The easiest way to understand backgammon probabilities
is to first look at all of the possible rolls of the dice.
With two dice, there are 36 possible rolls, as listed
in the table below:
1st
die |
2st
die |
Total
(doubles count 2x) |
1 |
1 |
4 |
1 |
2 |
3 |
1 |
3 |
4 |
1 |
4 |
5 |
1 |
5 |
6 |
1 |
6 |
7 |
2 |
1 |
3 |
2 |
2 |
8 |
2 |
3 |
5 |
2 |
4 |
6 |
2 |
5 |
7 |
2 |
6 |
8 |
3 |
1 |
4 |
3 |
2 |
5 |
3 |
3 |
12 |
3 |
4 |
7 |
3 |
5 |
8 |
3 |
6 |
9 |
4 |
1 |
5 |
4 |
2 |
6 |
4 |
3 |
7 |
4 |
4 |
16 |
4 |
5 |
9 |
4 |
6 |
10 |
5 |
1 |
6 |
5 |
2 |
7 |
5 |
3 |
8 |
5 |
4 |
9 |
5 |
5 |
20 |
5 |
6 |
11 |
6 |
1 |
7 |
6 |
2 |
8 |
6 |
3 |
9 |
6 |
4 |
10 |
6 |
5 |
11 |
6 |
6 |
24 |
OK, here are some points
to note regarding the table:
(1) There are 36 possible combinations of the dice. Remember
that fact.
(2) Each combination of two different numbers occurs twice--once
with the first number on the first die and the second
number on the second die, and another time with the second
number on the first die and the first number on the second
die. For example, the combination of 6-1 also occurs as
1-6.
(3) Each combination of the same number on two dice occurs
only once in the table. For example, 1-1, 2-2, 3-3, etc,
each occur only once.
(4) A probablilty for any given roll can be computed as
a simple fraction of the total number of possible ways
to get that roll, divided by the total number of rolls.
For example there are two ways to get a combination of
a 6 and a 1. The probability therefore is 2/36 (or 0.05555,
which also is 5.555 percent).
(5) As indicated in the previous point, the probability
of getting ANY specific pair of two different numbers
is 2/36.
(6) As indicated by point 3 and point 4, the probability
of getting a any specific double is 1/36. For example,
the probability of getting double 1s ones 1/36, the probability
of getting double fours is 1/36, etc.
(7) Important point here! If your eyes have glazed over,
WAKE UP. A point to be "hit" in backgammon that
is a total of 6 or less from the piece to be moved is
called a "direct shot". This is because the
numbers 1 through 6 occur on each die, and there are are
lots more combinations of die that can hit 1 through 6,
than points 7 or more away, which are referred
to as "indirect shots". To illustrate the probabilities,
lets look at the number of possibilities of hitting your
opponents blot (a blot is a single piece on a point) that
is four points in front of your piece. The rolls that
can hit 4 away, assuming intermediate points are not blocked
are: 1-4, 4-1, 2-4, 4-2, 3-4, 4-3, 4-4, 4-5, 5-4, 4-6,
6-4, 1-1, 1-3, 3-1, 2-2. The total number listed here
is 15, so the probability is 15/36 or about 42 percent.
Note the first 11 combinations
include a 4 on at least one of the dice. The last four
combinations rely on 4 being the total or an intermediate
place where the dice will allow the piece to land. The
probability of ANY direct shot is at least 11/36. Start
with that number and then add any combinations of two
numbers that add up to the direct shot number. Note that
a direct shot one
point away has no combinations that can add up to 1, so
the probability of hitting 1 away is 11/36.
(8) The probability of rolling a particular number or
set of numbers has nothing to do with numbers you got
on any previous roll. Each roll is independent, assuming
that the dice are fair (or computer dice algorithm is
fair).
QUIZ
For these questions, Assume that the dice are fair.
(1) What is the probability
of hitting your opponents blot, 7 points in front of your
piece, with no intervening points blocked? (Hint: add
up all of the combinations in the table with a total of
7)
(2) You have one piece on the bar and your opponent has
made all of the points in his home board except the 2
and the 5 points. What is the probability that you come
in on then next roll? (hint: add up all of the rolls in
the table above that have a 2 or a 5 on one die or both
dice.
(3) Your opponent, a_sly_guy, has rolled double 6s on
his two previous rolls. what is the probability that he
will roll a third pair of 6s on his next roll.
(4) You get the opening roll of a game and it is a 3-2.
You decide to play two pieces around the corner to the
10 and 11 points of your "outer board". These
two blots will be 9 and 10 points away from your opponents
back 2 pieces on your 1 point. Assuming your opponent
can't land on your 6 or 8 points (these points are "made"
with the original board setup), what is the probability
of your opponent
hitting one of the 2 blots?
Here is a link to download a FREE light version of the
Jelly Fish Backgammon Program
