How to calculate the ecspected winning-chances?
Erik M Sørensen
anyone could answer it, I would be very happy.
Somehow I would like to know, when I'm playing against somebody: who is the
best player and by how much ?
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?
To make it all easier lets assume:
Both players dont get tired etc. and dont get affected by misfortune etc.
I would like to be able to make expressions like:
I'm 60 % favorite, and in 100 games I will win 20 points plus minus 40
points
Of course, the purpose of this is to calculate my ecspected average income
per game (hour) and the dispersion (uncertainty) on this number.
Erik M Sørensen
Douglas Zare
> [...]
> 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 estimated your
advantage, you could assume that your luck was as good as it 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
André Nicoulin
Very brief answer: Snowie version 3 (www.oasya.com) will compute exactly
what you want, but it will do it in an even better manner, since it will
take out the luck present in your data. It will base its Favorite-By
estimator on the errors made, and not on the outcome of the games.
André Nicoulin, Oasya
"Erik M Sørensen" <e...@get2net.dk> a écrit dans le message news:
MHr45.224$n%5.6...@news.get2net.dk...
> I'd like to ask one of the most fundamental questions of backgammon! If
> anyone could answer it, I would be very happy.
>
> Somehow I would like to know, when I'm playing against somebody: who is
the
> best player and by how much ?
>
> 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?
>