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Showing posts from December, 2013

Solving IMO 1989 #6 using Probability and Expectation

A Stupid and Strange Way of Looking at Sports Power Ratings that could be Smart and Useful

As I've mentioned previously, a common method used in sports for estimating game outcomes known as log5 can be written \[p = \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\] where \(p_i\) is the fraction of games won by team \(i\) and \(q_i\) is the fraction of games lost by team \(i\). We're assuming that there are no ties. What's the easiest way to derive this estimate? Here's one argument. Assume team \(i\) has a probability \(p_i\) of beating an average team (a team that wins half its games). Now imagine that this means for any given game the team has some "strength" sampled from [0,1] with median \(p_i\) and that the stronger team always wins. Thus, the probability that team 1 beats team 2 is \[ p = \int_0^1 \int_0^1 \! \mathrm{Pr}(p_1 > p_2) \, \mathrm{d} p_1 \mathrm{d} p_2 .\] This looks complicated, but but with probability \(p_1\) team 1 is stronger than an average team and with probability \(p_2\) team 2 is stronger than an average team. From this perspective…