Opening book from a statistical point of view

[Edmund]

Thank you very much!

Here my output.
Score can be interpreted as the elo difference needed to on average score even after playing a certain move. I am sorting by score minus standard deviation.

Code: Select all

move		   score		    sd		    score-sd
Ng1-f3		 180,38		   0,02		  180,36
Pg2-g3		 173,36		   0,58		  172,78
Pd2-d4		 134,08		   0,00		  134,08
Pc2-c4		 122,61		   0,01		  122,59
Pe2-e4		  83,90		   0,00		   83,90
Pd2-d3		 136,07		  68,03		   68,03
Nb1-c3		  20,39		   0,29		   20,10
Pb2-b3		  18,37		   0,22		   18,15
Pf2-f4		 -74,67		   0,16		  -74,83
Pc2-c3		   0,00		  76,54		  -76,54
Pb2-b4		 -84,25		   0,58		  -84,82
Pe2-e3		-346,09		  53,24		 -399,34
Ng1-h3		 612,31		1224,63		 -612,31
Ph2-h3		1224,63		2449,26		-1224,63
Pf2-f3		1224,63		2449,26		-1224,63
Nb1-a3		na
Pa2-a3		na
Ph2-h4		na
Pg2-g4		na
Pa2-a4		na

And here the LOS table - ie. probability that the move in the first column is stronger than the move in the first row:

Code: Select all

	      Ng1-f3	Pg2-g3	Pd2-d4	Pc2-c4	Pe2-e4	Pd2-d3	Nb1-c3	Pb2-b3	Pf2-f4	Pc2-c3	Pb2-b4	Pe2-e3	Ng1-h3	Ph2-h3	Pf2-f3
Ng1-f3	    50	   100	   100	   100	   100	   100	   100	   100	   100	   100	   100	   100	    39	    na	    na
Pg2-g3	     0	    50	   100	   100	   100	   100	   100	   100	   100	   100	   100	   100	    39	    na	    na
Pd2-d4	     0	     0	    50	   100	   100	    43	   100	   100	   100	   100	   100	   100	    38	    na	    na
Pc2-c4	     0	     0	     0	    50	   100	    12	   100	   100	   100	   100	   100	   100	    38	    na	    na
Pe2-e4	     0	     0	     0	     0	    50	     0	   100	   100	   100	   100	   100	   100	    37	    na	    na
Pd2-d3	     0	     0	    57	    88	   100	    50	   100	   100	   100	   100	   100	   100	    38	    na	    na
Nb1-c3	     0	     0	     0	     0	     0	     0	    50	   100	   100	    93	   100	   100	    36	    na	    na
Pb2-b3	     0	     0	     0	     0	     0	     0	     0	    50	   100	    90	   100	   100	    36	    na	    na
Pf2-f4	     0	     0	     0	     0	     0	     0	     0	     0	    50	     0	   100	   100	    34	    na	    na
Pc2-c3	     0	     0	     0	     0	     0	     0	     7	    10	   100	    50	   100	   100	    35	    na	    na
Pb2-b4	     0	     0	     0	     0	     0	     0	     0	     0	     0	     0	    50	   100	    34	    na	    na
Pe2-e3	     0	     0	     0	     0	     0	     0	     0	     0	     0	     0	     0	    50	    29	    na	    na
Ng1-h3	    61	    61	    62	    62	    63	    62	    64	    64	    66	    65	    66	    71	    50	    na	    na
Ph2-h3	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Pf2-f3	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Nb1-a3	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Pa2-a3	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Ph2-h4	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Pg2-g4	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na
Pa2-a4	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na	    na

[Edmund]

Let me elaborate on how I got to the score and sd for move Ng1-f3 as an example.

The likelihood function to be maximized for a certain score:
L(score) = wins*LN(1/(1+EXP(-(delta/400+score))+ v * EXP(-(delta/400+score)/2))) + losses*LN(1/(1+EXP( (delta/400+score))+ v * EXP( (delta/400+score)/2))) + draws*LN(1/(1+EXP(-(delta/400+score)/2)*(1+EXP(delta/400+score))/v))

I set delta=0 (ie we assume winning and loosing side were equal strength throughout) and v=1.
wins=47608
losses=29403
draws=46584

I found that the likelihood function can be approximated by a quadratic equation (=ax^2+bx+c). Thus I need to calculate at 3 points. I calculate at L(0), L(-q) and L(q), where I set q=1:
L(-1)=-154978
L(0)=-135783
L(1)=-136773

Now I can derive the quadratic equation:
a=(L(q)-(L(q)-L(-q))/2-L(0))/q^2
b=(L(q)-L(-q))/(2q)
c=L(0) 

L(x) = -10092 *x^2 + 9103 *x -135783

Setting the first derivative = 0 I get the score that maximizes the equation. I multiply by 400 to convert to ELO scores:
score = -b/(2*a) * 400 = 180.38
The standard deviation I derive from the inverse of the second derivative:
sd = -1/(2*a) * 400 = 0.02

Note, the true Likelihood-maximizing score (not the one approximated by the quadratic equation) is 178.21

[stegemma]

Let me elaborate on how I got to the score and sd for move Ng1-f3 as an example.

The likelihood function to be maximized for a certain score:
L(score) = wins*LN(1/(1+EXP(-(delta/400+score))+ v * EXP(-(delta/400+score)/2))) + losses*LN(1/(1+EXP( (delta/400+score))+ v * EXP( (delta/400+score)/2))) + draws*LN(1/(1+EXP(-(delta/400+score)/2)*(1+EXP(delta/400+score))/v))

I set delta=0 (ie we assume winning and loosing side were equal strength throughout) and v=1.
wins=47608
losses=29403
draws=46584

I found that the likelihood function can be approximated by a quadratic equation (=ax^2+bx+c). Thus I need to calculate at 3 points. I calculate at L(0), L(-q) and L(q), where I set q=1:
L(-1)=-154978
L(0)=-135783
L(1)=-136773

Now I can derive the quadratic equation:
a=(L(q)-(L(q)-L(-q))/2-L(0))/q^2
b=(L(q)-L(-q))/(2q)
c=L(0) 

L(x) = -10092 *x^2 + 9103 *x -135783

Setting the first derivative = 0 I get the score that maximizes the equation. I multiply by 400 to convert to ELO scores:
score = -b/(2*a) * 400 = 180.38
The standard deviation I derive from the inverse of the second derivative:
sd = -1/(2*a) * 400 = 0.02

Note, the true Likelihood-maximizing score (not the one approximated by the quadratic equation) is 178.21

Wow!

It was a great work, I'm must study the base math involved to understand it but it can really help.

Thanks.

[Edmund]

Wow!

It was a great work, I'm must study the base math involved to understand it but it can really help.

Thanks.

Let me know if there is anything unclear.
But bear in mind that the real value added comes when you are considering elo-differences of the players per game. Otherwise I suppose the outcome will not be radically different to the non-chess specific solutions proposed in this thread.