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| Tags: been, checkers, has, solved |
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#11
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On Mar 17, 7:08*am, Sanny wrote:
* Am I missing something in my intuitive argument or is one of the calculations incorrect? The Trice argument looks solid to me. Are you the one who used to play with username "Bob" at GetClub? I was searching for Bob for long time. You left playing long before. Play a few games at GetClub and see how well it plays now. Play Chess at:http://www.GetClub.com/Chess.html Help Bot used to say that you use Computer's help while playing against GetClub is that True? Your games are remarkable. Only Zebediah matches your game style. Bye Sanny Play Chess at:http://www.GetClub.com/Chess.html A different Bob than me. Bob Koca |
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#12
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On Mar 17, 8:46*pm, bob wrote:
On Mar 17, 7:08*am, Sanny wrote: * Am I missing something in my intuitive argument or is one of the calculations incorrect? The Trice argument looks solid to me. Are you the one who used to play with username "Bob" at GetClub? I was searching for Bob for long time. You left playing long before. Play a few games at GetClub and see how well it plays now. Play Chess at:http://www.GetClub.com/Chess.html Help Bot used to say that you use Computer's help while playing against GetClub is that True? Your games are remarkable. Only Zebediah matches your game style. Bye Sanny Play Chess at:http://www.GetClub.com/Chess.html * *A different Bob than me. Bob Koca- Hide quoted text - - Show quoted text - Huge coincidence since Bob is an extremely uncommon name. Paul Epstein |
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#13
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My recollection is slightly different. I remember looking at results
from the highest level of checkers play and a player would win a match by 1 game to 0 with 20 draws. However, that's an impression from memory only -- I haven't been able to check it. Can you back up your claim with hard stats? If the world no. 1 plays the world no. 2, would they draw less than 95% of their games? (I doubt it.) I think it's a pretty dead game at the highest level. I could get exact stats from the American Checker Federation. I am pretty sure though that, for instance world championship matches, are something above 80% draws, if maybe not 95%, even with 3-move restriction. Whether that means it's a "dead" game is a matter of definition. Certainly at the NON-championship level, the percentage of draws is very much lower (you can verify this by looking at on-line play sites). High-level tournaments, when they achieve under 75% draws, are considered "lively." But my point was that, while computer analysis shows an absolute draw for unrestricted play and very likely will show the same for 3-move play, humans still win and lose. As you point out, it is at a relatively low percentage, but still, the game continues to be played. |
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#14
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samsloan writes:
One factor to be considered is that the number of possible moves in a backgammon games is infinite. The players could easily just keeping hitting each other to infinity. That doesn't matter, as long as the number of possible board positions is finite (which it is). The main difference between Backgammon and, say, Checkers is not the possibility of infinite play but the fact that Backgammon involves random elements, so few positions are definitely winning or definitely losing -- all you can say is the probability of winning with perfect play (i.e., always picking the move that gives you the best winning probability after moving). You can solve Backgammon by for each possible position have edges to every other position that it is possible to get to in one move, and label each edge with the dice outcome that allow this move). This can be translated into a set of equations that you can solve to find the probability of each possible position being winning or losing. The set of equations is huge, but finite. Torben |
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#15
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On Mar 16, 8:39*pm, " wrote:
First, what Schaeffer did is to show that freestyle (unrestricted) checkers is an absolute draw. He has not analyzed all of the 3-move restriction openings so he has not proven that tournament checkers is a draw. *He has proven that some of the 3-movers are a draw, and will likely eventually show that all 156 of the accepted tournament choices are a draw (the other few are almost certain losses and are not used). *Of course, there could be a deeply-buried surprise in one or more of the 156, but with the amount of other computer analysis done to date, it's not very likely --- but it hasn't been categorically proven yet. I wouldn't be all that surprised if instead he finds that some of the 156 accepted three-move tournament choices are not draws. After all, the ones that were rejected were obvious losses, so openings that provided one player sufficient advantage to narrowly force a win with perfect play might not have been noted. I realize that this is a fairly large advantage, though, and that might mean it would have been suspected, but then the fact that freestyle checkers was a draw wasn't known for certain until it was proved. John Savard |
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#16
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On Mar 25, 10:18 am, (Torben Ęgidius Mogensen)
wrote: samsloan writes: One factor to be considered is that the number of possible moves in a backgammon games is infinite. The players could easily just keeping hitting each other to infinity. That doesn't matter, as long as the number of possible board positions is finite (which it is). The main difference between Backgammon and, say, Checkers is not the possibility of infinite play but the fact that Backgammon involves random elements, so few positions are definitely winning or definitely losing -- all you can say is the probability of winning with perfect play (i.e., always picking the move that gives you the best winning probability after moving). You can solve Backgammon by for each possible position have edges to every other position that it is possible to get to in one move, and label each edge with the dice outcome that allow this move). This can be translated into a set of equations that you can solve to find the probability of each possible position being winning or losing. The set of equations is huge, but finite. Torben Even that is not obvious. There are 21 possible rolls of the dice (6! = 21) and three possible positions of the doubling cube plus 24 possible slots for each checker. The average chess position has 27 moves and most chess games are over in 50 moves. I have written a chess playing computer program and a shogi playing computer program. However, I once tried to write a backgammon playing computer program and I quickly gave it up as hopeless. It is much harder than it looks. Although backgammon seems to be an easier game than chess, I am not sure that this is really true. Sam Sloan |
