Page 3 of 4
Why Knight and (lone) Bishop are so nearly equal in value.
Posted: Thu Sep 27, 2012 9:09 pm
by Ajedrecista
Hi Adam:
Adam Hair wrote:Here is some data from high Elo, longer time control engine matches, using similar criteria Larry used in his material imbalance study:
Code: Select all
qr>=0p7 = 1 queen, 0 up to 2 rooks, 7 pawns
B2 vs N2 Total Games Centered Score Centipawns
total 9084 58.15 57.1
qr>=0p7 2842 55.15 35.9
r>=0p7 902 58.5 59.6
qr>=0p6 3220 56.2 43.3
r>=0p6 1191 59.5 66.8
qr>=0p5 1568 60.35 73.0
r>=0p5 1125 63.75 98.1
qr>=0p4 559 61.65 82.5
r>=0p4 738 64.3 102.2
qr>=0p3
r>=0p3 403 65.5 111.4
qr>=0p2
r>=0p2
B2N vs BN2 Total Games Centered Score Centipawns
total 76553 52.7 18.8
qr>=0p7 46066 52.05 14.3
r>=0p7 6465 55.4 37.7
qr>=0p6 24444 51.95 13.6
r>=0p6 4872 54.75 33.1
qr>=0p5 6157 54.2 29.3
r>=0p5 2407 57.6 53.2
qr>=0p4 1091 55.45 38.0
r>=0p4 927 60.65 75.2
qr>=0p3
r>=0p3
qr>=0p2
r>=0p2
B2 vs BN Total Games Centered Score Centipawns
total 46787 54.1 28.6
qr>=0p7 14073 51.6 11.1
r>=0p7 3474 52.4 16.7
qr>=0p6 16737 53.75 26.1
r>=0p6 6622 54.85 33.8
qr>=0p5 8476 55.7 39.8
r>=0p5 5999 58.4 58.9
qr>=0p4 2813 57.95 55.7
r>=0p4 3954 60.7 75.5
qr>=0p3 603 59.45 66.5
r>=0p3 2128 60.9 77.0
qr>=0p2
r>=0p2 1032 59.4 66.1
BN vs N2 Total Games Centered Score Centipawns
total 20479 52.4 16.7
qr>=0p7 6953 52.9 20.2
r>=0p7 1464 52.2 15.3
qr>=0p6 7571 52.35 16.3
r>=0p6 2630 51.65 11.5
qr>=0p5 3707 53.45 24.0
r>=0p5 2487 52.25 15.6
qr>=0p4 1215 52.6 18.1
r>=0p4 1717 52.8 19.5
qr>=0p3
r>=0p3 844 52.95 20.5
qr>=0p2
r>=0p2
B vs N Total Games Centered Score Centipawns
total 64957 51.45 10.1
qr>=0p7 8201 50.45 3.1
r>=0p7 3384 47.45 -17.7
qr>=0p6 15498 51.15 8.0
r>=0p6 8739 48.85 -8.0
qr>=0p5 13530 52.35 16.3
r>=0p5 12857 50.8 5.6
qr>=0p4 7699 52.35 16.3
r>=0p4 13758 51.75 12.2
qr>=0p3 3266 51.95 13.6
r>=0p3 11546 52.6 18.1
qr>=0p2 1146 51.3 9.0
r>=0p2 8052 52.45 17.0
Good work! It looks like you have used the relationship between score and centipawns given by
S. Fischer & P. Kannan 's study:
Code: Select all
Centipawns_i = 400*log[score_i/(1 - score_i)]
May I give a link about a recently published material imbalance study by Antonio Torrecillas (the author of Rocinante and Simplex IIRC):
Re: OT. revisiting Material Imbalance study.
Antonio Torrecillas wrote:...
I crush the likelihood probability with a logistic function, setting the constant for 100 = having an extra pawn.
