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

 From: Joseph Heled Subject: Re: [Bug-gnubg] A simple question about backgammons... Date: Thu, 26 Feb 2009 08:14:02 +1300

```Yes I hit the same pitfall long ago when I was looking to generate
data for the race network.

-Joseph

On Thu, Feb 26, 2009 at 8:09 AM, Øystein Johansen <address@hidden> wrote:
> 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
>>>> 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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