 Probability and Statistics Calculating winning chances

 From: Douglas Zare Address: zare@math.columbia.edu Date: 23 June 2000 Subject: Re: How to calculate the ecspected winning-chances? Forum: rec.games.backgammon Google: 3953E045.652CE4FB@math.columbia.edu

```Erik Sřrensen wrote:

> The answer I want should take these factors as input:
> - The number of games played so far
> - The number of points actually won so far
>
> This should be the output:
> - By how much in percent am I favourite?
> - What is the uncertainty on this number?

There are a lot of factors and complications one could include, but I
won't. The following will give a rough method that is reasonably accurate
if you are winning or losing less than .5 points per game.

A fair estimate for your advantage in points per game is the number of
points you have won divided by the number of games. You could assume that
you have neither been lucky nor unlucky, and that the results display the
skill difference. Of course, don't do this if the number of games you have
played is 2 and you are up 38 points.

If this is your advantage, you next want to know the variance,
approximately the expected value of the square of the result of a game.
This is depends strongly on your playing styles. It might even be infinite
if you are too loose with the cube. I would guess that a variance of 6 is
plausible, but it might be higher or lower for you. Then the variance for n
games is 6n, e.g., 600 for 100 games. The standard deviation is the
square-root of the variance, and the standard deviation is what we want. If
you play a large number of games, there is about a 2/3 chance that the
result will be within 1 standard deviation of the average, and just over a
95% chance that the result will be within 2 standard deviations of the
average.

Suppose you have a .1 points per game advantage, a variance of 6, and you
play 100 games. The average result will be that you win 10 points, with a
variance of 600 -- a standard deviation of sqrt(600)~24.5. So 2/3 of the
time your score would be between -14 and +34, and 95% of the time it would
be between -39 and +59.

One can use this to give a different notion of confidence. Instead of
assuming that there was 0 net luck in the set of games from which you
will be 1 time in 40, or as bad as it will be 1 time in 40. 95% of the time
the correct value will be between the two estimates, but as you can see, it
takes much more than 100 games to get a decent window. If you won 10 points
in 100 games, this method would suggest that you have between a -.39
deficit and a +.59 advantage, so one should not actually be very surprised
if you score -50 in the next 100 games. It would probably mean that you
were lucky in the first 100 games, and unlucky in the second, but not
amazingly so.

Douglas Zare
```

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