Re: [Bug-gnubg] A simple question about backgammons...

[Øystein Johansen]

Jonathan Kinsey wrote:
> Øystein Johansen (OJOHANS) wrote:
>>> How often will the player win a backgammon in this position:
>>>
>>> GNU Backgammon Position ID: dhsAaQwfAAAAAA
>>> Match ID : cAkAAAAAAAAA
> +13-14-15-16-17-18------19-20-21-22-23-24-+     O: gnubg
> |                  |   |    O  O  O  O    |     0 points
> |                  |   |    O  O  O  O    |     
> |                  |   |          O       |     
> |                  |   |                  |     
> |                  |   |                  |    
>v|                  |BAR|                  |     (Cube: 1)
> |                  |   |                X | XX 
> |                  |   |                X | XX  
> |                  |   |                X | XX  
> |                  |   | O        O     X | XX  On roll
> |          O     O |   | O        O     X | XX  0 points
> +12-11-10--9--8--7-------6--5--4--3--2--1-+     X: me

>>> This is such a simple question that I really feel silly
>>> asking. Can someone answer how many backgammons the player
>>> wins and why? (The "and why"-part of the question is important here.)
>>
>> Thinking one more minute, and the answer comes obviously: 0.4350
> 
> You didn't give your why part! Anyway I get something like:
> 
> X rolls a double in 2 rolls (11/36) and O doesn't escape (32/36) =
> (11*32/36*36)
> = 0.272
> X doesn't roll a double either time (25/36) and O doesn't escape two
> times in a
> row [by either rolls two high numbers either time or one high number
> twice]~(1/4) = (25*1) / (36*4) = .174
> 
> So I get a similar number .446 (although who knows if my reasoning is
> correct) -
> if only we had some kind of computer program to work these things out...


The point was that I could not figure out why the backgammon probability
was not the same as a bearoff:

 GNU Backgammon  Position ID: YwAA4A8AAAAAAA
                 Match ID   : cAngAAAAAAAA
 +13-14-15-16-17-18------19-20-21-22-23-24-+     O: gnubg
 |                  |   |       O        O | OOO 0 points
 |                  |   |       O        O | OO
 |                  |   |                  | OO
 |                  |   |                  | OO
 |                  |   |                  | OO
v|                  |BAR|                  |     (Cube: 1)
 |                  |   |                7 | X
 |                  |   |                X | X
 |                  |   |                X | XX
 |                  |   |                X | XX  On roll
 |                  |   |                X | XX  0 points
 +12-11-10--9--8--7-------6--5--4--3--2--1-+     X: Øystein


Go figure... why is the backgammon probabilities in the first position,
the same as the winning probability in the bearoff position? Think for a
while, before you read on......









it's because 2s and 3s can make the important crossover in the
backgammon position, however 2s and 3s can not be used to bear off in
the bearoff position. Quite obvious....

-Øystein

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