What is the chance that in an n-point match the loser will end up with m points?
Assume Crawford rule, no beavers/raccoons, equal playing strengths for the two players, any MET with common gammon/backgammon rates, n > m >= 0.
This sounds complicated at first since there are so many ways in which a certain final matchscore can come about. Then I remembered that it does not matter when you get your luck, only that you get it. So my guess is that my chance of winning a 5 point match with a final score of 5-2 should be the same as winning it from 0-2. If this is correct, then the problem reduces to looking up the probability for 5-away/3-away in a MET (which is where the assumption of equal playing strengths is needed). The general case would amount to looking up n-away / (n-m)-away.
I don't feel 100% confident yet about this answer, please correct if I have overlooked something.
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Correction:
My answer cannot be right since winning from 0-2 can result in other final scores than 5-2 and the probabilities for 5-3 and 5-4 are included in P_MET(5-away / 3-away).
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Correction 2:
Since P_MET(5-away / 3-away) gives my winning probability for the case of my opponent getting at least 2 points, deducting the probability of him getting at least 3 points should give the correct answer: P(I win 5-2) = P_MET(5-away / 3-away) - P_MET(5-away / 2-away).
For the general case we get: P(I win n-m) = P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)
Correction 3:
The original question didn't ask for P(I win n-m), but for P(either player wins n-m), which is 2*P(I win n-m).
Some numbers (using Neil Kazaross's MET KazarossXG2.xml bundled with gnubg):
Probabilities are given as P(the match ends 5-n).
P(5-0) = 15.4%
P(5-1) = 14.0%
P(5-2) = 19.2%
P(5-3) = 19.6%
P(5-4) = 31.6%
At first I thought this looks plausible: The players have equal strength, so close match scores should occur more often than more extreme ones. But can 5-4 really be expected to occur that much more often than 5-3? After all, at 3-3, cubing is mandatory. So I let gnubg play 15 5-pointers against itself, both players at world class strength. Result:
5-0 5 times
5-1 0 times
5-2 7 times
5-3 2 times
5-4 1 time
15 matches is not much, but still it seems there is something seriously wrong with my theory, back to the drawing board...
My first approach was to use a Markov chain: in an n point match, you start at position (n, n) in the MET. After each finished game, you either move up k_W positions (a won game) or you move k_L positions to the left (a lost game), with k_W and k_L being 1, 2 or 3 times the value of the cube. (Respecting the borders of the MET of course). A new game is started until position (0, m) or (m, 0) is reached.
If the probabilities for each k_W and k_L for each position in the MET were known, the problem can easily be solved using a Markov chain. Unfortunately, I don't know them, but their information should somehow be incorporated in the MET.
Pck's answer, P(I win n-m) = 2 * [P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)], fulfills two requirements: it uses the MET, and the n probabilities are positive and add up to 1. However, I am not convinced for three reasons:
- Using the Rockwell-Kazaross MET, I calculated the probabilities for a 25 point match and it showed that a 25-0 end score is 30 times as likely as a 25-23 end score (resp. 6.2% vs. 0.21%). This sounds a bit unrealistic for players of equal level;
- A MET consists of a pre-Crawford and a post-Crawford part and the latter is not used;
- P_MET(n-away / (n-m)-away) gives us the probability of final scores n-m, n-(m+1), ..., n-(n-1) from match score 0-m, and P_MET(n-away / (n-m-1)-away) gives us the probability of final scores n-(m+1), n-(m+2), ..., n-(n-1) from match score 0-(m+1). Since P_MET(n-away / (n-m)-away) and P_MET(n-away / (n-m-1)-away) are conditional on different match scores, I am not sure that probabilities in red will cancel out.
None of my arguments refute pck's answer though... I'll return to the drawing board as well!
I can't make a meaningful contribution to the mathematical argument, but I can count. Of the 258 5 point matches in my database, the breakdown of results is as follows.
5-0 103
5-1 33
5-2 49
5-3 35
5-4 38.
All the matches involve me of course against a wide range of humans. Is equal skill important to the answer?
Quote from: dorbel on January 11, 2012, 09:07:04 AM
All the matches involve me of course against a wide range of humans. Is equal skill important to the answer?
Good question. As long as results like 5-2 and 2-5 are not counted as the same, the distribution of scores in 5 point matches must certainly depend on the players' skill difference, otherwise there would be no advantage in consistently playing better than your opponent.
If however, as in the postings above, we conflate n-m and m-n scores, the number of 5-2's might or might not change. If the better player on average wins some additional percentage p of 5-2's (compared to the equal skills case) but loses the exact same amount of 2-5's less, the total number of matches ending in 5-2 would not change.
Whatever the answer may be, your numbers do confirm my own gnubg tests. I ran more matches of the same type as the 15 I listed above and 5-0 and 5-2 kept turning up as the most frequent scores in 5 point matches between two players of equal strength.
Quote from: boomslang on January 11, 2012, 03:47:35 AM
Pck's answer, P(I win n-m) = 2 * [P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)], fulfills two requirements: it uses the MET, and the n probabilities are positive and add up to 1. However, I am not convinced for three reasons:
I'm pretty sure now that my answer is not only wrong but off by a mile. The reasons you gave, plus dorbel's and my own numerical data make that clear. The confusion is this: I based everthing on the idea that with regard to winning, it is immaterial when you get your luck in a match. It is only important that you get a certain total amount. That is certainly correct, but it pertains to winning the match, not to the question of what the final score will be. The final score will indeed depend on
when you get your luck. You can be unlucky first, then very lucky and win 5-3, or you can be moderately lucky from start to finish and win 5-0. In both cases you win with the same amount of luck (in fact, and rather trivially, all matches are won with the same amount of luck given a certain fixed difference of skill of the players). I think this is compatible with what the third point in your list says (red probabilities not cancelling out).
Therefore, "
P_MET(5-away / 3-away) gives my winning probability for the case of my opponent getting at least 2 points" is wrong and the entire thing collapses.
Quote from: boomslang on January 11, 2012, 03:47:35 AM
If the probabilities for each k_W and k_L for each position in the MET were known, the problem can easily be solved using a Markov chain. Unfortunately, I don't know them, but their information should somehow be incorporated in the MET.
That was my initial thought too, and I could not come up with a quick way to calculate P(k_W) and P(k_L) either. Obviously they depend on the match score. Perhaps it is possible to derive them recursively in a manner similar to the way METs are created? I haven't thought this through in any way yet, but the probability of the cube being turned may be a big problem.
In search for some empirical data I simulated 133000 11-pointers in GNUbg. See the table for the results.
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n m A B
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11 0 9.9% 10.27%
11 1 9.9% 4.79%
11 2 10.6% 10.34%
11 3 10.2% 6.04%
11 4 10.7% 12.18%
11 5 9.6% 8.07%
11 6 10.6% 12.06%
11 7 8.0% 8.04%
11 8 8.8% 11.85%
11 9 5.3% 8.29%
11 10 6.4% 8.07%
__________________________________
A = 2 * [P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)].
B = observed probability.
As we suspected, the results (B) do not match probability A... Given the large sample size (>130k), the results are pretty accurate so they should give some guidance to an analytical solution.
Both A and B are based on the same MET (g11). Chequer play: Expert; cube play: Expert.
BTW, the source code for a match equity calculator can be found here: http://cvs.savannah.gnu.org/viewvc/*checkout*/gnubg/gnubg/mec.c?revision=1.2&content-type=text%2Fplain (http://cvs.savannah.gnu.org/viewvc/*checkout*/gnubg/gnubg/mec.c?revision=1.2&content-type=text%2Fplain)