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# end-game (bearoff)

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### Terran_yea

Nov 11, 2002, 5:58:13 AM11/11/02
to
Does anyone know how to find the winning propability based on number of
rolls left for player+opponent ?

Simon ;-]

### Terran_yea

Nov 11, 2002, 6:01:28 AM11/11/02
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I believe that Stein Kulseth once made such a formula (or program with
source code ?!) - but i can't find it on da' internet...)

plz help, man ;-D

"Terran_yea" <lar...@ostenfeld.dk> wrote in message
news:aqo23t\$bt\$1...@eising.k-net.dk...

### Gregg Cattanach

Nov 11, 2002, 8:18:40 AM11/11/02
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"Terran_yea" <lar...@ostenfeld.dk> wrote in message
news:aqo23t\$bt\$1...@eising.k-net.dk...
> Does anyone know how to find the winning propability based on number of
> rolls left for player+opponent ?
>
> Simon ;-]
>
Not sure this is exactly what you are asking, but if you have both sides
with N rolls to go (with no misses and all doubles working) this table is
accurate (CPW=Cubeless probability of winning):

These are the winning probabilities in an n-roll position:

Rolls
CPW

2
86

3
79

4
75

5
72

6
70

7
68

8
67

### Frank Berger

Nov 11, 2002, 4:23:00 PM11/11/02
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"Terran_yea" <lar...@ostenfeld.dk> wrote in message news:<aqo23t\$bt\$1...@eising.k-net.dk>...
> Does anyone know how to find the winning propability based on number of
> rolls left for player+opponent ?
>
> Simon ;-]

go to http://www.bgblitz.com In the *free* edition of BGBlitz is the
endgame database from Stein Kulseth build in with some improvements.
It will give you:
- the average number of rolls needed and
- the winning probabilty and
- the best move for a given roll.
It is cubeless and an approximation, but I guess that no value is off
more than 0,2% and in most cases far less.

ciao
frank

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