Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Michael's 432 rule

0 views
Skip to first unread message

micha...@yahoo.com

unread,
Feb 27, 1998, 3:00:00 AM2/27/98
to

Michael's 432-Rule
I have analyzed some positions when bearing off.
The positions were about beeing hit while you are bearing off,
and your opponent has a closed board. An example is shown below.
O is bearing off, and sometime in the bear-off, he was hit.
After that X managed to closed his board with O on the bar.
This sequence is quite common in BG today.

+------------------------------------------+
| | | 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 ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading

micha...@yahoo.com

unread,
Feb 27, 1998, 3:00:00 AM2/27/98
to

Michael's 432-Rule
I have analysed some positions when bearing off.
The positions were about being hit while you are bearing off,

and your opponent has a closed board. An example is shown below.
O is bearing off, and sometime in the bear-off, he was hit.
After that X managed to closed his board with O on the bar.
This sequence is quite common in BG today.

+------------------------------------------+
| | | 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

micha...@yahoo.com

unread,
Feb 27, 1998, 3:00:00 AM2/27/98
to

9 men, the formula isn't accurate enough. I've used this formula a lot, and I

Jean.Pierre SEIMAN

unread,
Mar 9, 1998, 3:00:00 AM3/9/98
to

micha...@yahoo.com wrote:
>
> Michael's 432-Rule
> I have analyzed some positions when bearing off.
> The positions were about beeing hit while you are bearing off,

> and your opponent has a closed board. An example is shown below.
> O is bearing off, and sometime in the bear-off, he was hit.
> After that X managed to closed his board with O on the bar.
> This sequence is quite common in BG today.
>
> +------------------------------------------+
> | | | 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)
>
> ------------------------------------------------------------------------
You can find a good and general study of this problem in "A Useful
Formula for Post-Ace-Point Games" by B. Robertie in INSIDE BACKGMMON
Volume 7, Number 4.

0 new messages