I remember a discussion on this forum many years ago where it was
claimed that, in non-contact positions, whenever a single ace can be
played 1/off, it is always optimal to do so. (The wording needs to be careful here. With double aces, 4/off might be worse than 4/3(4).)
Apparently, XG thinks that the below position is a counter-example.
After 3/off, it played 3/2 rather than 1/off.
Of course, with my previous little adventure, I'm less inclined to take XG's word for it.
On 12/6/2021 7:52 PM, peps...@gmail.com wrote:
I remember a discussion on this forum many years ago where it was
claimed that, in non-contact positions, whenever a single ace can be played 1/off, it is always optimal to do so. (The wording needs to be careful here. With double aces, 4/off might be worse than 4/3(4).)
Apparently, XG thinks that the below position is a counter-example.The rollout below (with no variance reduction, just in case) suggests
After 3/off, it played 3/2 rather than 1/off.
Of course, with my previous little adventure, I'm less inclined to take XG's
word for it.
that the two plays are equally good. I could do more trials and/or use
a stronger setting, but I don't think that's necessary.
I remember the discussion you're referring to, and a Google Groups
search found it (Google Groups search seems to be better than it used
to be, which is good news):
https://groups.google.com/g/rec.games.backgammon/c/brLijq7r8UI/m/8KeVP9lzILgJ
It seems that the claim has never been rigorously proved. Nowadays I
think it would be possible to prove the theorem by a brute-force check
of all possible non-contact positions, if one cared enough to do so and
had access to a sufficiently large computer.
XGID=---B-EE-B---------A---d-c-:1:1:-1:31:0:0:3:0:10
X:Player 1 O:Player 2
Score is X:0 O:0. Unlimited Game, Jacoby Beaver +12-11-10--9--8--7-------6--5--4--3--2--1-+
| X | | O O |
| | | O O |
| | | O O |
| | | O |
| | | |
| |BAR| |
| | | X X |
| | | X X |
| | | X X | +---+
| X | | X X X | | 2 |
| X | | X X X | +---+
+13-14-15-16-17-18------19-20-21-22-23-24-+
Pip count X: 95 O: 15 X-O: 0-0
Cube: 2, X own cube
O to play 31
1. Rollout¹ 3/Off 1/Off eq:+1.4982
Player: 100.00% (G:49.82% B:0.00%)
Opponent: 0.00% (G:0.00% B:0.00%)
Confidence: ±0.0017 (+1.4965..+1.4999) - [51.7%]
2. Rollout¹ 3/2 3/Off eq:+1.4982 (-0.0001)
Player: 100.00% (G:49.82% B:0.00%)
Opponent: 0.00% (G:0.00% B:0.00%)
Confidence: ±0.0017 (+1.4965..+1.4999) - [48.3%]
¹ 279936 Games rolled.
Dice Seed: 271828
Moves and cube decisions: XG Roller+
Search interval: Huge
eXtreme Gammon Version: 2.19.207.pre-release
I'm fairly certain it has been proven but don't care about it to try to dig up the details.
Koca also provided "Koca's paradox" to show why any proof would have
to be delicate: he showed that if you know in advance what the
remaining rolls are going to be, then bearing off with the ace is
not always going to be best. Details of Koca's paradox were not
provided in that thread; I just emailed him to find out if he still
remembers (or has a record of) the details of Koca's paradox.
On 12/7/2021 8:53 AM, I wrote:
Koca also provided "Koca's paradox" to show why any proof would haveBob replied quickly and provided two references:
to be delicate: he showed that if you know in advance what the
remaining rolls are going to be, then bearing off with the ace is
not always going to be best. Details of Koca's paradox were not
provided in that thread; I just emailed him to find out if he still remembers (or has a record of) the details of Koca's paradox.
Page 4 of http://chicagopoint.com/PTpdf/101_1997-08.pdf
Page 2 of http://chicagopoint.com/PTpdf/102_1997-09.10.pdf
But I should say that I slightly misstated "Koca's paradox" above.
I quote Kleinman: "No matter how you play it, having to move an ace
in a pure race can cost you an extra shake." In this form, it is
essentially the same as Chapter 56 of Backgammon Funfair, "Better
not to move in bear off race." But I'd lobby for a slightly
different definition of Koca's paradox. Consider the position
below, adapted from Art Benjamin's letter.
XGID=-AADBA--------------------:0:0:1:51:0:0:0:0:10
X:Player 1 O:Player 2
Score is X:0 O:0. Unlimited Game
+13-14-15-16-17-18------19-20-21-22-23-24-+
| | | |
| | | |
| | | |
| | | |
| | | |
| |BAR| |
| | | |
| | | X |
| | | X |
| | | X X |
| | | X X X X X |
+12-11-10--9--8--7-------6--5--4--3--2--1-+
Pip count X: 28 O: 0 X-O: 0-0
Cube: 1
X to play 51
Let's tell a little story. Tim says, "In this position, we would
obviously play 5/off 1/off no matter what the opponent's position
is (assuming no contact)." Bob replies, "Well, depending on what
rolls we get in the future, maybe 5/4/off is better." Tim says,
"You're crazy!" Bob says, "Okay, let me offer you a wager. We set
up two boards with the same starting position above. On your board,
you play 5/off 1/off, and on my board, I'll play 5/4/off. Then I'll
call a dice roll. I'll play the roll on my board you'll play the
same roll on your board. Then I'll call the next roll, and so on.
I bet you I finish bearing off before you do." Tim says, "You're on!"
Bob calls 41 and plays 4/off 1/off. Tim plays 4/off and thinks a
moment, then plays 3/2. Bob calls 33 and plays 3/off(4), and now
Tim sees that he's lost: he can bear off only three checkers and
Bob will win by calling 42 next.
Tim says, "Okay, you win, but let's try that again...I think I
misplayed the first roll." They start from the beginning again,
and when Bob calls 41, Tim tries 4/off 2/1, expecting Bob to call
33 next. But Bob instead calls 42 and plays 4/off 2/off, and Tim
loses again; he must miss, and Bob wins by calling 33 next.
Finally, if Tim tries 4/off 4/3, then Bob calls 22 and plays
4/off 2/off 3/1, allowing him to win by calling 33 next, since
Tim can bear off only one checker with the 22.
This little story doesn't disprove "Magriel's theorem" but it does
illustrate that any proof has to be subtle. The "obvious" way to
try to prove that Tim's initial choice of 5/off 1/off is better
than 5/4/off is to argue that Tim does no worse than Bob no matter
what the subsequent rolls are. But the above story shows that this
proof strategy fails.
I remember a discussion on this forum many years ago...
Wouldn't it be amazing if Ullrich showed up...?? :)
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