doubling window lower ledge
Robert Koca
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From: "Robert Koca" <ko...@bobrae.bd.psu.edu>
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Date: Fri, 22 Sep 1995 19:25:19 -0400
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Subject: doubling window
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In a couple recent posts from Kit Woolsey and Michael Zehr I saw a discussion
of how to calculate the lower pont of a "doubling window" for match play
and that if do not have at least these chances then a double is theoretically
wrong.
I've seen others make this statement also but it is false.
The obvious counterexample is at 2 away 2 away score.
Another more esoteric example was shown to me by Asger Kring (Albatross)
Suppose X is losing 5 away 2 away and has 3 on ace point and opponent
has a checker on 3 and a checker on 4.
Doubling window techniques give 65% as lower ledge and cubeless % in this position
is 61% so it would say no double.
However in this position doubling is correct. The key point is that O is good
enough to double after X's non doubles anyways so X should activate the cube for his winning rolls.
,Bob Koca
(bobk on FIBS)
>From koca Fri Sep 22 19:25:24 1995
From: "Robert Koca" <ko...@bobrae.bd.psu.edu>
Message-Id: <95092219...@bobrae.bd.psu.edu>
Date: Fri, 22 Sep 1995 19:25:19 -0400
X-Mailer: Z-Mail (3.2.0 26oct94 MediaMail)
To: rec-games-backgammon
Subject: doubling window
Mime-Version: 1.0
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In a couple recent posts from Kit Woolsey and Michael Zehr I saw a discussion
of how to calculate the lower pont of a "doubling window" for match play
and that if do not have at least these chances then a double is theoretically
wrong.
I've seen others make this statement also but it is false.
The obvious counterexample is at 2 away 2 away score.
Another more esoteric example was shown to me by Asger Kring (Albatross)
Suppose X is losing 5 away 2 away and has 3 on ace point and opponent
has a checker on 3 and a checker on 4.
Doubling window techniques give 65% as lower ledge and cubeless % in this position
is 61% so it would say no double.
However in this position doubling is correct. The key point is that O is good
enough to double after X's non doubles anyways so X should activate the cube for his winning rolls.
,Bob Koca
(bobk on FIBS)
John B Kelly
>Suppose X is losing 5 away 2 away and has 3 on ace point and opponent
>has a checker on 3 and a checker on 4.
>Doubling window techniques give 65% as lower ledge and cubeless % in this position
>is 61% so it would say no double.
>However in this position doubling is correct. The key point is that O is good
>enough to double after X's non doubles anyways so X should activate the cube for his winning rolls.
>,Bob Koca
>(bobk on FIBS)
I don't understand why w/ 4 and 3 one would double. You win less than
1/2 the time (17/36). Explain please.
--
John Bryan Kelly jbr...@world.std.com
--------------------------------------------------------------------
Kit Woolsey
: >From koca Fri Sep 22 19:25:23 1995
: Received: by bobrae.bd.psu.edu (940816.SGI.8.6.9/930416.SGI.AUTO)
: for rec-games-backgammon id TAA05882; Fri, 22 Sep 1995 19:25:23 -0400
: Return-Path: <koca>
: From: "Robert Koca" <ko...@bobrae.bd.psu.edu>
: Message-Id: <95092219...@bobrae.bd.psu.edu>
: Date: Fri, 22 Sep 1995 19:25:19 -0400
: X-Mailer: Z-Mail (3.2.0 26oct94 MediaMail)
: To: rec-games-backgammon
: Subject: doubling window
: Mime-Version: 1.0
: Content-Type: text/plain; charset=us-ascii
: In a couple recent posts from Kit Woolsey and Michael Zehr I saw a discussion
: of how to calculate the lower pont of a "doubling window" for match play
: and that if do not have at least these chances then a double is theoretically
: wrong.
: I've seen others make this statement also but it is false.
: The obvious counterexample is at 2 away 2 away score.
: Another more esoteric example was shown to me by Asger Kring (Albatross)
: Suppose X is losing 5 away 2 away and has 3 on ace point and opponent
: has a checker on 3 and a checker on 4.
: Doubling window techniques give 65% as lower ledge and cubeless % in this position
: is 61% so it would say no double.
: However in this position doubling is correct. The key point is that O is good
: enough to double after X's non doubles anyways so X should activate the cube for his winning rolls.
: ,Bob Koca
: (bobk on FIBS)
Good point, Bob. Essentially the equities have to be redefined to take
into account the opponent's doubling potentials at the match score. This
might be somewhat complex, but I think if it is done correctly then the
statement would be correct. What I was trying to do in my posting was to
show that being in the doubling window was not sufficient to make
doubling correct, which is a common misconception.
Kit
Ron Karr
>ko...@bobrae.bd.psu.edu (Robert Koca) writes:
>
>>Suppose X is losing 5 away 2 away and has 3 on ace point and opponent
>>has a checker on 3 and a checker on 4.
>>Doubling window techniques give 65% as lower ledge and cubeless % in this position
>>is 61% so it would say no double.
>
>>However in this position doubling is correct. The key point is that O is good
>>enough to double after X's non doubles anyways so X should activate the cube for his winning rolls.
>
>
>>,Bob Koca
>>(bobk on FIBS)
>
>1/2 the time (17/36). Explain please.
>--
>John Bryan Kelly jbr...@world.std.com
>--------------------------------------------------------------------
To explain:
This is an interesting quirk of match cube strategy, where it can be
correct for the leader in a match to double when an underdog in the
game!
To see this, we have to look at the match equities involved. The score
is 2 away 5 away. The relevant equities are:
O doubles & wins: 1.00
no double/ win .85 (-1:-5)
no double/ lose .68 (-2:-4)
double / lose .60 (-2:-3)
Unlike in a money game, where the potential gain and loss from doubling
are equal, O stands to gain almost twice as much as he
stands to lose by doubling.
O's gain from doubling is 1.00 -.85 = .15. His loss from doubling is
68 - .60 = .8. This means that it can be correct for O to double if
his winning chances are at least 8/(8+15) = .35. In other words, the
"doubling window" begins at 35%. Since this is the last roll of the
game (X has no "redouble equity"), it is correct to double.
Ron
(FIBS:ronkarr)