You, Casper and others seem interested in the thought process of retro-solving a backgammon sequence. I'll walk through an example, to give you an idea.
Rory asked, "What is the most likely combination of rolls played correctly that got to this position?"
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| Blue on roll, trails 𥑳 | |
How does one interpret "most likely"? Well, two Giants are playing, so it is improbable that a blunder occurred this early in the game. In such a case, I tend to invalidate sequences with blunders and discount sequences with non-trivial errors.
Always a prime consideration is that the shorter the sequence, the higher the probability that it occurred (each additional roll/play decreases the likelihood by a factor of 18 or 36). As on the surface it looks feasible that each player moved only twice, we'll start by looking for a 4-roll sequence that is "well-played" (by that, I generally mean that no error is worse than -.02).
Blue is on roll. If only two moves for each side have occurred, White went second and fourth (last). She has played two 6s and one 2 from her midpoint; thus, her move portions must have come from a 2, a 6, and another 6 being the sum of two smaller numbers (4 + 2, or 5 + 1). Possible roll combinations, then, are 62 + 42, 62 + 51, 61 + 52, and 65 + 21.
As opening 62D, 42D, 42d, 51D, 65D and 21d are blunders, 61P + 52D (in either order) is White's only viable two-play combination. If 52D came first, White's last play with 61 (making the 7pt instead of the 5pt) was a blunder. If 61P came first, then 52D (instead of 52H) was a blunder. [If either or both are not obvious blunders to you, check with a bot and you've already learned something.]
Conclusion: White has no combination of two plays that work; she must have moved at least three times.
On to five-roll sequences. Blue is on roll, so White started. Without being hit, could she have moved a mere 14 pips in three moves? It doesn't seem possible without slotting the 7pt, and that would only make sense if White were hitting there. However, if Blue has split with a 6, he lacks enough move portions to also enter, come down, and make the 4pt with his three remaining move portions. Blue cannot be hit if he moves only twice. The inescapable conclusion is that White must have been hit.
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| Blue on roll, trails 𥑳 | |
If White entered bar/24 on the fifth roll (her third and last turn), it would not have been accompanied by 13/11 (as 6/4* would have been an obvious hit). Could it instead have been with 13/8? No; even if that were the right move, it means that she must have misplayed her 61P (7pt instead of 5pt) or her 32S (stripping with 13/11 instead of coming up or pointing on the 4pt) the turn before. Therefore, she entered on the third roll (her second turn), which means that (in order to get hit) she split to Blue's 7pt or 4pt on the opening move. [She didn't open with 63R because Blue would not play 13/10*/4 or 13/10* 6/4, and cannot legally play 13/10* 24/21.] White must have played opening 62S or 32S.
After getting hit, White must enter with bar/24 and come down with a 6, 5 or 2. The right 6 is clearly 24/18, so forget 13/7. If 13/8, the final roll would have to be 61 and again making the 7pt (instead of the 5pt) is a blunder. By process of elimination, that leaves 13/11. White's second turn must be 21P (i.e., bar/24 13/11, covering her 11pt) and her third turn 64P (i.e., 13/7 11/7).
Blue's two moves must include 13/4 as two portions (the sum of either 6+3 or 5+4), and the other portions were a 2-cover (i.e., 6/4) and a 3-up (i.e., 24/21). Assuming he played exactly twice, and knowing that Blue must hit first, we're down to seven possibilities: 63X + 32p, 32H + 63C, 52H + 43C, 13/4* + 32E, 42P + 53S, 32X + 13/4, and 43X + 52N.
The ones with 32X and 43X don't work because White would surely reenter and hit on her 4pt rather than cover her 11pt. (If you're unsure of this or anything else, check with a bot; another lesson learned.) Interestingly, the evals show 32S-32H-21P wrong by .047 and 32S-52H-21P by .041 (I would have guessed about half that), so I excluded them too even though it seems conceivable a Giant could falter with 21P in either position.
That leaves three reasonable possibilities, or four if hitting 13/4* with 63 and 54 are deemed to be different. The five-roll sequences I listed here are repeated below.
- 32S-42P-21P-53S-64P
32S-63H-21P-32E-64P
32S-54H-21P-32E-64P
62S-63X-21P-32p-64P
For the first sequence, Snowie says (for the third play) that 21@ (i.e., bar/23 24/23) is best by .005, but that looks like a donkey evaluation to me. Unless someone demonstrates otherwise, I'm assuming 21P (i.e., bar/24 13/11) is best.
For the second and third sequences, 21W (i.e., bar/23 6/5) evaluates better than 21P by .017, a bigger difference than I thought. If that holds up, it meets the "well played" standard, and anyway 21P looks like a move a Giant could easily make.
For the last one, I figured White should hit (bar/24 6/4*) with 21, and that Blue should make his 5pt instead of 4pt with 32; the Snowie eval agrees but respectively only by .008 and .019 (both closer than I thought), so in absence of further data the fourth sequence ekes in as "well played."
The fifth play of all the sequences is Snowie-evaluated [R P7 p11], and someone else noted that XGR++ evals P at -.004, so the odds seem against P being the best move. It seems close enough to be well played, though, and in any case a Giant might make any of those three moves.
Rory requested a sequence that is "played correctly." Uncertain how strictly he meant that, I explored a roll deeper to find a perfecto (even though I think the chances are much higher the game sequence was five rolls). The perfectly played six-roll sequence is given here. The logic is a bit more involved, but having worked on the shorter sequences I was able to build on what I had already learned and deduced.
Hope that helps.
Nack