Here are some results:
Code: Select all
data: CCRL-4040.[379894].pgn + CCRL-404.[731822].pgn + cegtallblitz.pgn
P1N0B0R0Q0 -> +50.54 -14.80 =34.66 -> 67.87 -> 100
P-3N1B0R0Q0 -> +37.75 -30.43 =31.83 -> 53.66 -> 19
P-3N0B1R0Q0 -> +45.38 -24.52 =30.10 -> 60.43 -> 56
P0N1B0R0Q0 -> +92.43 - 2.52 = 5.04 -> 94.96 -> 392
P0N0B1R0Q0 -> +94.70 - 1.57 = 3.73 -> 96.57 -> 446
P0N-1B1R0Q0 -> +35.37 -27.84 =36.79 -> 53.76 -> 20
P0N-2B2R0Q0 -> +45.16 -24.29 =30.54 -> 60.43 -> 56
P0N0B-1R1Q0 -> +72.06 -12.55 =15.39 -> 79.75 -> 183
P0N-1B0R1Q0 -> +78.40 - 8.91 =12.69 -> 84.75 -> 229
P0N0B0R1Q0 -> +97.81 - 0.93 = 1.26 -> 98.44 -> 554
P0N0B0R-2Q1 -> +29.01 -30.17 =40.82 -> 49.42 -> -3
P0N0B-1R-1Q1 -> +64.03 -11.22 =24.74 -> 76.41 -> 157
P0N-1B0R-1Q1 -> +68.48 -10.09 =21.43 -> 79.20 -> 178
1 => passed_2 -> + 130178 - 84222 = 71593
1 => passed_2 -> +45.52 -29.45 =25.03 -> 43
2 => passed_3 -> + 150534 - 119345 = 99116
2 => passed_3 -> +40.80 -32.34 =26.86 -> 22
3 => passed_4 -> + 207096 - 153461 = 140791
3 => passed_4 -> +41.31 -30.61 =28.08 -> 28
4 => passed_5 -> + 220429 - 111173 = 126192
4 => passed_5 -> +48.15 -24.28 =27.57 -> 65
5 => passed_6 -> + 194629 - 61672 = 85535
5 => passed_6 -> +56.94 -18.04 =25.02 -> 109
6 => passed_7 -> + 111988 - 29737 = 44441
6 => passed_7 -> +60.15 -15.97 =23.87 -> 126
I am not totally sure, but I think that as he mentioned the logistic function, the equivalent from scores to centipawns is something like this (using his data with the reference point at centipawns = 100):
Code: Select all
Centipaws_i = 100*ln[score_i/(1 - score_i)]/ln(0.6787/0.3213)
Given a same score, if I match Antonio's centipawns with yours, I obtain the following:
Code: Select all
(Adam's centipawns)/(Antonio's centipawns) = 4*log(0.6787/0.3213) ~ 1.2991
Given the same value of centipawns:
Code: Select all
(Adam's score) = 1/[1 + 10^(-centipawns/400)
a = (0.6787/0.3213)^(centipawns/100); (Antonio's score) = a/(1 + a)
I hope no typos and/or math blunders.
Regards from Spain.
Ajedrecista.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 1:53 am
by lkaufman
hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 2:34 am
by Adam Hair
Ajedrecista wrote:Hi Adam:
Good work! It looks like you have used the relationship between score and centipawns given by
S. Fischer & P. Kannan 's study:
Code: Select all
Centipawns_i = 400*log[score_i/(1 - score_i)]
Hi Jesús,
Yes. I had previously corroborated their findings with my own:
http://www.talkchess.com/forum/viewtopic.php?t=43323
Ajedrecista wrote:
May I give a link about a recently published material imbalance study by Antonio Torrecillas (the author of Rocinante and Simplex IIRC):
Re: OT. revisiting Material Imbalance study.
Antonio Torrecillas wrote:...
I crush the likelihood probability with a logistic function, setting the constant for 100 = having an extra pawn.
Here are some results:
Code: Select all
data: CCRL-4040.[379894].pgn + CCRL-404.[731822].pgn + cegtallblitz.pgn
P1N0B0R0Q0 -> +50.54 -14.80 =34.66 -> 67.87 -> 100
P-3N1B0R0Q0 -> +37.75 -30.43 =31.83 -> 53.66 -> 19
P-3N0B1R0Q0 -> +45.38 -24.52 =30.10 -> 60.43 -> 56
P0N1B0R0Q0 -> +92.43 - 2.52 = 5.04 -> 94.96 -> 392
P0N0B1R0Q0 -> +94.70 - 1.57 = 3.73 -> 96.57 -> 446
P0N-1B1R0Q0 -> +35.37 -27.84 =36.79 -> 53.76 -> 20
P0N-2B2R0Q0 -> +45.16 -24.29 =30.54 -> 60.43 -> 56
P0N0B-1R1Q0 -> +72.06 -12.55 =15.39 -> 79.75 -> 183
P0N-1B0R1Q0 -> +78.40 - 8.91 =12.69 -> 84.75 -> 229
P0N0B0R1Q0 -> +97.81 - 0.93 = 1.26 -> 98.44 -> 554
P0N0B0R-2Q1 -> +29.01 -30.17 =40.82 -> 49.42 -> -3
P0N0B-1R-1Q1 -> +64.03 -11.22 =24.74 -> 76.41 -> 157
P0N-1B0R-1Q1 -> +68.48 -10.09 =21.43 -> 79.20 -> 178
1 => passed_2 -> + 130178 - 84222 = 71593
1 => passed_2 -> +45.52 -29.45 =25.03 -> 43
2 => passed_3 -> + 150534 - 119345 = 99116
2 => passed_3 -> +40.80 -32.34 =26.86 -> 22
3 => passed_4 -> + 207096 - 153461 = 140791
3 => passed_4 -> +41.31 -30.61 =28.08 -> 28
4 => passed_5 -> + 220429 - 111173 = 126192
4 => passed_5 -> +48.15 -24.28 =27.57 -> 65
5 => passed_6 -> + 194629 - 61672 = 85535
5 => passed_6 -> +56.94 -18.04 =25.02 -> 109
6 => passed_7 -> + 111988 - 29737 = 44441
6 => passed_7 -> +60.15 -15.97 =23.87 -> 126
I am not totally sure, but I think that as he mentioned the logistic function, the equivalent from scores to centipawns is something like this (using his data with the reference point at centipawns = 100):
Code: Select all
Centipaws_i = 100*ln[score_i/(1 - score_i)]/ln(0.6787/0.3213)
I somehow missed Antonio's post. Thanks for pointing it out.
