Frank’s Position Analysis — September 2005
by Frank Frigo
The Situation:
- match to 9 points
- white has 2 (Roger Hickman)
- blue has 4 (Brandon Debes)
- white on-roll
- pip count: white = 23, blue = 29
This bear-off problem is submitted by Brandon Debes from one of his matches during the recent 53rd Indiana Open held over Labor Day weekend in Indianapolis.
To assess this decision properly, both Brandon and his opponent have multiple considerations:
- What is the match score? And how does this doubling decision differ from a standard “money” doubling decision?
- What are blue’s (and white’s) winning prospects if this game is played to completion?
- Does the cube have any future value to the doubler or the taker?
- If the decision is close, should the players be considering any skill differentials?
First let’s assess all of the possible match scores that affect the decision. For those who haven’t memorized a match equity table, a simpler method exists for calculating equities over the board. Several years ago, Neil Kazaross developed the “Neil’s Numbers” shortcut. It is really easy to use once you have committed a few number correlations to memory. Here’s how it works:
| Trailer’s points needed | 3 | 4 | 5 | 6 | 8 | 11 | 15 |
|---|---|---|---|---|---|---|---|
| Corresponding multiplier | 10 | 9 | 8 | 7 | 6 | 5 | 4 |
To assess the match equity for the leader, determine the number of points the trailer needs to win the match, find the corresponding multiplier, and multiply that number by the score differential. Then simply add that total to 50%.
In this case, if Brandon passes the double he will be trailing 6-4 to 9. Therefore, as the trailer he will be 5 points from victory. The corresponding multiplier is 8. (You may interpolate for numbers not posted in the sequence.) The score differential is 2. This gives the leader a MWC of (8 × 2) + 50 = 66%. Conversely, Brandon would have a 34% MWC. This number may vary slightly from some of the match equity tables, but for practical purposes it works just fine.
To determine the trailer’s take point we also need to know what happens when he takes and wins. That’s easy since the 8 cube will put him over the 9 points needed to win the match. Thus, 100%.
The opposite is true if he takes and loses, as the 8 cube will now put his opponent over the 9 points. Thus, 0%.
Therefore, Brandon risks 34% to gain 66%. His take point is simply 34%. This means he should accept the double when he believes he can win the game more than 34% of the time and pass otherwise. If the cube had some future value to Brandon, there would be other considerations on the take point. But here it is a dead cube at the 8 level.
The risk/reward analysis for his opponent’s doubling decision is a bit different. If he doesn’t double and wins, he will be leading 6-4 (again 66% MWC). If he doesn’t double and loses he will now be trailing 8-2, Crawford. It is better to try to memorize the MWCs around Crawford scores as Neil’s Numbers don’t work as well in these situations. The trailer will need to win a maximum of 4 games (fewer if he gammons along the way). For 7-away, Crawford, I use a MWC of 9% for the trailer.
From the doubler’s perspective he gains 34% (100% − 66%) when he doubles and wins, and risks 9% when he doubles and loses (9% − 0%). This is a fairly attractive proposition for letting the cube loose. Putting the doubling point at the surprisingly low level of 9 / (9 + 34) or about 21%!
Our doubling window has now been established. The doubler can begin to consider redoubling at about 21%, and Brandon can take if he is no worse than 34%. From the doubler’s perspective his window is 21% – 66%. This differs greatly from a typical money doubling window (no gammons considered) of 50% – 81%. I say the doubler can “consider doubling” at 21% because, in fact, he may do better if he waits. Backgammon is a very dynamic and volatile game, and each roll is a new decision. If this were the last roll of the game, then all we would need to consider is whether we were above the doubling point. When it isn’t the last roll of the game (as is the case here), we must assess where we lie within the doubling window and how volatile the position is. The likelihood that we “lose our market” (i.e. exceed the take point before our next roll) can greatly affect a cube decision.
Now let’s see exactly where this position lies within the established doubling window. To do this, we must determine the actual game winning chance of the position. This problem is particularly tricky because of the somewhat unusual configuration of the checkers (gaps, extras, etc.). A wonderful formula developed by Jeff Ward more than 20 years ago can be used to assess these types of positions fairly accurately. For more on this type of analysis, I strongly suggest getting a copy of Backgammon Boot Camp by Walter Trice. Walter begins to address the basics of these situations on pages 144-146.
Here is how the Ward Formula works:
- Start with the pipcount.
- If a player has more checkers on the board than his opponent, add 2 pips for each additional checker.
- If there are more than 2 men on the 1-point, add 2 pips for the third and each additional checker.
- If there are more than 2 men on the 2-point, add 1 pip for the third and each additional checker.
- If a player has fewer vacant home board points than his opponent, subtract one pip for each extra occupied point.
- Add 1/2 pip for each extra checker outside the home board.
- Subtract 4 pips for being on roll.
In this problem the actual pip counts are 23 for the doubler (white) and 29 for Brandon (blue). Using the method outlined above, we can determine the adjusted pipcount for each player. White has 3 more checkers than Brandon (add 6). White has 3 extra checkers on the 1-point (add 6). White is on-roll (subtract 4). Therefore, his adjusted pipcount is 23 + 6 + 6 – 4 = 31.
Brandon has one additional checker on his 2-point (add 1), and none of the other conditions apply. Therefore, his adjusted pipcount is 29 + 1 = 30.
Now for the slightly trickier calculation. To determine GWC we use the following formula (I believe this was originally developed by Danny Kleinman):
D2 ÷ (A + B)
where:
- D = the adjusted pipcount differential
- A = player A’s adjusted pipcount
- B = player B’s adjusted pipcount
The resulting number is then found in the following table, and a corresponding GWC is determined:
| Above formula result | 0 | .05 | .13 | .30 | .55 | .91 | 1.42 | 2.15 | 3.3 | 7.0 |
|---|---|---|---|---|---|---|---|---|---|---|
| Game winning chances | 50% | 55% | 60% | 65% | 70% | 75% | 80% | 85% | 90% | 95% |
(Again, as in Neil’s Numbers, interpolations can be used.)
To get the GWC for our bear-off problem: D = 31 – 30 = 1, A = 31, B = 30 (resulting in 1/61 = .016).
Using interpolation we end up with approximately 51% GWC. Since Brandon has the lower adjusted pip count, the doubler is actually the underdog with only 49% GWC! Independent computer rollouts from gnubg confirm this position to be 50.5% GWC for Brandon.
Finally we have all the pieces to the puzzle. This position lands well inside the doubling window:
Conclusion: The take is absolutely clear. In fact, declining this redouble would take Brandon’s MWC down to 34% from 51%, a whopping difference of 17%.
There are certainly many market losers for the doubler (double 4s, 5s, and 6s for starters, along with many other poor roll combinations for Brandon), and the position is, of course, well within the doubling window. Therefore, with respect to the volatility, the doubler is significantly improving his MWC by pushing over the 8 cube right now.
Finally, when a doubling decision is this clear, it is very unlikely that skill differentials should be considered. To put it in perspective, the more skillful player would have to believe his winning chances trailing 4-6 to 9 would exceed 51% (his GWC) as opposed to the 34% MWC figure we calculated for 2 equal opponents.
Redouble — Take.
– Frank Frigo