Quote:
Originally Posted by grndmstr_c
Okay, Ill take the crow. I think I must have been getting myself crossed on winning possessions instead of winning games, but I just worked out a simple proof and dirno and CD are correct on this. Should've taken the time to look at it systematically first, but for those who are interested, here's an asymptotic argument that illustrates why:
Suppose you have Team A and Team B playing a game that proceeds in discrete fashion, i.e., in possessions, and suppose Team A has constant probability of winning a single possession = p. Now, what we're interested in is what if we extend the game so that it has N possessions, and winning the game is defined as winning a majority of the possessions, i.e., winning (N+1)/2 or more possessions. We can use the normal approximation to the binomial if we assume N is large enough, and doing so it's straightforward to show that the probability of winning (N+1)/2 contests is aproximately equal to the cumulative distribution function (cdf) of the standard normal distribution evaluated at the square root of N times a function of p that is positive if p > .5, and negative if p < .5. The standard normal cdf is strictly increasing from 0 to 1 across the real number line, and it follows that the probability of winning (N+1)/2 contests will go to 0 as N goes to infinity if p is even slightly less than .5, and will go to 1 as N goes to infinity if p is even slightly greater than .5. Obviously, the reverse occurs as N goes from large to small.
My apologies gentlemen, and may this serve as a reminder to all that there is real value in striving even to disagree with civility.
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I guess I look at it differently and make different assumptions. Instead of winning a possession, in basketball you'd have two teams with a possession each trying to score. Let's assume we have good team A and bad team B. And those are the only teams in the league. Now, each team has a probability to score respectively p and q, where p>q. If those probabilities are normally distributed, you'd get various possibilities for winning scores with one single possessions each. If you play the game with an infinite number of possessions, team A would always end up winning, and it's expected score would approach N*p where N is the number of possessions. But the mean would always be p=!q regardless.