+------------------------------------------+
| | | O |
| | | O |
| | | O |
| | | O |
| | | O |
| | O | |
| | | |
| | | X X X |
| | | X X X X X X |
| | | X X X X X X |
+------------------------------------------+ X on roll
12 11 10 9 8 7 6 5 4 3 2 1
The question is now:
What is the probality of X winning the game?
I have found out a simple rule that can give you the neccesarry information,
and I've named the rule "Michael愀 432 rule"
The rule is as follows: When O have 4 men left on his ace point, the
probality of X winnning is between 30 and 20% (4,3,2), dependent of where his
extra builders are placed (cubeless). For an optimum distribution of spares
on the 6-,5- and 4 point (see figure) will give X 30% of winning chances,
while having all the spares on the ace-point, which is the worst condition,
will give X 20% probality of winnning the game. You just have to remember the
4,3,2 sequence: When the opponent has N builders left , your chances of
winning are between 10*(N-1)% and 10*(N-2)%. The formular can be extended up
to O having 9 men on the ace-point. Then X's probality of winning is between
80% and 70% (9,8,7). The formular is accurate within 2-3%, which is accurate
enough for human players. When O has below 4 men and beyond 9 men, the
formular isn't accurrate enough. The formular also work "in reverse". This
means, that if you are hit while bearing off, you have a take (in MG) when
you have a maximum number of 7 men on the ace point. Then the opponent's
winnning chances are between 60 and 70%. The 7 men is also what Bill Robertie
consider to be the turning point. I've used this formular a lot, and I found
it quiet easy to use. I hope it can help other players around the world.
Hi from
Michael Bo (snog at FIBS)
--------------------------------------------------------------------------------
There are three kinds of backgammon players in the world:
Those who can count and those who cannot.
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
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+------------------------------------------+
| | | O |
| | | O |
| | | O |
| | | O |
| | | O |
| | O | |
| | | |
| | | X X X |
| | | X X X X X X |
| | | X X X X X X |
+------------------------------------------+ X on roll
12 11 10 9 8 7 6 5 4 3 2 1
The question is now:
What is the probability of X winning the game?
I have found out a simple rule that can give you the necessary information,
and I've named the rule "Michael's 432 rule"
The rule is as follows: When O have 4 men left on his ace point, the
probability of X winning is between 30 and 20% (4,3,2), dependent of where
his extra builders are placed (cube less). For an optimum distribution of
spares on the 6-,5- and 4 point (see figure) will give X 30% of winning
chances, while having all the spares on the ace-point, which is the worst
condition, will give X 20% probability of winning the game. You just have to
remember the 4,3,2 sequence: When the opponent has N builders left , your
chances of winning are between 10*(N-1)% and 10*(N-2)%. The formula can be
extended up to O having 9 men on the ace-point. Then X's probability of
winning is between 80% and 70% (9,8,7). The formula is accurate within 2-3%,
which is accurate enough for human players. When O has below 4 men and beyond
9 men, the formula isn't accurate enough. The formula also work "in reverse".
This means, that if you are hit while bearing off, you have a take (in MG)
when you have a maximum number of 7 men on the ace point. Then the opponent's
winning chances are between 60 and 70%. The 7 men is also what Bill Robertie
consider to be the turning point. I've used this formula a lot, and I found
9 men, the formula isn't accurate enough. I've used this formula a lot, and I