Annotated Game

Malcolm Davis vs. Chris Peterson
Finals of the B.S.O. Benefit Tournament

 
Bill Robertie, 1982

From Backgammon Times, Volume 2, Number 1, Winter 1982.

Over the weekend of October 30 to November 1, the New England Backgammon Club hosted its premier tournament, the Boston Symphony Orchestra Benefit, at the Hyatt Regency in Cambridge. Normally held in the spring, the tournament was moved to the autumn this year to avoid a possible scheduling conflict with the Turnberry Isle event.

The Championship Division was won by Malcolm Davis of Texas who made his first appearance in the New England area. Davis beat Chris Peterson, formerly of Boston but now residing in New York, in the finals. Gino Scalamandre and I were semifinalists.

The prize winners in the other sections were a collection of former champs. Ellen Jacoby-Lee, runner-up in 1979, beat Leslie Stone, the winner in 1980, in the consolation. Al Hodis, last spring's winner, and Kent Goulding were the semifinalists. In the Last Chance, Mike Valentine edged Doug Mayfield (runner-up last spring).

On his way to the championship, Davis put together one of the most amazing runs of good dice ever seen in a major tournament. In his 17-point match with Uli Koch in the round of eight, Davis trailed 16–4 entering the Crawford game. He then proceeded to:

  • take 13 points in a row from Koch to win the match, 17–16,
  • demolish Gino Scalamandre 19–3 in the semifinals, and
  • grab a 15–1 lead against Chris Peterson in the finals before coasting to a 21–7 win.

Although Peterson played well in the finals, Davis's dice never gave him a chance to get in the match. For analysis, I've selected the penultimate game of the finals, which began with Davis sitting on a hefty 16–7 lead.

  Davis (Black) Peterson (White)

1.     2-1: 13/11, 6/5
2.     5-2: 13/8, 13/11
12 11 10 9 8 7 6 5 4 3 2 1
 
   
13 14 15 16 17 18 19 20 21 22 23 24
White to play 6-1.

2.     . . . 6-1: 24/18, 6/5

An excellent play. Trailing 16–7 in the match, Peterson wants to create complex, double-edged positions with gammon possibilities. Moving to the bar point puts immediate pressure on Davis's position. If Davis hits on the bar, he won't be able to make an inner point and will be vulnerable to several return hits.

Compare the position after Peterson's play with the position that would have resulted if he had played 11/5, 24/23 instead:
12 11 10 9 8 7 6 5 4 3 2 1
   
 
     
13 14 15 16 17 18 19 20 21 22 23 24
After Peterson's play of 24/18, 6/5.
12 11 10 9 8 7 6 5 4 3 2 1
   
 
     
13 14 15 16 17 18 19 20 21 22 23 24
After alternate play of 11/5, 24/23.

Notice in this position that Davis's rolls of 3-1, 4-2, 5-3, and 6-1 are all much better for him than in the actual game position. He can use them to make inner points without worrying about his vulnerable blot on the 11-point.
13 14 15 16 17 18 19 20 21 22 23 24
 
   
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 5-4.

3.     5-4: 11/7*, 6/1*

One of Davis's worst rolls. He makes the best play possible.
12 11 10 9 8 7 6 5 4 3 2 1
 
 
13 14 15 16 17 18 19 20 21 22 23 24
Should White double?

3.     . . . Double to 2

Another good play. Objectively, this is not a particularly strong double. In a money game, for instance, it would be quite premature. When trying to come from behind in a match situation, however, handling the cube requires a bit of creativity and psychological insight. The basic strategy is to offer an early double at the precise moment when your opponent most wants (incorrectly) to pass.

Here Peterson guages the position perfectly. Right now, Davis's position looks scary. He has no points, two blots, back men still trapped on the 24-point, and his opponent already in possession of the crucial 20-point. A player could easily talk himself into dropping this cube. Next turn, however, Davis's game might look much stronger. If Peterson enters without hitting and Davis makes an inner point, or covers the bar and splits his back runners, he would have reached a position that almost no one would pass. Peterson correctly doubles at his optimal bluffing opportunity.

4.     Take

Davis is a very experienced player, and if anything he tends to err on the side of aggression rather than caution. He properly scoops up the cube.

4.     . . . 2-1: bar/24*, bar/23
5.     4-3: bar/21, 24/21 4-3: 24/20, 23/20
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 5-3.

