Calculating probabilities

I'm glad to see there were no problems with the positions from the last time.
The only "trick" was to notice the cube has been turned in 1st and 3rd example, so those positions are no redoubles (but they would be initial doubles).
Second example was double/pass.

Before we continue to our next problems, couple of words about backgammon probabilities.





As you probably know, backgammon is played with two six-sided dice, so there are 6x6 = 36 possible outcomes.

It is illustrated in the image below, with one white and one blue die:


(Image from bkgm.com)

As you can see, each double (e.g. 1-1) can appear only once (1 on the white die + 1 on the blue die). Each non-double (e.g. 2-1) appears twice (as white-2 and blue-1, or as white-1 and blue-2).

When calculating probabilities (for entering from the bar, for hitting a blot, for bearing off, for escaping a prime, etc.) you must calculate how many rolls (of 36 total) are good for your goal, or how many rolls fail to make your goal.


Fortunately there are some shortcuts worth remembering:

1. When you need one number, 11 rolls are good (slightly less than 1/3)

is Player 2 score: 0 pip: 70

Unlimited Game

pip: 78 score: 0 is Player 1

XGID=-BBBBCC-----a----a-ccabbbA:1:1:1:00:0:0:0:0:10

You are on the bar and you need to roll a four to enter from the bar.

You can calculate this with several methods:
a) counting all good rolls - 41, 42, 43, 45 and 46 (all counting twice, because they are the same as 14, 24, 34, 54 and 64), plus 44 counting once = 2x5 + 1 = 11 rolls (you can use the above image to help you counting)
b) counting bad rolls and subtracting them from 36 - if you remember from the earlier post, for every closed point in the home board you square that number to get the number of rolls that dance - here it would be 5x5 = 25 rolls dance, 36-25 = 11 rolls that enter

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2. When you need two numbers, 20 rolls are good (slightly more than 1/2)

is Player 2 score: 0 pip: 58

Unlimited Game

pip: 43 score: 0 is Player 1

XGID=-BBBB--------------cdcbAb-:1:-1:1:00:0:0:0:0:10

One of your checkers was sent back and now it is trapped in your opponent's home board. You need to roll either 5 or 6 to escape.

Again you can count this with two methods:
a) counting all good rolls - 51, 52, 53, 54 and 56 counting twice, 55 counting once, and 61, 62, 63 and 64 counting twice (65 is already included from earlier!) and 66 counting once. We have (2x5+1) + (2x4+1) = 11+9 = 20 rolls that escape
b) counting bad rolls and subtracting them from 36 - it is similar to the coming in from the bar - four small numbers don't escape so we have 4x4 = 16 bad rolls, 36-16 = 20 rolls that escape

Try to figure out how many good rolls there are if you need to roll one of three good numbers and one of four good numbers.
I recommend you to use a pen and a paper, and to have the image from the top in front of your eyes to help you counting.

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In previous posts we talked about doubling in straight races when bearing in.
Let's say you doubled, your opponent took the double and the game continues. We arrive at this position:

is Player 2 score: 0 pip: 5

Unlimited Game

pip: 3 score: 0 is Player 1

XGID=-AA-------------------aa--:1:-1:-1:00:0:0:0:0:10

on roll, cube action?

It's his turn and he decides to double.

For correct take, you need to have 25% winning chance. (I will explain why 25% in the next post)

Can you take this, or is it better to drop? Please explain your choice.