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SMITH and JONES: The Undoubled Game
Posted By: Nack Ballard In Response To: SMITH and JONES: The Undoubled Game (Timothy Chow)
Date: Wednesday, 31 May 2017, at 2:44 a.m.
Oops, Jim Stutz is right! Clearly from your comments you understand the same thing: that the modified opening position cannot be used for the next game (if there is one), or it disqualifies the use of the standard MET in this game.
You gave me too much credit for assuming that the starting position applied only to the –2–2 score/game. It's certainly the way I should have written the story. What actually happened is that somewhere along the way I forgot that any 24–23 (Crawford) game would also be played from the modified starting position (the way I told it), which of course nullifies the standard MET.
If I reprint the story, I will change it so that it is only this game that is played with the modified starting position (the charitable assumption you felt forced to make, otherwise Nack is a dolt), or in any case make it clear that the next game (if there is one) is to be played from the standard opening position. (There are many ways to stage this that can sound reasonable.)
The second half of my post (marked "Epilogue"), illustrating the undoubled game with only one error of .00005, appears to still be sound. I don't think I need to change anything there.
But let's get back to the discussion of the modified opening position (now assuming that the next game, if there is one, will be played from the standard opening position). You said:
If we assume this (and therefore a standard MET), then Bob Koca's proof seems to apply to your starting position. That is, when it's your first opportunity to double, you will not have already lost your market. The strategy "double at first opportunity, always take if doubled, and play checkers as if it were DMP" will achieve at least 50% winning probability against any player that always accepts your double. Since you haven't lost your market when you offer your double, the opponent can't hold you to less than 50% by dropping the double, so this strategy must be optimal.
We seem to disagree on this aspect.
From the modified position, Blue opened with 65D. Suppose White doubles and does NOT roll 21 or 66. Any other roll will lead to the same result (as White doubling now) because Blue will double, White will take, and the game will be played to conclusion at the 2-level. On that, I think we agree.
But what if White doubles and does roll 21? She regrets having doubled, because now the cube is on 2 in a position that otherwise would have gone Blue-double, White-pass. She has squandered the chance to pass a 1.0223 position. It costs her equity.
Put another way, if White blindly adopts the double-automatically strategy, she is NOT going to be 50% against someone like me, because if in a parallel universe she opens with 65D, I will keep the cube centered for a roll and with my 21 I'll get doubled and pass and come out slightly ahead between the two universes.
White can also gain something by having doubled, due to some 66-xx market-losing sequences. But let's ignore the offsetting gains for now. I just want to keep it simple and make sure you see that White can lose something by doubling. (If not, perhaps you are making some other point and I would ask you to please rephrase your comment.)
Nack
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