With everything else normal, if White had 4 knights and Black had 4 bishops, which side would have better chances of winning?
In the following set-up, which side do you think has better chances?
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Re: In the following set-up, which side do you think has better chances?
I couldn’t understand the question at first. I thought it was perhaps an endgame situation. Then I found that the experiment has been discussed before and it is in the initial setup of a game with white having knights replacing his bishops and black with bishops replacing his knights.
See:
http://www.chess.com/forum/view/gene...hts-a-new-test
The consensus is that the bishops will win every time.
Something to think about when one is shut in on these cold winter days.
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Re: In the following set-up, which side do you think has better chances?
Originally posted by Wayne Komer View PostI couldn’t understand the question at first. I thought it was perhaps an endgame situation. Then I found that the experiment has been discussed before and it is in the initial setup of a game with white having knights replacing his bishops and black with bishops replacing his knights.
See:
http://www.chess.com/forum/view/gene...hts-a-new-test
The consensus is that the bishops will win every time.
Something to think about when one is shut in on these cold winter days.
Trying out a few opening like positions with the knights having the move on Rybka suggests equality, other things being equal. (I say opening like positions because my software will not allow 4 knights without 2 pawns being gone - presumably to account for the under promotions to the extra knights and bishops.)
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Re: In the following set-up, which side do you think has better chances?
I put the position on the ICC interface and ran Stockfish on it, where Black has the 4 Bs. After about five minutes it evaluated it at depth 24 as 0.60 pawn better for Black. The general trend appeared to be that the more it looked the greater Black's advantage. So if you ran it all day maybe you would get 0.75 or so I would guesstimate. A pleasant but not winning advantage."Tom is a well known racist, and like most of them he won't admit it, possibly even to himself." - Ed Seedhouse, October 4, 2020.
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Re: In the following set-up, which side do you think has better chances?
Originally posted by Tom O'Donnell View PostI put the position on the ICC interface and ran Stockfish on it, where Black has the 4 Bs. After about five minutes it evaluated it at depth 24 as 0.60 pawn better for Black. The general trend appeared to be that the more it looked the greater Black's advantage. So if you ran it all day maybe you would get 0.75 or so I would guesstimate. A pleasant but not winning advantage.
This is similar to Q + N vs. Q + B, where the Knight is favoured by many textbooks and local masters I know because the B duplicates a function already possessed by the Queen, whereas the Knight does not.
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Re: In the following set-up, which side do you think has better chances?
I think the best strategy would be to trade off one B of each colour for the opponent's two best knights and try to win 2B v 2N positions. One of the reasons that Bs are more highly coveted on average is that it is generally easier to trade them for Ns than vice-versa.
As an aside, I input the same position into Stockfish except left squares c1, f1, b8 and g8 vacant (so normal starting position but Black having 2Bs vs White's 2Ns). The computer gave Black a tiny .16 advantage after 25 ply. Inserting another B on b8 and a N on c1 (so now Black has two dark-squared Bs but only one light-squared one) actually increased Black's advantage to .36 at 25 ply. The computer even preferred situations where Black has two dark-squared Bs on b8 and f8 vs Ns on b1 and f1 with no other minor pieces on the board again by .16. This last fact I found rather surprising but maybe the idea is to trade one of the Bs for a N and try to win various B vs N positions.
It seems that at least to a computer the more Bs the better."Tom is a well known racist, and like most of them he won't admit it, possibly even to himself." - Ed Seedhouse, October 4, 2020.
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Re: In the following set-up, which side do you think has better chances?
Originally posted by Tom O'Donnell View PostI think the best strategy would be to trade off one B of each colour for the opponent's two best knights and try to win 2B v 2N positions. One of the reasons that Bs are more highly coveted on average is that it is generally easier to trade them for Ns than vice-versa.
As an aside, I input the same position into Stockfish except left squares c1, f1, b8 and g8 vacant (so normal starting position but Black having 2Bs vs White's 2Ns). The computer gave Black a tiny .16 advantage after 25 ply. Inserting another B on b8 and a N on c1 (so now Black has two dark-squared Bs but only one light-squared one) actually increased Black's advantage to .36 at 25 ply. The computer even preferred situations where Black has two dark-squared Bs on b8 and f8 vs Ns on b1 and f1 with no other minor pieces on the board again by .16. This last fact I found rather surprising but maybe the idea is to trade one of the Bs for a N and try to win various B vs N positions.
It seems that at least to a computer the more Bs the better.
Certainly, the computer analysis of tactical sequences will still be much stronger than you or I but it is quite possible that new long term positional considerations for these piece configurations is completely missing from the program's evaluation.
I don't trust hand waving explanations such as Ed's. How we currently justify the advantage of the B's in regular chess (more mobility, long range action, no superfluous Bs) is after the fact rationalization of what we observe over a large number of games. Kaufman (e.g. at http://home.comcast.net/~danheisman/..._imbalance.htm ) notes that a B is worth roughly twice as much as N in Shogi but in Chinese chess, a N is worth twice as much as a B so other rules and the board configuration have a large impact on the relative value of the pieces. Kaufman even argues that there is no advantage of the B over N per se, only if you have two Bs is there an advantage.
Really, the only way to assess this is to establish a large database of games - perhaps let the programs play automatically in shoot out mode and see what the distribution of results is.
That link above to Kaufman's article is interesting reading by the way, at least if you like data analysis as an approach to figuring these things out.
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Re: In the following set-up, which side do you think has better chances?
I don't put much weight in computer evaluations from such an early stage in the game. The horizon effect is a big factor which should not be underestimated.
A good example of the horizon effect in preparing openings is the Kramnik - Leko Spanish Marshal championship game from 2004.Gary Ruben
CC - IA and SIM
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Re: In the following set-up, which side do you think has better chances?
If the proposed start position is the same as for standard chess's, except White has 4Ns (and the first move) and Black has 4Bs then how to evaluate the proposed start position is not in any way clear to me.
In normal chess I used to like playing the Nimzo-Indian line 1.d4 Nf6 2.c4 e6 3.Nc3 Bb4 4.e3 b6 5.Ne2 Ne4 6.Qc2 Bb7 7.a3 Bxc3+ 8.Nxc3 Nxc3 9.Qxc3 with Black as often as possible. Timmin called the whole line starting with 5...Ne4 "controversial" in his book Art of chess analysis.
ECO evaluates the position after move 9 as ultimately slightly better for White, based on variations given, but I've seen other evaluations more friendly to Black given elsewhere. White has the B pair, but I think what allows room for argument is that all 16 pawns are still on the board, and have not come into contact, and thus the nature of the pawn structure (open? closed? any outpost(s) for the lone N?) has not been determined yet. That's what the coming middlegame battle often will about, unless opposite-sides castling happens (then a brawl might result).
I think a similar controversy might occur if Black had a N pair instead of B+N but all 16 pawns still were on the board. Thus I don't think, in a battle of 4Ns vs. 4Bs, the Bs will automatically be better if one B of each colour is traded for a N, provided the battle over establishing a favourable pawn structure for the Ns or Bs respectively remains to be fought.Anything that can go wrong will go wrong.
Murphy's law, by Edward A. Murphy Jr., USAF, Aerospace Engineer
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