Author: Dieter Buerssner
Date: 12:22:37 10/03/03
Go up one level in this thread
On October 03, 2003 at 10:05:06, Tord Romstad wrote:
>After 200 games, the score was 111.5-89.5.
Which does add up to 201.
>Could you do a similar
>calculation with these numbers (or better yet, teach me how to do it
>myself)?
I was too lazy (or unable :-) to figure it out analytically. However, a Monte
Carlo simulation of a 200 game match between equal opponents, where the games
are considered independant (no learning, wide book to avoid repeated games,
...), and assuming white wins 40% of the games, black wins 30% of the games, and
30% are drawn, shows:
C:\e\dcrand>cmatch 200 40 30 30 1000000
Result of chess matches between equal opponents
White wins 40.0%, black 30.0% and 30.0% draws
A match of 200 games was simulated by 1000000 Monte Carlo tries
result probability <= this > this
100.0 - 100.0 ( 50.0%): 3.398% 3.398% 96.602%
100.5 - 99.5 ( 50.3%): 6.800% 10.198% 89.802%
101.0 - 99.0 ( 50.5%): 6.649% 16.847% 83.153%
101.5 - 98.5 ( 50.8%): 6.565% 23.412% 76.588%
102.0 - 98.0 ( 51.0%): 6.420% 29.832% 70.168%
102.5 - 97.5 ( 51.3%): 6.157% 35.988% 64.012%
103.0 - 97.0 ( 51.5%): 5.954% 41.942% 58.058%
103.5 - 96.5 ( 51.8%): 5.719% 47.661% 52.339%
104.0 - 96.0 ( 52.0%): 5.404% 53.065% 46.935%
104.5 - 95.5 ( 52.3%): 5.063% 58.128% 41.872%
105.0 - 95.0 ( 52.5%): 4.734% 62.862% 37.138%
105.5 - 94.5 ( 52.8%): 4.417% 67.280% 32.720%
106.0 - 94.0 ( 53.0%): 4.053% 71.333% 28.667%
106.5 - 93.5 ( 53.3%): 3.663% 74.996% 25.004%
107.0 - 93.0 ( 53.5%): 3.313% 78.309% 21.691%
107.5 - 92.5 ( 53.8%): 2.985% 81.294% 18.706%
108.0 - 92.0 ( 54.0%): 2.708% 84.002% 15.998%
108.5 - 91.5 ( 54.3%): 2.412% 86.415% 13.585%
109.0 - 91.0 ( 54.5%): 2.103% 88.518% 11.482%
109.5 - 90.5 ( 54.8%): 1.823% 90.341% 9.659%
110.0 - 90.0 ( 55.0%): 1.587% 91.927% 8.073%
110.5 - 89.5 ( 55.3%): 1.367% 93.295% 6.705%
111.0 - 89.0 ( 55.5%): 1.171% 94.465% 5.535%
111.5 - 88.5 ( 55.8%): 0.981% 95.447% 4.553%
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
112.0 - 88.0 ( 56.0%): 0.837% 96.284% 3.716%
112.5 - 87.5 ( 56.3%): 0.711% 96.995% 3.005%
113.0 - 87.0 ( 56.5%): 0.584% 97.579% 2.421%
113.5 - 86.5 ( 56.8%): 0.485% 98.064% 1.936%
114.0 - 86.0 ( 57.0%): 0.393% 98.457% 1.543%
114.5 - 85.5 ( 57.3%): 0.329% 98.786% 1.214%
115.0 - 85.0 ( 57.5%): 0.269% 99.055% 0.945%
115.5 - 84.5 ( 57.8%): 0.212% 99.267% 0.733%
116.0 - 84.0 ( 58.0%): 0.169% 99.436% 0.564%
116.5 - 83.5 ( 58.3%): 0.130% 99.566% 0.434%
117.0 - 83.0 ( 58.5%): 0.108% 99.674% 0.326%
117.5 - 82.5 ( 58.8%): 0.083% 99.757% 0.243%
[...]
The above line means: in 95.4% of such matches between equal opponents the
result will be less extreme or equal to 111.5-89.5. In 4.6% of such matches, one
can expect a more extreme result. Or in other words, it is quite likely, that
the winner is better.
The Monte Carlo simulation can be done easily. Just roll a dice with numbers
1-100. If the result is between 1 and 40, white wins. If the result is between
41 and 70, black wins. If the result is between 71 and 100, it is draw (for the
numbers above). Roll the dice 200 times, and add up the results (don't forget to
switch colors). Repeat this many times, and calculate an average.
Writing a good routine to roll a dice is more tricky, than it may seem at first
sight. Obviously one will use a pseudo random number generator (PRNG). But the
straightforward methods (all the ones I have read about in books or for example
in the comp.lang.c FAQ) still have a bias, when the maximim return of the PRNG+1
is not devidable by the number of faces of the dice without remainder.
I suggest something like
/* returns 0 <= r < range equally distributed */
unsigned long rand_range(unsigned long range)
{
unsigned long rmax, r, d;
/* find the largest number rmax <= MY_RAND_MAX, for which
(rmax+1) % range == 0.
All returns from rand() > rmax will be skipped, to guarantee
equal probability for all return values. */
d = (MY_RAND_MAX+1-range) / range + 1; /* Ignore possible compiler warning */
rmax = d * range - 1; /* -1 to avoid "overflow to zero" */
do
r = MY_PRNG();
while (r > rmax);
return r/d;
}
Remi has written a nice article, where he looks at it analytically; you will
find it easily, when you follow the link Peter has suggested.
Regards,
Dieter
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