win-loss records in a 4-game season (assuming ties are not possible): 4-0, 3-1, 2-2, 1-3, and 0-4. Therefore, if all we know is that the team finished 3-1, we can't solve for a fixed P, because even a team with a P=.01 (1%) has a chance (0.0004%) of finishing 3-1. So I think I have to look into confidence intervals (ugh). I am even worse at statistics than at math in general. (sigh)

I think this is a malformed question. The original question refers to "the probability that they would win any one game" but then you appear to calculate the odds of actually winning ONLY one game. Does it not seem odd that a team winning MOST of its games would only stand a 42% chance of winning a particular one? If the team has a winning percentage of 75% then the odds of winning any particular game (without any more information) is 75%.

Ha-ha. Very likely. And it wouldn't be the first time. I think this is a malformed response. (Sorry, I couldn't resist.) I didn't say that the chance of winning one game is 42%. I said that the odds of finishing 3-1 is 42%. I don't think so. Suppose I have a collection of black boxes, each with a Play button and two lights: a Green "Win" light and a red "Lose" light. Each time I press the Play button, one of the lights lights up according to some internal probability between 0% and 100%. If I press the button 4 times on one box and get 3 green lights and 1 red one, I cannot conclude that the internal probability is 75%. It could be anything from just over 0% to just under 100%. No?

Agreed! But now you're talking about confidence intervals, as in, "the internal probability is X% with a confidence of Y%" and Y gets larger with more data. Like I said, 75% represents the most likely answer (i.e. highest confidence) when announcing the odds of winning a particular game. If you think this is a malformed response, it's possible that I misinterpreted what you wrote or didn't understand what you're after. Perhaps you could rephrase it? I can't imagine what YOU think 42% represents. If it's a given that the team won 3 games and lost 1, then there is a 25% chance that they lost any specific game. This should be obvious.