This puzzle has its origins in the Mystery of the Big Stack post (solutions to which can be found here). Bob suggested an alternative puzzle could be to start from backgammon's opening position: What is the fewest number of perfectly played moves (according to XGR++ evaluation) by which 15 checkers can be stacked on one point?
These big-stack puzzles can be divided by point number (and I believe it's possible, even with perfect play, to manipulate 15 checkers onto any point). In my first attempt, I tried to stack 15 checkers onto the 3pt; that 15-roll solution (8 rolls for the first player) can be found in the second half of this post.
Next, I went after the 6pt, which turns out to be somewhat easier. Here is my 13-roll solution:
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| XGID=-----bE-C---cF--bc-e----A-:0:0:1:66:0:0:3:0:10 | |||||
Blue opened with 65R, White played 44B, Blue rolls 66 | |||||
If you don't know the answer to this one (65R-44B-66), you will learn something valuable. Without easy access to my rollouts at the moment, I'll give you XGR++'s opinion for Blue's 66 roll: For Money, Z (1:3 Split, 24/18 13/7(3)) beats D (Down, 13/7(4)) by .046.
For 65R-xxx, here is a list of XGR++ perfectly-played White moves against which Blue's 66Z answer is "best:".. 31P 42P 53P 64P 21$ 51S 52D 33B 44B, while against 64P it is too close to call. (Z isn't legal against 61P 11N 66B 55A; play D against those and 32S 41S 43Z 54S 22N. Against 62S 63S play P (Point, 13/1*(2).) 66Z is also best against other replies you may see; e.g., 65R-41$, 65R-51$, 65R-52S, 65R-21S, 65R-43D, 65R-32D, 65R-54D, etc.
In short, 65R-xxx-66Z is the plurality play (for Money, and especially at DMP); you would do well to remember it.
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| XGID=-----bECC---cC--bcAe------:0:0:-1:54:0:0:3:0:10 | |||||
Blue played 66Z, White rolls 54 | |||||
[White's biggest roll that neither hits nor runs is 54. Although it is not a condition of the puzzle, I enjoy the additional challenge of keeping the cube centered. White needs two big rolls to counter Blue's 66 that will be coming up in a couple of rolls.]
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| XGID=-----bECC---bC--bcAe-a----:0:0:1:51:0:0:3:0:10 | |||||
White played 54$, Blue rolls 51 | |||||
Blue now rolls 51. If instead we assign 41 or 32, he'll later be a pip shy of achieving the full-stack setup on his 7th turn. If we assign him 52, he'll double-cross us by slotting the 4pt. Finally, if we assign 65, he'll have too few checkers on the midpoint to carry out our diabolic plan.
Going back to the beginning, it may seem like Blue could simply play 65R twice and 66D. However, that leaves only three checkers on the midpoint. His fastest follow-up of 61 + 65D puts him 5 pips behind the solution sequence, which costs him two rolls! The virtue of the feature sequence lies in the 66Z finesse, enabling Blue to put a fourth checker back on the midpoint and roll 66 a second time.
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| XGID=-----bFBC---bD--bc-e-a----:0:0:-1:33:0:0:3:0:10 | |||||
Blue played 51C, White rolls 33 | |||||
For Blue's 51, I would have played the ace 8/7, but XGR++ prefers 7/6 by .002. I won't lose sleep over it. The solution works with either ace.
White now rolls 33. This is the largest roll we can assign without her running.
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| XGID=-----bFFC---b---ba-cbab---:0:0:-1:55:0:0:3:0:10 | |||||
White played 33A, Blue played 66D, White rolls 55 | |||||
I added two moves to the above diagram: White's 33A and Blue's 66D. [Oddly, 33A (8/5(2) 6/3(2)) is not on the XGR++ list; I had to add it.]
With 55, White stays in the race (down 8). Indeed, with the one number that's larger (66), White should run even though behind 4 pips after the roll.
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| XGID=-----bFFC--------c-cbcb---:0:0:1:11:0:0:3:0:10 | |||||
White played 55A, Blue rolls 11 | |||||
The remainder of the sequence is trivial, as long as White does not hurdle the short prime. While Blue finishes off with three double 1s, I decided to give White 33 and 22, which equalizes the pipcount (at Naccel zero). The final position is...
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| XGID=-----bO------------bccbba-:0:0:-1:00:0:0:3:0:10 | |||||
65R-44B-66Z-54$-51C-33A-66D-55O-11O.33a_11O-22I-11T | |||||
All of Blue's checkers are home, while White still has two checkers back. Nevertheless, according to XGR++, White's winning chances are over 61%.
Incidentally, the above was the second 13-roll solution I found. I was not entirely satisfied with the first, because the cube did not stay centered (though technically that doesn't matter). I believe I can encapsulate that first solution well enough with the aid of just one diagram:
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| XGID=-----bE-D--BcBb--cBbab----:0:0:1:55:0:0:3:0:10 | |||||
62S-22N-62S-41W-32D-31@-55 | |||||
Blue played 62S (24/18 13/11) twice, followed by 32T (13/8). Meanwhile, White played 22N (13/11(2) 6/4(2), 41W (24/20 6/5) and 31@ (24/20). These 17 pips were the fewest I could assign White while making sure (a) she did not hit, and (b) she anchored on her 20pt.
From here, Blue rolls 55 twice, playing C (18/13(2) 11/6(2)) and D (13/8(4)), followed by 22T (8/6(4)) twice. Sounds simple enough. The problem is that Blue will opt to play a couple/few of his deuces to the 4pt for better distribution (and/or for a little containment equity against the anchor) unless he is so far ahead in the race that safety becomes his overriding concern; indeed, so far ahead that he would have already cashed the cube!
To salvage the sequence, it is therefore necessary to sandbag White's rolls pipwise in order to allow Blue to give a (takeable) cube before he rolls 22. The earliest opportunity will be after Blue's 55C (noting that White's deficit in the race also helps make it right for Blue to clear his 11pt). I gave White a puny roll of 21, and Blue is able to cube with his 21-pip lead (having only his midpoint and 8pt to bring home).
Assigning White two more rolls of 21 then ensures that Blue will stack all eight deuces onto his already-tall 6pt (though by a mere .005 over moving three checkers to the 4pt on the second set). In the final big-stack position (not shown), White trails by 51 pips and her winning chances aren't much greater than her chances to be gammoned on the run.
Nack