Over the decades, there has been much speculation, some valid but also some invalid, about the –2–2 (need 2, need 2) score. Many of the attempted elucidations that I've read seem convoluted.
For example, it is true that it is never a theoretical mistake to double at the first opportunity given the current opening position, but it may surprise some readers (based on some explanations going around) to find out that it could be a mistake if backgammon had a different (but still symmetrical) opening position.
Imagine that backgammon were instead played with the opening position shown below.
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| XGID=------B-AAAAiIaaaa-b------:0:0:1:00:0:0:0:25:10 | |||||
Modified Opening Position | |||||
Sir Smith and Dr. Jones, who had spent a lot of time studying race formulas, played a 25-point match using the above opening position (the pipcount being identical to that of the standard opening position) for every game. This match was played for the club championship.
When Smith and Jones reached the score of 23–23, quite a crowd had gathered, including press reporters. Hence the players moved into a private room, but a camera displayed their board on a big screen and two experts broadcast their comments.
This championship game was being billed (among other things) as "virtual DMP" and "the final game." Little did they know...
To start the game, each player rolled a single die, in traditional fashion. Smith (Blue) rolled a 6 and White (Jones) rolled a 5, which of course gave Smith an opening roll of 65. His play is diagrammed below.
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| XGID=------BABAAAiGaaaa-b------:0:0:-1:00:23:23:0:25:10 | |||||
White on roll. Cube action? | |||||
The score is –2–2 and this is White's first opportunity to double. Murmurs in the crowd amplified to animated debates with sweeping gestures, then competitive shouting to be heard over the commentators and others. Prevailing wisdom seems to say that doubling is optional, but unless you're playing a clearly weaker player, that you should just do it automatically, before you forget or make a mistake, just get it over with. Right?
No. Granted, the error is tiny (at this point) but it would wrong to double. To see this, we must weigh the impact of the market gainers (incentive NOT to double) against the impact of the market losers (incentive TO double).
A "market gainer" (which you can think of it as a market loss to the opponent) is a number so poor that it causes the player to (correctly) pass on the 1-level (thereby escaping a bad game played to conclusion on the 2-level). Here, if White rolls 21, he will trail 8 pips after his play, and a lead of 156–164 is sufficient for Blue to cash at this score.
A "market loser" (or "market-losing sequence") is a combination of two rolls, one by the player and one by his opponent, such that it will allow the player to cash. Here, if White rolls 66 and Blue's roll is 5 pips or fewer, White will have a cash.
For the breakdown, XGR++ evaluated equities should be reliable enough for our purposes:
White's market gainer (represented by his rolling 21 and playing 13/10) occurs with a probability of 2/36 = 72/1296, and realizes a gain/savings of .0223. (If White had doubled before rolling 21, it would lead to the same result as not doubling, rolling 21, and making a bad take of 1.0223.) The probability times the gain equals .0012.
White's market losers (represented by his playing 13/7(3) 11/5 and Blue making a small-roll play) are 66-41, 66-32, 66-31, 66-21, and 66-11, and occur with a probability of 2/1296 or (in the last case) 1/1296. Multiplying the probabilities times the various losses comes to 2/1296 (.0514 + .0531 + .1320 + .2027) + 1/1296 (.1306) = .0008.
In summary, White's net gain from not doubling is .0012 – .0008 = .0004. (XGR++'s assessment of the cube decision is no-double by .0019, but I suspect it is doing something procedurally wrong in generating the evaluations.)
The mere .0223 impact of the 21 roll exceeds that of all five 66-xx sequences combined, even though the latters range from .0514 to .2027. (One-roll sequences are much more likely to occur than two-roll sequences.)
After Jones (White) correctly does NOT double, he rolls 66 and reaches the position below.
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| XGID=------BABAAAfG-aaacba-----:0:0:1:00:23:23:0:25:10 | |||||
Blue on roll, cube action? | |||||
Should Blue double? Definitely NOT. He is in even worse shape (being down 13 pips) than White was (down 11). His market gainers here are the same rolls in the back half of White's market-losing sequences in the previous position: namely, 41, 32, 31, 21 and 11.
