Why Knight and (lone) Bishop are so nearly equal in value

[Ajedrecista]

Hi Adam:

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.

...

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.

[lkaufman]

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.

[Adam Hair]

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

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.

...

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.

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.

[Adam Hair]

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.

[lkaufman]

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?

[Adam Hair]

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.

[lkaufman]

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.

[Adam Hair]

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.

[lkaufman]

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.

[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.