Ajedrecista wrote:
Given a same score, if I match Antonio's centipawns with yours, I obtain the following:
Code: Select all
(Adam's centipawns)/(Antonio's centipawns) = 4*log(0.6787/0.3213) ~ 1.2991
Given the same value of centipawns:
Code: Select all
(Adam's score) = 1/[1 + 10^(-centipawns/400)
a = (0.6787/0.3213)^(centipawns/100); (Antonio's score) = a/(1 + a)
I hope no typos and/or math blunders.
Regards from Spain.
Ajedrecista.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 2:41 am
by Adam Hair
lkaufman wrote:hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
I found that it is dependent on the number of pawns. The difference fades away as the number of pawns on the board decreases.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 4:58 am
by lkaufman
Adam Hair wrote:lkaufman wrote:hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
I found that it is dependent on the number of pawns. The difference fades away as the number of pawns on the board decreases.
Are you saying that BN vs NN is always better than B vs N, but by a decreasing amount as the pawns come off, or does it "cross over" at some point beyond which BN vs NN is inferior to B vs N?
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 5:11 am
by Adam Hair
lkaufman wrote:Adam Hair wrote:lkaufman wrote:hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
I found that it is dependent on the number of pawns. The difference fades away as the number of pawns on the board decreases.
Are you saying that BN vs NN is always better than B vs N, but by a decreasing amount as the pawns come off, or does it "cross over" at some point beyond which BN vs NN is inferior to B vs N?
From my data, it appears to me that there is a point where there is no measurable difference between the two (possibly when there is only 2 to 3 pawns for each side):
Code: Select all
qr>=0p7 means 1 queen, 0 to 2 rooks, 7 pawns
BN vs N2 Total Games Centered Score Centipawns
total 20479 52.4 16.7
qr>=0p7 6953 52.9 20.2
r>=0p7 1464 52.2 15.3
qr>=0p6 7571 52.35 16.3
r>=0p6 2630 51.65 11.5
qr>=0p5 3707 53.45 24.0
r>=0p5 2487 52.25 15.6
qr>=0p4 1215 52.6 18.1
r>=0p4 1717 52.8 19.5
qr>=0p3
r>=0p3 844 52.95 20.5
qr>=0p2
r>=0p2
B vs N Total Games Centered Score Centipawns
total 64957 51.45 10.1
qr>=0p7 8201 50.45 3.1
r>=0p7 3384 47.45 -17.7
qr>=0p6 15498 51.15 8.0
r>=0p6 8739 48.85 -8.0
qr>=0p5 13530 52.35 16.3
r>=0p5 12857 50.8 5.6
qr>=0p4 7699 52.35 16.3
r>=0p4 13758 51.75 12.2
qr>=0p3 3266 51.95 13.6
r>=0p3 11546 52.6 18.1
qr>=0p2 1146 51.3 9.0
r>=0p2 8052 52.45 17.0
I do not believe that BN vs NN becomes worse than B vs N.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 5:40 am
by lkaufman
Adam Hair wrote:lkaufman wrote:Adam Hair wrote:lkaufman wrote:hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
I found that it is dependent on the number of pawns. The difference fades away as the number of pawns on the board decreases.