Yes, Blue has market losers: 66-31, 66-21 and 66-11. But just having market losers isn't enough (as some people seem to believe). The impact of the market gainers (see previous paragraph) has to be accounted for on the other side of the ledger.
(According to XGR++, it is an error of .0285 for Blue to double, but I trust my calculation more. Gainers this roll are 36 times what the losers were last roll, which comes to .0448; and the losers are merely 2/1296 (.0097 + .0909) + 1/1296 (.0054) = .0002. Thus, doubling is about a .045 error.)
Note that to cash, Blue needed an 8-pip lead last turn (after a projected 65-21), but this turn only a 7-pip lead (after a projected 66-31). This tightening of the pass threshold is due to the shorter length of the race (Blue's lead-count dropping from 156 to 132).
After Smith (Blue) correctly does NOT double, he rolls 66 and reaches the position below.
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| XGID=----AABCBA--fE-aaacba-----:0:0:-1:00:23:23:0:25:10 | |||||
White on roll, cube action? | |||||
Should White double? Nope. Gainers are roughly .00016 * 36 = .0057 (borrowing from my previous calculation), and I estimate 66-xx losers at .0015 (work not shown), which nets to a doubling error of .0042. (XGR++ says .0052.)
After Jones (White) correctly does NOT double and plays 66O (13/7(2) 10/4 9/3), Blue should again NOT double (to do so would be a blunder).
This game of leapfrog went on, with Smith and Jones continuing to alternate double 6s. Eventually, the market gainers grew (hugely) to any non-doublet, and the market losers grew (merely) to doublet vs non-doublet, and on the final roll the error that would have been committed by doubling escalated to a duodecuple (12-times) whopper.
The stunned audience was completely hoarse and exhausted from seeing so many spectacular lead changes. When Smith (Blue) triumphantly ripped off his last three checkers in this perfectly-played undoubled game, he won a single point and therefore went into the Crawford game leading 24–23, defying all expectations that the game just illustrated would end the match.
The final chapter of that story has yet to be written.
Starting from the standard opening position, is it possible to construct a complete undoubled game at –2–2 that is either can-never-double or error-free? Answers: Can-never-double -- no way. Error-free -- almost. First, permit me to make a couple of observations.
(1) Some positions at the –2–2 score are optional double. That can no longer be the case after contact is broken or minimized to the degree it was at the start of the Smith-Jones game. That game throughout was "never can double." I'm clarifying a distinction implied in the above question: "Error-free" is a more liberal condition than "can-never-double."
(2) If the first diagram of this post is (by hook or by crook) reached from the standard opening position, the pipcount is even, so (at –2–2) whichever side is on roll has a clear double because the next two rolls might be exactly 65-21. The can-never-double condition was perfectly satisfied only because neither player is allowed to double in backgammon just before the opening roll, and that position was agreed upon by Smith and Jones to be their opening/starting position.
Well, then, why can't the same leapfrog method be applied to the standard opening position, which starts with the identical 167–167 pipcount? In a vacuum, it could. The only factor preventing it is contact.
When contact exists, a positional advantage can replace or augment a race lead, but it is even harder to flip a substantial advantage from one player to the other in one roll even once, let alone multiple times. Even at the –2–2 score, for a position to have both market gainers and market losers, it needs to (essentially) be a straight race.
If an opening 65R (24/13) could somehow turn the game into a straight race, a can-never-double game at –2–2 is possible.
Once the opening roll has been played, if there are no market gainers (at least one self-roll causing the player to pass), it cannot be theoretically wrong to double. It can only be proper or optional.
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. [The sole exception to that is if the previous roll was the opening roll, when the cube could not legally be turned -- see the Smith-Jones game.]
With that information soberly delivered, let us nevertheless see how close we can come to constructing an error-free undoubled game at –2–2 from the standard opening position.
The position below was reached via 42P-44B-66B, and to this juncture, White and Blue have each encountered one optional cube decision.
