Pip Count percentage
Mogath3
if you're more than twelve percent behind in the pip count you should probably
drop. How do you figure this percentage? Any help is appreciated. Thanks
Regards,
Jeff
hi
If you are playing online usually there is a pip count displayed, so if the
pip count for you is 100 and you are more than 12 pips behind you should
drop (.12 x 100 = 12 you bonehead). If your pip count is 60 and you are 8
pips behind you should drop (.12 x 60 = 8 rounded up, you bonehead).
Hopefully, the preceding is not what you wanted to know, or you really are a
bonhead, and it is a wonder that you even figured out how to use a keyboard.
If there was more to your question than that, read on.
If the pip count is not displayed, then you have to learn how to count pips,
and there are many ways to do that. Look around online for the various
methods. In Paul Magriel's "Backgammon" he recommends taking if you are
behind 10 pips or less in a 60 pip race, and 13 pips or less in a 100 pip
race. How do you know how long a 60 pip or 100 pip race is approximately?
If all your men are in your inner board and equally distributed, you are in
a 60 pip race (15 men, average position is on the 4pt-->15 x 4 = 60). A 100
pip race is one in which there are about 4 or 5 rolls before the bear off
begins (average roll of the dice is 8 pips, 5 x 8 = 40).
Gregg Cattanach
easier).
Take the leader's pip count and multiply by 10%. Then take that number,
subtract 2, and that's the pip difference for an initial double. Subtract 1
from the 10% number, and that's the difference for a redouble. Add 2 to the
10% number and that's the pip difference where the opponent should pass.
This applies if both sides have similar high variance positions, (not a lot
of checkers piled up on the low points.) This turns out to be remarkably
accurate for any pip count from around 65 up to any large number. And the
math is easier :) The 8%, 9%, 12% system overestimates the differences
necessary when you get up to 120 pips or more, and underestimates it when
you are down below 80 pips.
This 10%-2, 10%-1, 10%+2 system is based on long computer simulations of the
'single-checker model'.
This method does require getting an accurate pip count for both sides. I
recommend Jack Kissane's Cluster Counting method:
http://www.northcoast.com/~mccool/cluster.html
--
Gregg Cattanach
Zox at GamesGrid, Zone
http://gateway.to/backgammon
ICQ# 2266410
gcattana...@prodigy.net
hi
Why wouldn't the redouble be the same as the double, i.e. 10%-2?
Martin Jensen
Because owning the cube is a very valuable asset. Therefore giving up the
cube is going to cost you some equity.
Before you double your opponent already has access to the cube but before
you redouble only you have access to the cube. That's why you have to have a
slightly stronger position before redoubling than doubling.
regards
Martin
Nick Wedd
<noneofyou...@none.com> writes
>Why wouldn't the redouble be the same as the double, i.e. 10%-2?
You need better odds for a redouble than for a double because you are
giving away more. When you redouble, you are putting the cube where
your opponent can use it. With a double, it was already where he could
use it.
This effect is stronger towards the end of the game. The "10%-2" rule
is only an approximation, but the best one that can be easily
remembered.
Nick
--
Nick Wedd ni...@maproom.co.uk
Mogath3
>
>If you are playing online usually there is a pip count displayed, so if the
>pip count for you is 100 and you are more than 12 pips behind you should
>drop (.12 x 100 = 12 you bonehead). If your pip count is 60 and you are 8
>pips behind you should drop (.12 x 60 = 8 rounded up, you bonehead).
>Hopefully, the preceding is not what you wanted to know, or you really are a
>bonhead, and it is a wonder that you even figured out how to use a keyboard.
>If there was more to your question than that, read on.
Heheheh! I guess I really am a bonehead then. What's your point??? Heheheheh
Anyway, from one bonehead to another, thanks for the info.
Regards,
Jeff
Mogath3
Thanks for the information and the link to the nifty site. I never really
thought all that much about the count and just kind of winged it. It is a part
of my game that I really would like to improve. Maybe my goal of being able to
get a count at little more than a glance may be unrealistic, but I'd like to
give it a try. Thanks again.
Regards,
Jeff
Donald Kahn
wrote:
>Hi,
>
>Why wouldn't the redouble be the same as the double, i.e. 10%-2?
>
>
Think about it this way: if you own the cube you only have to get to
~78% winning in order to "win" the game, because you don't care if
opponent takes or passes. If you do not have access to the cube, you
have to get to 100%. Big difference.
dk