BACKGAMMON - Probabilities with Dice

BACKGAMMON
Probabilities with Dice

In Backgammon a player's men are often exposed to the possibility of being hit. A good player chooses where and when to expose such a man according to the degree of risk involved, taking into account the probable outcome of his opponent's next throw and the player's own chances of immediate re-entry from the bar if such a hit should occur. A bad player simply trusts to luck.

THE OUTCOMES OF A THROW

Each die has 6 faces, such that any one of 6 different numbers of pips may show when thrown. Therefore with 2 dice the total number of possible outcomes that can occur when thrown is 6 x 6, or 36 combinations, as follows:

Outcomes with 2 dice
1+1 2+1 3+1 4+1 5+1 6+1	1+2 2+2 3+2 4+2 5+2 6+2	1+3 2+3 3+3 4+3 5+3 6+3	1+4 2+4 3+4 4+4 5+4 6+4	1+5 2+5 3+5 4+5 5+5 6+5	1+6 2+6 3+6 4+6 5+6 6+6

From the table, it can be seen that:

1.	For any one doublet (for example 6-6), only one combination gives the result required. Therefore the chances of rolling a particular doublet are 1 in 36.
2.	For any unmatching pair of numbers (for example 3 and 4), there are always two combinations which will give the result required (ie. 3+4 and 4+3). Therefore the chances of rolling any two particular different numbers together are 2 in 36.
3.	For any one specific number (eg. 1), there are eleven outcomes which include that number (ie. 6+1, 5+1, 4+1, 3+1, 2+1, 1+1, 1+2, 1+3, 1+4, 1+5, 1+6). Thus the chances of a roll that includes any one particular number are 11 in 36.

THE CHANCES OF ENTERING FROM THE BAR

To enter from the bar, a number must be thrown matching an open point on an opponent's inner table. The probability of immediate success is governed by the number of (a) open points and (b) men on the bar.

One's chances increase with the number of open points at a rate easily calculated, for just as the total of dice combinations that include one specific number showing on at least one die is 11, so the total of those which include either one of two specific numbers is 20 (being 11+9 more), that for any one of three numbers is 27 (being 11+9+7 more) etc. Each addition is equivalent to "2 less than the previous". This can be tabled as:

No. of Open Points	No. of combinations which enable entry from the bar for:
No. of Open Points	One man	Two men
1 2 3 4 5 6	11/36 20/36 27/36 32/36 35/36 36/36	1/36 4/36 9/36 16/36 25/36 36/36

For two or more men on the bar, the total of useable dice combinations that will bring in two men at once is equal to the no. of open points, squared (for example with 4 points, they are (4)²=16 out of 36 possible).

Another quick way of calculating for one man on the bar is to know that the total of useable combinations is (36 minus the no. of closed points squared), eg.
with 1 open point, the total is 36-(5 closed)²=36-25=11 out of 36
with 4 open points, the total is 36-(2 closed)²=36-4=32 out of 36

SUCCESSFUL PERCENTAGE PROBABILITY

Overall, for any situation:

the chances of a successful roll = (the total of useable combinations/36) %

And whenever the useable combinations are greater than 18/36, then the probability is more than 50% that the outcome will be successful - in other words the player will know that there is a more than even chance of an immediately successful roll and that over a sufficient period of time (given several such occurrences in every game) successful outcomes must occur more often than not.

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