RE: [Bug-gnubg] real match winning chances

[Christopher D. Yep]

At 10:33 PM 12/6/2003 +0100, Misja Alma wrote:
> Sorry, I think I didn't express very clearly what I meant, the example I
> gave could be more complete. Here is my explanation again:
> So 2 players start a match, and at that point both have 50% chance.
> When player 1 makes a mistake which costs him half his mwc, this costs him
> .5 * 50% = 25% mwc. Gnubg will find this also.
> But now let's assume that the player gets lucky and comes back to 50% mwc
> again. At that point he makes again a mistake which costs him half his mwc;
> so 25%. What gnubg does is it addes up error1 and error2, for a total of
> 50%. If these were all the errors both players made this match, it will then
> estimate player 1 to have 0% mwc in an analysis.
> The luck rate will do something similar: After his first error player 1 must
> have had luck worth for 25% mwc. Suppose that player1 won the match anyway
> after his second mistake, he must have had another portion of luck worth for
> 75% mwc this time. Both are added up for a total of 100%. final - initial =
> netLuck + netSkill; Filling the numbers in gives that netSkill must be zero.

For Player 1:

initial mwc = 50%, final mwc = 100%
netSkill = -50% (Player 1 gave up 50% mwc, Player 2 gave up 0% mwc in this 
hypothetical example)
net luck = +100%
result = +50%

> Btw I checked this by playing a 1pt match against gnubg with manual dice,
> where I tried to make all possible errors but gave gnubg such poor dice that
> it lost anyway; An analysis gives me a total error rate of -128% and a luck
> adjusted result of -137%.

(Note: the total error rate = luck adjusted error rate if gnubg's match 
analysis is perfect; since gnubg's match analysis is not perfect, the above 
result seems reasonable.)

Suppose that Player A (playing perfectly) plays Player B (an extremely weak 
player, e.g. a computer making all checker and cube decisions 
randomly).  In general Player A is > 99.9999% favorite (let's round this to 
100%).  However, if these matches are analyzed, sometimes Player A will get 
very good dice and will quickly win the match.  In these matches, player B 
won't have many opportunities to make errors and may only make, say, 25% 
mwc worth of errors, resulting in a luck adjusted result of -25% mwc.  In 
other matches though, Player B will get very good dice which will prolong 
the match, giving him more opportunities to make errors.  In these matches 
he might make, say, 75% mwc worth of errors, resulting in a luck adjusted 
result of -75% mwc.

On average player B will make 50% mwc errors per match.  In a given match 
he may make less than or more than 50% mwc worth of errors.  Player A's 
expected result is +50% mwc (i.e. he expects to win 100% of the matches).

> I think that those numbers are not right. If I would do a prediction of the
> mwc of player 1 against player 2 based only on the match above, I would say
> that player 1 apparently gives away half his mwc twice during a match,
> regardless of what the situation or matchscore is at those times. So his mwc
> will be on average 50% * .5 *.5 = 12.5%.

I disagree.  In the extreme example that I gave, I'd be annoyed if gnubg 
didn't report Player A's luck adjusted result as > +50% mwc in some of its 
matches.  Gnubg is just reporting what happened in the individual match (I 
like how it does it).

It's important to analyze a large number of matches.  What matters is the 
average luck adjusted result.  In extreme examples, as I gave above, the 
length of the match (which is positively correlated with Player B's net 
luck) significantly affects the luck adjusted result of an individual match.

Chris