6.     5-3: 8/3, 6/3

Picking up the blot with 8/3, 7/4 leaves Peterson free to maneuver in the outer boards without fear. Davis needs some teeth in his game and the 3-point is a start.

6.     . . . 6-6: 20/14(2), 13/7(2)
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 5-2.

7.     5-2: 21/14*

This is not quite as berserk a play as it might appear. Twos, threes, and fours are Peterson's only working numbers. Still, it doesn't appear to gain very much for the potential long-term risk of losing the anchor. I generally advocate making big plays to win the game even if gammon chances are increased. In this position, however, Davis remains the underdog even if his play works, which makes the extra risk hardly worth the trouble. I would move 13/8, 7/5.

7.     . . . 5-4: bar/20, 8/4*
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 5-1.

8.     5-1: bar/24, 13/8

14/9 looks like a more natural play with the 5.
12 11 10 9 8 7 6 5 4 3 2 1
 
 
13 14 15 16 17 18 19 20 21 22 23 24
White to play 3-1.

8.     . . . 3-1: 8/4

Another fine play. 14/11*, 14/13 is tempting, particularly since Peterson needs a gammon given the match score. However, sending the third checker back doesn't increase his gammon chances. In fact, the extra checker actually makes Davis more likely to form an advanced anchor. Peterson has the right idea: lock in the back checkers securely, make the win certain, then look around for a gammon.
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 6-3.

9.     6-3: 14/5*

Certainly no one can accuse Davis of faintheartedness. This play is tremendously optimistic: his chances of priming a checker of Peterson's are miniscule at best. Furthermore, if Peterson can somehow hit both loose blots, his gammon chances will be very real. Just 14/8, 7/4 looks simple and best.

9.     . . . 6-2: bar/23, 14/8
10.     5-1: 6/5, 7/2* 6-4: bar/21, 14/8
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 6-2.

11.     6-2: 24/16

Excellent. This move gives some real winning chances at the cost of a very slightly increased risk of a gammon. Only 15 numbers hit the blot, and Davis would then be a favorite to re-anchor before being closed out. This play is much stronger than the passive 8/2, 8/6.

11.     . . . 5-1: 21/16, 8/7
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 6-2.

12.     6-2: 24/16

Effortlessly, Davis moves his anchor from the 24 to the 16-point. This parlay (6-2, miss, 6-2), by the way, was 837 to 1 against.

12.     . . . 3-1: 16/13, 7/6
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 3-3.

13.     3-3: 16/10(2)

And now Davis takes the lead in the race! The pip count right now is Davis 108, Peterson 111.

13.     . . . 4-1: 13/9, 7/6
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 4-3.

14.     4-3: 8/4, 8/5

Good technique. Davis moves to fill the gap on his 4-point rather than stack more checkers on the 6-point.

14.     . . . 1-1: 8/6, 7/5
15.     3-2: 8/6, 8/5 6-5: 13/7, 13/8
16.     6-5: 10/4, 13/8 3-1: 8/5, 7/6
17.     6-2: 10/4, 13/11
12 11 10 9 8 7 6 5 4 3 2 1
 
 
13 14 15 16 17 18 19 20 21 22 23 24
White to play 5-4.

17.     . . . 5-4: 9/4, 8/4

Both sides have played the bear-in well, giving themselves a good distribution of builders across the 4, 5, and 6-points.

18.     1-1: 8/6, 11/9 3-2: 5/off
19.     4-2: 9/5, 2/off
12 11 10 9 8 7 6 5 4 3 2 1
 
 
13 14 15 16 17 18 19 20 21 22 23 24
White to play 2-1.

19.     . . . 2-1: 6/3

A slight inaccuracy. Since Peterson can't bear a man off anyway, he should even out his distribution on the higher points with 6/4, 6/5. This would give him plenty of fours and fives to play in the bearoff. His three will play efficiently from 6/3 in any case.

20.     5-3: 5/off, 3/off 5-4: 5/off, 4/off
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black to play 5-2.

21.     5-2: 5/off, 6/4

4/2 is equally good.

21.     . . . 4-2: 19/off
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Should Black
redouble to 4?

Davis has finally moved into the lead with a 52 to 55 lead in the pip count and the roll. Premature to double, of course, even for money.

22.     6-2: 6/off, 4/2 2-1: 3/off
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Should Black
redouble to 4?