Are you saying that BN vs NN is always better than B vs N, but by a decreasing amount as the pawns come off, or does it "cross over" at some point beyond which BN vs NN is inferior to B vs N?
From my data, it appears to me that there is a point where there is no measurable difference between the two (possibly when there is only 2 to 3 pawns for each side):
Code: Select all
qr>=0p7 means 1 queen, 0 to 2 rooks, 7 pawns
BN vs N2 Total Games Centered Score Centipawns
total 20479 52.4 16.7
qr>=0p7 6953 52.9 20.2
r>=0p7 1464 52.2 15.3
qr>=0p6 7571 52.35 16.3
r>=0p6 2630 51.65 11.5
qr>=0p5 3707 53.45 24.0
r>=0p5 2487 52.25 15.6
qr>=0p4 1215 52.6 18.1
r>=0p4 1717 52.8 19.5
qr>=0p3
r>=0p3 844 52.95 20.5
qr>=0p2
r>=0p2
B vs N Total Games Centered Score Centipawns
total 64957 51.45 10.1
qr>=0p7 8201 50.45 3.1
r>=0p7 3384 47.45 -17.7
qr>=0p6 15498 51.15 8.0
r>=0p6 8739 48.85 -8.0
qr>=0p5 13530 52.35 16.3
r>=0p5 12857 50.8 5.6
qr>=0p4 7699 52.35 16.3
r>=0p4 13758 51.75 12.2
qr>=0p3 3266 51.95 13.6
r>=0p3 11546 52.6 18.1
qr>=0p2 1146 51.3 9.0
r>=0p2 8052 52.45 17.0
I do not believe that BN vs NN becomes worse than B vs N.
Thanks. I think where you say "centipawns" you mean "elo difference". It appears to me from your data that the effect is much more noticeable with queens off than with queens on, which makes good sense. I think we'll have to try this term again, perhaps modified as indicated by your data.
Could you print the similar comparison table for BBN vs BNN compared to BB vs BN? It's not at all obvious which is better, and I recall that the answer also depends significantly on the number of pawns present. Thanks.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 2:39 pm
by Adam Hair
lkaufman wrote:Adam Hair wrote:lkaufman wrote:Adam Hair wrote:lkaufman wrote:hgm wrote:This raises some questions about the B-N difference (for Larry!). This has been shown to corrlate with the number of Pawns present. But, as we all know, correlation does not necessarily mean a causal relationship. The number of Pawns correlates also quite well with total material. If we suppose the bare value of the (lone) Bishop is higher than that of Knight, the presence of opponent Knight(s) pulls its value down to Knight level, as it is doomed to be traded. But the later the game stage, the more likely it is that the number of Knights it faces is reduced, making it suffer less danger it will have to be traded for one. So that it is more likely to rise to its bare value.
So my question is: does the value of a B-N imbalance correlate in any way with the presence of the second Knight?
The general belief among chess masters is that BN vs NN is better than B vs N, since the bishop "needs" a knight more than a knight does. I was not able to substantiate that to any meaningful degree in my initial study. In Rybka we did make that distinction based on weak evidence. We're not making it in Komodo now but perhaps I'll revisit the evidence on this point. It's definitely not a major effect.
I found that it is dependent on the number of pawns. The difference fades away as the number of pawns on the board decreases.
Are you saying that BN vs NN is always better than B vs N, but by a decreasing amount as the pawns come off, or does it "cross over" at some point beyond which BN vs NN is inferior to B vs N?
From my data, it appears to me that there is a point where there is no measurable difference between the two (possibly when there is only 2 to 3 pawns for each side):
Code: Select all
qr>=0p7 means 1 queen, 0 to 2 rooks, 7 pawns
BN vs N2 Total Games Centered Score Centipawns
total 20479 52.4 16.7
qr>=0p7 6953 52.9 20.2
r>=0p7 1464 52.2 15.3
qr>=0p6 7571 52.35 16.3
r>=0p6 2630 51.65 11.5
qr>=0p5 3707 53.45 24.0
r>=0p5 2487 52.25 15.6
qr>=0p4 1215 52.6 18.1
r>=0p4 1717 52.8 19.5
qr>=0p3
r>=0p3 844 52.95 20.5
qr>=0p2
r>=0p2
B vs N Total Games Centered Score Centipawns
total 64957 51.45 10.1
qr>=0p7 8201 50.45 3.1
r>=0p7 3384 47.45 -17.7
qr>=0p6 15498 51.15 8.0
r>=0p6 8739 48.85 -8.0
qr>=0p5 13530 52.35 16.3
r>=0p5 12857 50.8 5.6
qr>=0p4 7699 52.35 16.3
r>=0p4 13758 51.75 12.2
qr>=0p3 3266 51.95 13.6
r>=0p3 11546 52.6 18.1
qr>=0p2 1146 51.3 9.0
r>=0p2 8052 52.45 17.0
I do not believe that BN vs NN becomes worse than B vs N.
