by Phil Simborg
29 July 2008
Here's a good example of how complex this game can be.
There are two positions below where Black holds a 2-cube and is on roll.
See if you can guess the proper cube action for both positions.
Scroll down and see what the answers are:
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LOL a georgian friend of mine told me if u put the checker on the 6 point u should cube since u have a take if u do not roll 52 or better.....
Assuming this is a money game and white can redouble to 8. I disagree with the analysis as black fails to take 2 off 44.5% of the time and then white will redouble to 8 in either position. So if black redoubles they lose some of there winning chances by allowing white to redouble to 8 and not risk rolling bad numbers.
I agree with the analysis. In both cases the equity after doubling is 8/36 since white cashes anyway after Black misses. On the other hand, the equity after not doubling is clearly worse in the second case, since White has no cube access and has to play the game to the end. I have both equities as 13.44/36 and 7.77/36, which makes position 2 a very thin double (I could be wrong with the numbers). Nice example, but the math is more convincing than putting the position into a bot.
This is an example of the Jacoby Paradox. The "paradox" comes from the fact that a decrease in black's cubeless winning chances turns the position from no-redouble into a redouble.
This principle repeats frequently in more complex positions. Sometimes you'll come across positions during bot analysis where the correct play is the one that wins fewer games and fewer gammons and gammon losses are about the same, and you'll be trying to figure out how the heck that can be so. The answer is that the seemingly inferior position results in an efficient double the following turn (i.e. take equal to cash) in exchange for no risk, while the play that wins more does so by trading big market losses when things go well in exchange for small immediate risk of loss. It helps to think of the object of the game of backgammon as not being the first to take your checkers off, but to be able to give an efficient double. An efficient double is the same as a win.
This is old material (the fact that it has a paradox named for a player who has been dead twenty-five years hits at how old). Nothing wrong with old material; we all need a reminder, and the new generation needs a heads up. But it points to a limitation of relying on the bots: they don't explain why something is right or wrong. The math is easy, and should have been included. A ten-cent #2 pencil will provide a more lucid explanation than a four-hundred dollar software program.
Can somebody explain the math, as if you're explaining it to a third grader please? Much appreciated.
This is indeed very old material and I agree it would have been better to present the maths (not difficult). There are five Jacoby Paradoxes where it is correct to double but not redouble. In each case the player on roll has checkers on 5pt and 2pt. The opponent has a checker on [5pt] or [4pt, 1pt] resulting in a recube to 8 which has to be dropped; or [6pt] with an optional take or drop of recube to 8; or [2pt, 2pt] or [2pt, 3pt] with a recube to 8 which should be taken.
Rich: can you give an example of the more complex positions to which you refer ? Thanks.
It is quite normal for Kamyar to ask someone else to do the work for him. Before, I might have explained it to him, but now his free lessons are over.
In response to Jake's comments, what's a "No. 2 pencil?" Many of us are not that old.
As for this being old material, the fact is that I discovered this myself, over the board. (A few weeks ago I also came up with a great idea for putting wheels on suitcases.)
As for the math, I agree with Coolray...if you do it yourself instead of having someone else do it for you, you will understand it better and it will be more meaningful.
But there is another answer that is less mathmetical...if you don't double one position you will win more because of the extra misses your opponent has, but he doesn't have those extra misses if he can cube you out.
Thank you for the non-math explanation Phil.
Its normal for Ray to be needlessly insulting, so I forgive him; he can't help it.
If I knew how to do the math I would. Last I checked, online forums were for sharing ideas and learning from eachother, no?
I teach people what I know on a daily basis, without asking for anything in return. If everyone did the same, the game would flourish.
In position 1, if black does not redouble, (a) black bears off 19 times out of 36 and wins 2; (b) black does not bear off 17 times out of 36 but then white misses 5 times out of 36 and black wins 2 (black doubles white out if black's previous roll was 21); (c) black does not bear off 17 times out of 36 but white does bear off 31 times out of 36 and black loses 2. On average, black wins [(19/36) x 2] + [(17/36) x (5/36) x 2] - [(17/36) x (31/36) x 2] = 0.373 points per game.
In position 1, if black does redouble, (a) black bears off 19 times out of 36 and wins 4; (b) black does not bear off 17 times out of 36 but then white doubles black out and black loses 4. On average, black wins [(19/36) x 4] - [(17/36) x 4] = 0.222 points per game.
In 1296 games, by redoubling black wins 4 instead of 2 points 684 times (19 x 36), loses 4 instead of 2 points 527 times (17 x 31) and loses 4 instead of winning 2 points 85 times (17 x 5).
Thank you for the math! Much more clear now.
P.S. My above comment was directed at Coolrey, not you, but I'm sure you figured that out. =)
Actually, putting it in any good program reveals that not recognizing the first position as a no redouble is a real mistake (about 0.07 mistake),
while not recognizing the second position as a redouble can hardly be categorized as a mistake, since it's only 0.003 off from no redouble to redouble/take.
Not understanding the second position will have little statistical significance on your play.
It is a very nice illustration of a very nice idea though.
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