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#17
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samsloan wrote:
Even that is not obvious. There are 21 possible rolls of the dice (6! = 21) and three possible positions of the doubling cube plus 24 possible slots for each checker. BZZZT! Are the checkers in your backgammon set marked in some way? -- Kenneth Sloan Computer and Information Sciences +1-205-932-2213 University of Alabama at Birmingham FAX +1-205-934-5473 Birmingham, AL 35294-1170 http://KennethRSloan.com/ |
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#18
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On Mar 25, 11:18*am, (Torben Ęgidius Mogensen)
wrote: samsloan writes: One factor to be considered is that the number of possible moves in a backgammon games is infinite. The players could easily just keeping hitting each other to infinity. That doesn't matter, as long as the number of possible board positions is finite (which it is). The main difference between Backgammon and, say, Checkers is not the possibility of infinite play but the fact that Backgammon involves random elements, so few positions are definitely winning or definitely losing -- all you can say is the probability of winning with perfect play (i.e., always picking the move that gives you the best winning probability after moving). You can solve Backgammon by for each possible position have edges to every other position that it is possible to get to in one move, and label each edge with the dice outcome that allow this move). This can be translated into a set of equations that you can solve to find the probability of each possible position being winning or losing. *The set of equations is huge, but finite. * * * * Torben Agree for backgammon played without a cube, for backgammon played with a cap on the cube, or for match play. If you are talking about money backgammon though then the cube position and value makes the number of positions infinite. Now one might say that its position is all that matters since if you know the correct theoretical play holding a 2 cube then you also know the correct theoretical play holding a 4 or any higher value cube. There is a problem though in that the equations might not have a solution. As a simple example of how this might come about suppose we are betting on the flip of a coin. The first person to toss a head wins. Before each flip a doubling cube may be used as in backgammon. Solving the equations gives that every turn is a double and take but this leads to undefined equities. How to prove there is no situation like that possible in backgammon? Bob Koca |
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#19
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On Mar 26, 1:45*am, bob wrote:
On Mar 25, 11:18*am, (Torben Ęgidius Mogensen) wrote: samsloan writes: One factor to be considered is that the number of possible moves in a backgammon games is infinite. The players could easily just keeping hitting each other to infinity. That doesn't matter, as long as the number of possible board positions is finite (which it is). The main difference between Backgammon and, say, Checkers is not the possibility of infinite play but the fact that Backgammon involves random elements, so few positions are definitely winning or definitely losing -- all you can say is the probability of winning with perfect play (i.e., always picking the move that gives you the best winning probability after moving). You can solve Backgammon by for each possible position have edges to every other position that it is possible to get to in one move, and label each edge with the dice outcome that allow this move). This can be translated into a set of equations that you can solve to find the probability of each possible position being winning or losing. *The set of equations is huge, but finite. * * * * Torben * Agree for backgammon played without a cube, for backgammon played with a cap on the cube, or for match play. If you are talking about money backgammon though then the cube position and value makes the number of positions infinite. Now one might say that its position is all that matters since if you know the correct theoretical play holding a 2 cube then you also know the correct theoretical play holding a 4 or any higher value cube. There is a problem though in that the equations might not have a solution. As a simple example of how this might come about suppose we are betting on the flip of a coin. The first person to toss a head wins. Before each flip a doubling cube may be used as in backgammon. Solving the equations gives that every turn is a double and take but this leads to undefined equities. How to prove there is no situation like that possible in backgammon? Bob Koca- Hide quoted text - - Show quoted text - Interesting example but I'm surprised by "How to prove there is no situation like that possible in backgammon?" Almost surely, there _are_ situations like that in backgammon. Can't you mimic the above scenario in backgammon by postulating 18 monster rolls for each side? But, suppose there are such situations in backgammon, why does it then follow that the value of a cube, as well as its position, can affect theoretical money play? This seems to be a hole in your argument. So let us assume that your scenario is exactly replicated in backgammon where there's a stalemate but each side has 18 monster winning rolls, and no gammons are possible. Please explain why this scenario leads to the conclusion that the scenario with a 2 cube is essentially different than the scenario with a 1024 cube. Paul Epstein |
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#20
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samsloan wrote: Even that is not obvious. There are 21 possible rolls of the dice (6! = 21) and three possible positions of the doubling cube plus 24 possible slots for each checker. I haven't been following this thread so maybe I've missed something... BUT from what I read above: What do you mean by "possible rolls of the dice"? Combinations or Permutations? There are 21 of the former but 36 of the latter. Also 6! (6 factorial) is 720 not 21. Now, where did I go wrong??? LRB |
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