Thanks. I think where you say "centipawns" you mean "elo difference". It appears to me from your data that the effect is much more noticeable with queens off than with queens on, which makes good sense. I think we'll have to try this term again, perhaps modified as indicated by your data.
Could you print the similar comparison table for BBN vs BNN compared to BB vs BN? It's not at all obvious which is better, and I recall that the answer also depends significantly on the number of pawns present. Thanks.
Well, I did mean centipawns due to the approximate relation 1 Elo = 1 centipawn. But Elo difference is correct.
All of the bishop data is located a little earlier in the thread.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 5:44 pm
by lkaufman
Adam Hair wrote:
Well, I did mean centipawns due to the approximate relation 1 Elo = 1 centipawn. But Elo difference is correct.
All of the bishop data is located a little earlier in the thread.
OK, I found it. It seems from your data that having both knights is bad when no one has the bishop pair, but good when one player does have it. That's probably why my initial study showed no significant plus or minus to having the knight pair. Maybe the underlying principle is that an unpaired bishop needs knights to cover the other color, and two is better than one, while with the bishop pair there is less need for knights.
Re: Why Knight and (lone) Bishop are so nearly equal in valu
Posted: Fri Sep 28, 2012 7:56 pm
by Antonio Torrecillas
Here I leave my data and conclusions:
Code: Select all
200 => KnightBishop -> + 25022 - 28337 = 36599
200 => KnightBishop -> +27.82 -31.50 =40.68 -> 48.16 -> -10
Two pawns each side.
201 => KnightBishop2PP -> + 540 - 930 = 5134
201 => KnightBishop2PP -> + 8.18 -14.08 =77.74 -> 47.05 -> -17
Three pawns each side.
202 => KnightBishop3PP -> + 1506 - 2310 = 7155
202 => KnightBishop3PP -> +13.73 -21.06 =65.22 -> 46.34 -> -21
203 => KnightBishop4PP -> + 3512 - 4297 = 8937
203 => KnightBishop4PP -> +20.97 -25.66 =53.37 -> 47.66 -> -13
204 => KnightBishop5PP -> + 5517 - 6062 = 10087
204 => KnightBishop5PP -> +25.46 -27.98 =46.56 -> 48.74 -> -7
205 => KnightBishop6PP -> + 6217 - 6075 = 9068
205 => KnightBishop6PP -> +29.11 -28.44 =42.45 -> 50.33 -> 1
206 => NNvsNB -> + 3963 - 4700 = 6127
206 => NNvsNB -> +26.80 -31.78 =41.43 -> 47.51 -> -14
207 => NNvsBB -> + 1511 - 3037 = 2378
207 => NNvsBB -> +21.82 -43.85 =34.33 -> 38.98 -> -65
208 => NBvsBB -> + 8393 - 12743 = 14187
208 => NBvsBB -> +23.76 -36.08 =40.16 -> 43.84 -> -36
209 => NQvsBQ -> + 790 - 763 = 2642
209 => NQvsBQ -> +18.83 -18.19 =62.98 -> 50.32 -> 1
210 => NRvsBR -> + 3391 - 3913 = 10339
210 => NRvsBR -> +19.22 -22.18 =58.60 -> 48.52 -> -8
211 => NRQvsBRQ -> + 2900 - 3328 = 5095
211 => NRQvsBRQ -> +25.61 -29.39 =45.00 -> 48.11 -> -11
212 => NRRvsBRR -> + 2603 - 2818 = 5094
212 => NRRvsBRR -> +24.76 -26.80 =48.45 -> 48.98 -> -6
213 => NRRQvsBRRQ -> + 4880 - 5510 = 6132
213 => NRRQvsBRRQ -> +29.54 -33.35 =37.11 -> 48.09 -> -11
Given the number of items involved the error should be around 8 cp if not more.
Being the values so similar to the error, I would not take too far into consideration.
The difference (-10) probably comes from the mobility and is not affected appreciably by the phase.(removing queens 2 or 4 cp)
The only interesting configurations are queen and knight vs queen and bishop,
and bishop against knight when each side has 6 or more pawns.Here the Knight has enough compensation to match the slight advantage of the bishop.
Interestingly, the bishop pair loses half of its strength by the presence of a bishop on the side of the knight.