Backgammon Bear Photo The Backgammon Bear Sherlock Holmes
Backgammon
in Austin
Analysis
Blunders
Problems
Links
About the
bear
Mail the
bear
Home >>> Analysis >>> 01/31/2005 1 previous next
Wolfram (Green) vs. Jay (White)
7 point Match

Game 01
Match to 7. Score Green-White: 5-4


Move 1 White





 
Pip: 101
Game 1
7 point match

Green-White:
Score 5-4
Pip: 58
 

# Ply Move Equity
* 1 M 13/9 13/10 -0,102
    0,2%   7,8%  43,5%    56,5%   2,9%   0,1%
    Live cube rollout: -0,105
  2 M 13/6 -0,102 (-0,000)
    0,2%   8,1%  42,8%    57,2%   3,7%   0,1%
    Live cube rollout: -0,100
  3 M 13/10 5/1 -0,128 (-0,027)
    0,2%   6,9%  42,4%    57,6%   2,9%   0,0%
    Live cube rollout: -0,096
  4 2 13/9 4/1 -0,408 (-0,306)
    0,2%   3,7%  36,8%    63,2%   6,7%   0,2%
  5 2 13/9 5/2 -0,426 (-0,324)
    0,1%   3,7%  36,1%    63,9%   6,2%   0,2%

Checker play
All reasonable plays are almost equivalent, you almost cannot go wrong here.


Move 2 Green





 
Pip: 94
Game 1
7 point match

Green-White:
Score 5-4
Pip: 58
 

# Ply Move Equity
* 1 1 13/3 -0,176
    0,1%   4,0%  48,6%    51,4%   7,8%   0,3%







 
Pip: 94
Game 1
7 point match

Green-White:
Score 5-4
Pip: 48
 

  White Double
3. Green Pass

Cube action equityAlert: Wrong pass
3-Ply Money equity: 0,292
  0,4%  14,7%  59,6%    40,4%   5,1%   0,2%
1. Double, take 0,636
2. No double 0,278 (-0,358)
3. Double, pass 1,000 (+0,364)
Proper cube action:   Double, take


Cube action
Easy double, easy take - but I dropped after thinking about this for a long time!
I tried to figure out this problem by estimating what will happen. There are 24 hitting numbers for him. They range from the perfect 2-2, hitting and closing the board to 5-1, which does not accomplish more than just hitting. After being hit, I will be in a very difficult position, even if I hit back. I have no board and need to get two men around. There is also significant gammon danger, which would result in a match win for White. My estimat was: 3 gammons for White, 18 simple wins, 3 wins for me. Those 18 wins would lead to White leading 6 to 5 in the match and making him a 70% favorite. Thus I would win the match in about 5.4 out of the 18 rolls. The 12 rolls where he does not hit seem to be relatively equal. He will always be able to keep his par point block and have a good chance af getting another chance (only 1/1, 2/2, 4/4, 5/5, 4/1, 4/2, 5/1 do not leave a shot after my next roll. How many of those 12 games, where he does not hit do I need to win? Well, if I drop it is 5 to 5 and I have a 50% chance of winning. Given my earlier evaluation I will win about 8.5 of his 24 hitting rolls. In order to get an even chance I will have to win 9.5 out of the 12 non-hitting rolls. That seemed like a stretch to me, even though I thought I might win close to 75% of these, and I dropped.
Where is the flaw in this argument? I badly miscalculated the number of wins after his hitting numbers. he gets 99% for 2-2 and about 90% for rolls, that hit and cover the 5-point. For rolls that do not cover any point though, he wins only 65% to 70%. On average I will win close to 20% of those games, i. e. 7 out of 24 hits. The number of gammons is also higher (let's say 5). That leaves 12 wins for him, out of which I can expect 30% match wins (3.6). My estimate of winning 75% of the non-hits seems ok. Thus I will achieve 7 + 3.6 + 9 = 19.6 wins out of 36. That is significantly better than 50% and makes the drop a big blunder.

This file has been generated by Snowie Professional Edition Version 4.1, a product of SnowieGroup SA
Output date: 19:45:16, 01.02.2005(Export v2.10)