After Blue's opening 42P, White had no market gainer (if he makes a decent play with any roll, he'll have an easy take). White also had no market loser (his best sequence was 66B-61A, which is .836).
I leave it to the reader to verify that Blue had neither a market gainer nor a market loser after 42P-44B (prior to his roll of 66).
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| XGID=----BbDBB---cC--bcBe------:0:0:-1:00:0:0:0:2:10 | |||||
White on roll, cube action? | |||||
What now: should White double? Answer: It's still optional. Clearly, White has no roll bad enough to cause her to pass. At the other end of the spectrum, her best sequence is 66B-21, which puts her up 7 pips in the count.
As we learned from reviewing the Smith-Jones game, at this score (–2–2), 7 pips is a large enough lead to cash in a straight race of this length. If we trade 6 pips in the diagram by giving Blue 18/15(2) and White any pair of 3's in exchange, White should double because 66-21 enters (what is essentially) a straight race and she loses her market, and there is no way to gain her market (with a small roll she can still take because she retains adequate contact from the 20pt).
However, 7 pips is NOT enough when bar-point contact remains: 66-21 proffers .888, which is short of a cash. Therefore, White's cube action in the above diagram is still optional.
Anyway, White leaves the cube centered, rolls 66 and plays 20/8(2), and now it is Blue's turn to decide on the cube:
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| XGID=----B-DBB---cC--beBe------:0:0:1:00:0:0:0:2:10 | |||||
Blue on roll, cube action? | |||||
We already determined there is no market gainer for Blue. (That would have meant White previously had a market loser.) However, Blue does have one market loser, which is 55B-21. If that occurs, Blue will have a cash of 1.0691. Not doubling, therefore, costs Blue 2/1296 (.0691) = .00005, which is one half of the tiniest amount XG can display. (XGR++ reports that this is a double by .0000.)
However, Blue makes that miniscule error by not doubling, then rolls plays 55 and plays it as shown below. It is now White's turn to act:
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| XGID=----B-DBD---cC--be-e------:0:0:-1:00:0:0:0:2:10 | |||||
White on roll, cube action? | |||||
We know White has a market gainer (21) that can save him .0691 (review text above the diagram if you don't remember). Multiplying that by 2/36 yields .00384.
Against that, White has market losers of 66-xx, where xx can be as many as 8 pips. (You see, White was down 11 pips in the Smith-Jones game, but only 10 pips here, and the race is shorter.) I have carefully researched the costs, which are 2/1296 (.0437 + .0433 + .1485 + .1361 + .1282 + .2397 + .2158 + .3140 + .2976 + .3894) + 1/1296 (.0024 + .3828) = .0033.
As in so many other instances, a single gainer (of the 21 roll) is bigger than the sum of the losers, though this time by only .0038 – .0033 = .0005. White has a very close NO-double. (XGR++ reports this to be a double by .0001, but I don't believe it -- I've triple-checked my logic and math.)
White enjoyed a bigger edge (+14 pips) after her roll than Blue enjoyed after his (+10 pips). That factor combined with gaps in the boards and White's overly efficient end-parity produces significant imbalances, and I could see that my intended mutual all-66 exchange would ultimately cause one of the players to double.
To solve these problems, I limited the next few rolls to 55. Starting from the above diagram (where the race is already short enough to absorb the reduction), the remaining rolls of the game, for anyone wishing to follow along with XG, are 55-55-55-55-66-66-66-66-55 (vital finesse)-44-33-33-22-22 (or any of those last five rolls can be larger).
Here, as in the Smith-Jones game, the margin by which not doubling is the right decision widens until the end. On Blue's final roll (where he has 2pt(2)+1pt vs 1pt(2)), turning the cube would be a tredecuple (13-times) whopper.
In conclusion, an entirely error-free undoubled game at –2–2 is not possible from the standard opening position (due to contact), but it IS possible to achieve it with a single microscopic error (in this game, .00005). It's a matter of getting past the inflection point with minimal disturbance.
Nack