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SMITH and JONES: The Undoubled Game
Posted By: Nack Ballard In Response To: SMITH and JONES: The Undoubled Game (Bob Koca)
Date: Wednesday, 31 May 2017, at 4:26 p.m.
"The very FIRST time a market gainer appears (which must happen eventually, if the game is to ever break contact), it automatically means that the other player had a market-losing sequence on the roll prior (without an opposing market gainer, by definition), and therefore had a double"
Does "had a double" equal "had a mandatory double"? Assuming that is what is meant I do not believe a proof of what is being said in that paragraph has been given. In particular suppose the game changes in the following way: The first several rolls do not change the game from 50-50 and then there is a roll (let's say on A's turn) which decides the game. A has a market losing roll but the double is optional with gains and regrets exactly balancing. Two turns before that A did not a have a market losing sequence (two rolls causing no change in win chance are coming) The turn before B has a market losing sequence but it is optional violating the conjecture.
I did consider the possibility of gains and losses (regrets) exactly balancing, but it seems to me that at best it would be extremely unlikely. For example, the regret might be failing to avoid a pass that is 1.0223160472... (further digits omitted) and the gains might be some group of 66-xx and 55-xx that happen to add up very close to the aforementioned number. But "exact"? Even if all pass variants involved happened miraculously to be 1 plus a multiple of .1296, those equities themselves would no doubt be rounded.
If equities are rounded to (say) four decimal places and then everything is compared, it would still be very rare that gains and losses would balance to four decimals. That is based on intuition, and also having calculated many examples. I saw one that was .0001 (.000052, actually), a .0002, a couple at .0003, one at .0004, but typically they were in the .01 to .05 range, not counting the ones that were obviously much larger and not worth figuring. But even if I'd seen a bunch at .0000, I would be inclined to say that they are extremely close to balancing rather than they exactly balance. That said, you may be better qualified to judge the mathematical meaning of "exact" in context than I am.
My inclination is to call such an occurrence (whether to three, four, five or six decimal places, whatever) a borderline double decision, and to save the term "optional double" for a situation that has zero gains and zero losses. At the least, having two terms makes a useful distinction.
I think it's analogous to identifying a 1.0000 take/pass in backgammon as "borderline." While one checker on the 1pt vs one on the 6pt in the bearoff could correctly be identified as being on the exact fence of the take/pass decision, I'm skeptical that there is an early game or middle game position that neatly lands on 1-decimal with an unlimited number of zeros after it. If I rolled out 51U-55A-F-C a billion trials and the result is 1.0000 (that's all the digits we can see), I wouldn't declare it to be exactly on the take/pass fence, and in any case I would prefer to call the take (or pass) "borderline" rather than "optional."
By the way, IF it were possible to exactly balance doubling losses and gains in an early/middle game position at the n2n2 score, then it might well be possible to construct an entirely error-free undoubled game. The game I constructed had a single error (at the unavoidable inflection point) that I calculated to be .00005, but even if that were to round to .0000 to four or more decimal places, I wouldn't claim the game to be error-free.
Nack
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