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Bearoff Tips at DMP
by François Tardieu - 19 June 2007
François Tardieu
FRANÇOIS TARDIEU
François Tardieu of France, world's #3 Giant of Backgammon, provides valuable info on how to calculate the correct play while bearing off at Double Match Point (DMP) when a player must strive to maximize the chances of winning the game.

Figure 1 - DMP Red to play 6-1

Figure 1 is a common situation where you have two alternatives:

- minimize shots but bear off only one checker out by playing 5/0 5/4*


- take two men out but leave more shots by playing 5/0 1/0

In this article, we will discuss only the play at Double Match Point (DMP) where a player must maximize his chances of winning the game without considering the gammon effect.

Over the board it is not always easy to find the right play, especially nowadays where clocks are in use and a player must be prepared by doing some homework, thus saving crucial time for calculations like this over the board.

First of all to compute chances of each play a player must have some knowledge - a player must know how to count shots, the easy part… and then he must know what the chances are to win once he has been hit.

For example, in Figure 2 (below), Red has 8 men out and he is on the bar against a closed board where the three spares are on the 6, 5 and 4 point. This distribution of spare is called “the dream position” because it is the best way White can place his checkers in order to take the maximum off while retaining a strong board and thus inhibit Red from re-circulating his checker on the bar.

Computer rollouts indicates that with 8 men out, Red is a tiny favorite (51%) without the cube in play.

For the sake of simplicity, we examine the reference positions where White has a closed board and the dream position and Red has one checker trapped on the bar, some number of men off and the remaining of the checkers on the lower point (i.e. 1 and 2 or 1, 2 and 3).

After extensive rollouts we have the following tables with the cubeless winning chances for both sides.

% to win in reference positions

N off

1

2

3

4

5

6

7

8

9

10

11

12

Red

4.5

7

11

15

22

30

39.5

51

58

69

76.5

84

White

95.5

93

89

84

78

70

60.5

49

42

31

23.5

16

d

 

2.5

4

4.5

7

8

9.5

11.5

7

11

7.5

7.5

Those numbers must be learnt by heart, first of all because they represent very common situations in match play and they permit to derive assessments of similar positions; for example, when all checkers in Red’s home board are on the 2 point or when White have borne off all his spares… such positions might be the subject of another article…

For example, with 6 men out and one man on the bar, Red wins 30% and White 70% without considering any cube action. The last line of the table, d, indicates the extra chances added by taking out one more man. For example from 5 to 6 men out Red increases his chances to win by 8% from 22 to 30 and decreases his chances to loose from 78 to 70.

Let us go back to Figure 1...

.... and let’s propose the following parameters:

S = number of shots when minimizing shots and taking only one man out

L = chances to lose after minimizing shots and being hit

Y = number of extra shots left while bearing off two men

L’ = chances to lose after taking two checkers out and being hit

We can observe that L’ = L–d   where d is the gain in chances of winning to bear off an extra checker.

We can observe, from our table, that d varies a lot and its range is from 2.5% to 11.5%.

The biggest value of d is when we pass from 7 to 8 men out. Actually, with 7 men out we have 40.5% to win and with 8 men out we have 51%.

We compute the chances to lose when we minimize shots as follow:

Wm = S * L

And the chances to lose while taking two men out as follow::

Wo = (S+Y) * L’ = (S+Y) * (L-d)

 

It is justified to leave more shots when:

Wo > Wm <=> S*L > (S+Y) * (L-d)

This can be simplified as follow:

(I)             (S+Y)*d > Y*L

The inequality (I) means that the benefit of the extra checker out when hit must compensate the games lost because of the extra hits we give when taking two checkers out.

 

Let’s apply this to our example in Figure 1.

S = 17/36 (i.e. All 4, 3-1, 5-1 and 6-1 hits Red’s checker)

Y = 3/36 (i.e. All 1 and 2 hits Red’s checker)

L = 42 (because Red has 9 checkers out when minimizing shots)

And d = 11 (the benefit to go from 9 to 10 checkers out is 11%, the difference between 42 and 31)

So (S+Y)*d > Y*L

<=>

we forget the denominator of 36 and we have

(17+3)* 11 > 3*31 <=> 220 > 93

So (I) is verified and we conclude that we must take two checkers out leaving 20 shots instead of 17 in order to maximize our chances at DMP.

Actually, rollouts indicate Red wins after taking two checkers out 83.5% and only 80.7% if not.

If we go through all the possibilities for L, S and d, we can find some interesting rules of thumb to handle these positions.

Here are some that are easy to remember and will avoid some mistakes of evaluation over the board.

R1: when you have 5 or less checkers out whatever your play you must always minimize shots.

R2: once you have at least 9 checkers out, it is always right to take out an extra checker if you don’t leave more than 4 extra shots.

Here an extreme example to leave 4 more shots!

6-2: 6/0 2/0 Red leaves 21 shots and has 10 checkers out wins 82.2%

6-2: 6/0 6/4* Red leaves 17 shots and has 9 checkers out wins 80.8%

Here the parameters are: S = 17, Y = 4, L = 31 and d = 11

(S+Y)*d = 21*11 = 231   and   Y*L = 4*31 = 124

So we have (I) verified to justify to leave more shots.

R3: Between 6 and 8 checkers out it is always right to leave one extra shot to bear off an extra man.

Here Red gains about 1.5% in winning chances by bearing off two checkers.

For all others cases you have to compute the figures to find out the right play…

Below, we present five examples…

4-1: 5/1 5/4* or 5/4* 4/0

 

The parameters are:

 

S = 17, Y= 2, d = 7, L = 49

 

So we have (S+Y)*d = 19*7 = 113 and Y*L = 2*49 = 98

 

We have (S+Y)*d > Y*L so the right play is to bear off 1 checker with a gain of 2.5% at DMP..

6-2: 6/0 6/4 or 6/0 2/0

The parameters are:

S = 11, Y= 3, d = 8, L = 78

So we have (S+Y)*d = 14*8 = 112 and Y*L = 3*78 = 234

We have (S+Y)*d < Y*L so here minimize is the right play with a gain of 2.5% at DMP.

 

Here, we can apply R3 because d = 1 and we can have 6 checkers out. So, taking two checkers out is worth the extra shot with a gain of 1.35%.

Here we apply R2 and it is correct to bear off two men by more than 1%.

Here it seems that R2 might be applied but it leaves a blot on the ace point, a liability that White might exploit during his bear off by perhaps bearing off more aggressively because of this weakness.

 

Actually it is slightly wrong to apply R2 here and it is better to minimize  shots because we can reach positions like the following:

In the two above, White must, by a wide margin, play aggressively and bear off a checker from the ace point in the first position and bear off a man from the 2 point in the second in order to pick up a second Red’s checker. So to dodge these scenarios it is better to minimize shot when you leave a checker in your home which might be picked up later.

Here you have four contenders:

A: 1/0 1/0 1/0 1/0 leaving 15 shots and taking off 4 checkers

B: 1/0 1/0 1/0 6/5 leaving 14 shots and taking off 3 checkers

C: 1/0 1/0 6/4 leaving 12 shots and taking off 2 checkers

D: 1/0 6/3 leaving 11 shots and taking off 1 checkers

 

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