Fortunes change quickly in the bearoff. Now, in a money game, Davis would redouble and Peterson would have to pass. At this score, it is still too soon to turn the cube back.

23.     6-1: 6/off, 4/3 5-3: 5/off, 6/3
24.     5-3: 5/off, 3/off 5-4: 5/off, 4/off
25.     Redouble to 4
13 14 15 16 17 18 19 20 21 22 23 24
 
 
12 11 10 9 8 7 6 5 4 3 2 1
Black redoubles to 4. Should White take?

This position merits close examination. Although Peterson has caught up by a couple of pips over the last two rolls, he still trails in the pip count by 29 to 35. With each side having only seven checkers left, this is too much of a discrepency to take a money double.

Here, however, we have a match situation where the notion of overage comes into play. Peterson has the option of taking and immediately redoubling to the 8 level. All 8 points are useful to Peterson, but only 5 of the 8 points are useful to Davis. The other 3 points are wasted for him, since he already has 16. Theoreticians refer to this situation as overage. When overage happens, the odds change rapidly from normal money play conditions. Let's see just what the correct strategy for this position really is.

First, let's decide if Peterson should take. Note that if he does take, he should immediately redouble to 8, since his chance of winning the match from a 20–7 deficit is minute (255 to 1 against, not counting gammons). To decide if he should take, we need to know three probabilities:

  1. The raw probability that he can win this position if it is played out to the finish.
  2. The probability that he could win the match if he dropped this cube (the score would then be 18–7, Davis's favor).
  3. The probability that he could win the match if he takes, redoubles, and wins (the score would then be 16–15, Davis).

To determine probability (1), I asked Kent Goulding to input the position to his bearoff computer program. After 11,000 trials, it revealed that Davis was an 84 to 16 favorite (all games played to a finish, no cube involved).

To determine probabilities (2) and (3), I referred to match probability charts. (The calculations to generate these charts are too tedious to explore here.) They revealed that an 18–7 lead is equivalent to a 97 to 3 likelihood of winning the match, while a 16–15 lead means a 58 to 42 likelihood of winning.

Knowing these figures, we can calculate Peterson's chances of winning the match in each case:

  1. Peterson drops. He trails 18–7 and his winning chances are 3%.
  2. Peterson takes and redoubles to 8. His probability of winning the match is the likelihood he wins this game (16%) times the probability he wins the match from 16–15 down (42%), or 7%.

So Peterson would more than double his chances of winning the match by taking.

Given that Peterson's best strategy is to take and redouble, what about Davis's double? Was it correct or not?

We know that Davis's winning chances if he doubles (and Peterson takes and redoubles) are 93%. It's no trivial matter to calculate his exact winning chances if he holds the cube, since his future cube actions depend on the position. Suppose, however, we compute his winning chances if he never doubles, no matter how good his position gets. Obviously, this is less than his optimal strategy, so his real winning chances must be higher than this number.

If he never doubles, his chances of winning the match are:

  1. the chance he wins the game (84%) times the chance he wins from an 18–7 lead (97%) = 81.4%, plus
  2. the chance he loses this game (16%) times the chance he wins from a 16–9 lead (87%) = 13.9%,

for a total winning probability of 95.3%. His real winning chances, using the cube effectively, are slightly higher than this, let's say around 96%. So by doubling from 2 to 4, he gives Peterson about a 50% better chance to win the match. It pays to be conservative with the cube when overage is involved. (Actually, this still isn't the whole story. A complete analysis of Davis's double would have to take into account two additional considerations: (1) the possibility that Peterson would incorrectly drop the double, and (2) the possibility that Peterson will take but fail to redouble next turn. These two possibilities make Davis's double much less incorrect than the pure mathematics would indicate. In fact, watch what happens.)

25.     . . . Take
26.     5-1: 5/off, 2/1 2-1: 3/off

Peterson neglects to double to 8, an incredible oversight on his part. Even though 6-6 is his only market-losing number, he must redouble now, since his chances for winning from 20–7 down are so slim. In fact, his previous take makes no sense without a redouble.

27.     6-4: 6/off, 4/off 5-2: 6/1, 6/4
28.     4-1: 4/off, 1/off 3-1: 1/off 6/3
29.     3-1: 3/off, 5/4 4-1: 4/off, 6/5

Peterson needs a 2-1 by Davis, followed by 6-6 or 5-5 on his part, a 323-to-1 long shot.

30.     6-3: Wins

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