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ELO

Posted By: Bob Koca
Date: Wednesday, 16 December 2009, at 11:14 p.m.

In Response To: ELO (Rich Munitz)

The short answer is no. The system being self correcting does not mean that its predictions are accurate over a range of ELO ratings differences.

Consider one point matches. Suppose we know the ratings difference between A and B and also between B and C. The formula then predicts a win % for the favorite for each of the possible matchups. Suppose that due to playing each other that the true ELO ratings difference for A with B and B with C is determined. This can be self correcting. That also determines though a ratings difference for A with C, and we just sort of need to just hope that this also works out well. There is a model that guarantess this though:

Suppose we play the following game. Everyone is assigned a positive number. When you go to play someone the result is determined randomly with your probability of winning being (your number)/(sum of your number and opponent's number). If the situation is like this then ELO works well. Note here that if you know the probability of A defeating B and also the probability of B defeating C then the probability of A defeating C is also known. For example suppose that A defeats B 3/5 of the time (60)% of the time and that B defeats C (3/5) of the time (60%). Work out for yourself that A defeats C 9/13 of the time (about 69.2%)

That this matches up well with FIBS ELO can be easily checked. For 1 point matches the 60% win chance is about 352 ELO difference. Doubling that to a 704 ELO difference gives a 69.2% win chance matching the prediction from the model above. Does the numbers model fit backgammon? No one has really checked as far as I know.

Now suppose we modify the game slightly. First with probability p you decide if the game will be decided by a coin flip or will be played out as above (with probability 1 -p). Now suppose that we use p = .5 and that relative skills are such that A defeats B 60% and that B defeats C 60%. This means that the half of the time where the above game is played that A must defeat B 70% of the time and B must defeat C 70% of the time. Then when the above game is played A defeats C 49/58 of the time (about 84.5%). This means that in our modified game that A defeats C 25% (from coin flip version) + (1/2) (49/58) from playing it out version = 67.25%

Here if ELO work out correctly for A vs B and for B vs. C then it gives too high a chance to the underdog when C plays A. If backgammon is more like the second model then big underdogs win more often the ELO predicts. My hunch is that the second model but with the coin toss version happening much less then 1/2 of the time is more accurate for backgammon.

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