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### Achim Müller

Feb 18, 1999, 3:00:00 AM2/18/99
to
Hi folks,

at the moment I'm playing a few longer money sessions against two guys
and (for training) 1000 money games against Snowie 1.3. While thinking
about some longer winning and loosing streaks I was wondering, if there
is reliable statistical material available about what may happen in a
given amount of games.

What I mean:

1) Let's assume you play against someone 1000 money games. You and your
opponent are ranked between expert and world class level (if that
matters). You play using the jacoby rule, beaver and raccoons, but no
automatics. Let's further assume, that the gammon probability is round

What is the interval of possible results with a given 95% probability?

2) Nearly the same, but now you play only 100 games.

3) And now you stop at 50 games.

4) To make it more difficult, the same scenario as shown above, but now
you are 5% better than your opponent (equity = +0.01 per game).

Years ago I studied economics including two semesters statistic. I know
that there is a possibility to calculate it, but I'm too short of time
to figure it out again.

Can anyone help?

Ciao

acepoint (Achim)

If that helps: A friend played 770 games against jellyfish 3.0. Points
per game are 2.063.

### stig...@yahoo.com

Feb 18, 1999, 3:00:00 AM2/18/99
to
Hi Achim.
Some time ago I made a formula which might be useful to you:
If you play n games, a 96% confidenceinterval for two
equally good players is 0 +/- 5*sqrt(n) points.
That is, if you play 100 games, you should be prepared
to lose 50 points.
If the players are of inequal strength, the 0 in the formula
can be replaced with ppg*n. But I don't think it's an easy
thing to calculate the ppg. But hey, David Montgomery made a formula
once to translate between fibs rating and moneygame.
Found it!
50 FIBS points are worth about 0.1 ppg.

Stig Eide

-----------== Posted via Deja News, The Discussion Network ==----------

### Gary Wong

Feb 18, 1999, 3:00:00 AM2/18/99
to
Achim Müller <acep...@deltacity.net> writes:
> 1) Let's assume you play against someone 1000 money games. You and your
> opponent are ranked between expert and world class level (if that
> matters). You play using the jacoby rule, beaver and raccoons, but no
> automatics. Let's further assume, that the gammon probability is round
>
> What is the interval of possible results with a given 95% probability?

A reasonable model for the distribution of points after a long series of
money games is that points per game after n games are normally distributed
with mean pn and variance 9n (where p is the advantage you have in
points per game). This is just an approximate model and is not tailored
to the parameters you specified, but it seems to fit the data I have
fairly well.

A 95% prediction interval for the score after 1000 games between evenly
matched players is from -186 to +186 points.

> 2) Nearly the same, but now you play only 100 games.

Now the interval is from -59 to +59 points.

> 3) And now you stop at 50 games.

-42 to +42 points.

> 4) To make it more difficult, the same scenario as shown above, but now
> you are 5% better than your opponent (equity = +0.01 per game).

I'm not sure what you mean by 5% better (expecting to win 55% of the games
would be worth well over 0.01ppg). 0.01ppg really is pretty tiny, but if
you had an advantage of that amount, the three intervals above would become
-176 to +196, -58 to +60 and -41 to +42 points respectively.

Cheers,
Gary.
--
Gary Wong, Department of Computer Science, University of Arizona
ga...@cs.arizona.edu http://www.cs.arizona.edu/~gary/

### David desJardins

Feb 18, 1999, 3:00:00 AM2/18/99
to
Gary Wong <ga...@cs.arizona.edu> writes:
> A 95% prediction interval for the score after 1000 games between evenly
> matched players is from -186 to +186 points.

Yes, but note that another "95% prediction interval" for the same
hypothesis (and your mathematical model) is that the score is anything
higher than -156 points. And another 95% prediction interval is that
the score is anything less than +156 points. And another 95% prediction
interval is that the score is anwhere outside the range from -6 points
to +6 points.

All of these prediction intervals are equally valid a priori. If you
played 1000 games and at the end the score were +3 points, then you
could "reject the null hypothesis at the 95% confidence level" on the
basis of the last confidence interval mentioned above.

Whether you believe that means the null hypothesis is wrong depends, at
least in part, on how plausible you think it is that there is a
mechanism operating which causes either high or low scores to be more
likely than scores near zero.

David desJardins

### Achim Müller

Feb 18, 1999, 3:00:00 AM2/18/99
to
Gary Wong wrote:
>
> > 4) To make it more difficult, the same scenario as shown above, but now
> > you are 5% better than your opponent (equity = +0.01 per game).
>
> I'm not sure what you mean by 5% better (expecting to win 55% of the games
> would be worth well over 0.01ppg).

Ooops! Of course I meant 0.1 ppg. Sorry for that mistake, tough night
last night ;-)

But now I have two models:

1. Stig Eide

> If you play n games, a 96% confidenceinterval for two
> equally good players is 0 +/- 5*sqrt(n) points.

2. Gary Wong

> A reasonable model for the distribution of points after a long series of
> money games is that points per game after n games are normally distributed
> with mean pn and variance 9n (where p is the advantage you have in
> points per game).

I also thought about the normal distribution, but wasn´t able to get the
formula. Could you, Gary, explain where the 9 in "9n" comes from?

Ans a last question to Stig: How did you get your formula?

Ciao

acepoint (Achim)

### Gary Wong

Feb 18, 1999, 3:00:00 AM2/18/99
to
Achim Müller <acep...@deltacity.net> writes:
> But now I have two models:
>
> 1. Stig Eide
>
>> If you play n games, a 96% confidenceinterval for two
>> equally good players is 0 +/- 5*sqrt(n) points. =

>
> 2. Gary Wong
>
>> A reasonable model for the distribution of points after a long series of
>> money games is that points per game after n games are normally distributed
>> with mean pn and variance 9n (where p is the advantage you have in
>> points per game).
>
> I also thought about the normal distribution, but wasn't able to get the
> formula. Could you, Gary, explain where the 9 in "9n" comes from?

Sure. There's no deep theoretical reason for picking that value, but an
observation of several hundred games of mine showed a sample variance
of 8.4. This result and others I've occasionally seen posted here
lead me to estimate that the true variance is somewhere around 9.
(Obviously it will depend on the players, the rules, etc. etc., but
9 is a convenient rule of thumb.)

The variance I estimate agrees fairly closely with Stig's; a variance of
9 is a standard deviation of 3 so I would predict a +/-2 sigma interval
of 0 +/- 6*sqrt(n) points -- essentially identical to his result, at this
level of accuracy.

(The normal distribution is a bad fit for small numbers of games of course,
because backgammon game distributions have very long tails because of the
relatively high probability of obtaining a score a long way from the mean,
because of the cube. It's not until you add up a lot of games that the
central limit theorem kicks in and this approximation becomes reasonable.
In practice I wouldn't be too worried about weird cube effects once there
were, say, 50 or more games.)

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