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Re: [Bug-gnubg] Missed doubles problem ?


From: Holger
Subject: Re: [Bug-gnubg] Missed doubles problem ?
Date: Mon, 01 Sep 2003 01:13:55 +0200

At 21:44 29.08.03 +0000, Joern Thyssen wrote:
>However, there is still the point about modifying the logic to use DP
>and TG instead of the crude "<= 0.95" logic.

I'm on my way to implement this.

Here's something I want to ask:

As an example in the same game from MaX, his 8th roll:

Cube analysis
2-ply cubeless equity  +0,908 (Money:  +0,840)
  0,852 0,149
0,002 - 0,148 0,013 0,000
Cubeful equities:
  2-ply cubeful 100% speed
[world class]
1. Double, pass         +1,000
2. Double, take         +1,180
 ( +0,180)
3. No double            +0,959  ( -0,041)
Proper cube action:
Double, pass

In MWC this is:
2-ply cubeless MWC  82,02% (Money:  +0,840)
1. Double, pass
83,08%
2. Double, take         85,17%  (  2,09%)
3. No double
82,61%  ( -0,47%)

Looking at the market window MaX's DP is 50,7% (for dead cube), but when
opp. redoubles it's 81,4% (which gnubg will do if it takes because score is
2-away - 4-away)
TGP is 88,9%.

This leads to:
DP1: 50,7% < Middle1: 69,8% < NoD: 82,6% < DPass: 83,0% < DTake: 85,2% <
TGP: 88,9%

DP2: 81,4% < NoD: 82,6% < DPass: 83,0% < Middle2: 85,15% < DTake: 85,2% <
TGP: 88,9%

Unfortunately, I still don't know enough bg theory. So what I'm not sure
about is which window to take to compute its middle and after I know the
middle, which cube action to compare it with.

I tend to prefer the 2nd window in this example, because the cube isn't
dead at this score. Either gnubg drops or it takes and redoubles. If this
is correct, then how can I determine whether to take aaarPointsMatch[ 1 ][
1 ][ 0 ] or aaarPointsMatch[ 1 ][ 1 ][ 1 ] ?  Also, what does a value in
aaarPointsMatch[ 1 ][ 3 ][ 1 ] mean if there is one? The logic is this:
    if ( ! afDead[ i ] )
      aaarPoints[ i ][ 3 ][ 1 ] = arCP2[ i ];

The current code in updateStatcontext() always compares NoDouble with
+0.95eq. But I'm wondering if maybe the equity of the current correct cube
decision is better. (to compare it with the above computed middle)

Regards,

        Holger




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