# Two dice but two rolls - probability

#### Uxinox

##### New member
I recently saw a table of probabilities in a book on backgammon, a game in which one has to remove checkers or men from the board. You have two "men" at position 6. You have two dice. What are the chances that you can win in a single roll? To win you would normally require two sixes and as we know, the chances of rolling two sixes are 1 in 36. But in backgammon, if you roll a double, it counts twice, so double 5 allows you to roll 20. With 20, you exceed the 12 required to move the two men at position 6, so a double 5 and a double 6 are winning rolls. It is clear, then, that double 4 and double 3 are also winning rolls, so the odds of winning in the above scenario are 4 in 36 or an 11% chance or odds of 1 in 9.

The table of probabilities then showed that the odds of winning if the player had TWO rolls instead of one was 78% and in another part of the book, that this represented 169/47, odds of around 3.5 to 1. I have been wracking my brains over this for some weeks and have finally sucumbed to the maths forum. Any smart guys or guyesses out there? How did they calculate the odds?

#### Uxinox

##### New member
An update

I recently saw a table of probabilities in a book on backgammon, a game in which one has to remove checkers or men from the board. You have two "men" at position 6. You have two dice. What are the chances that you can win in a single roll? To win you would normally require two sixes and as we know, the chances of rolling two sixes are 1 in 36. But in backgammon, if you roll a double, it counts twice, so double 5 allows you to roll 20. With 20, you exceed the 12 required to move the two men at position 6, so a double 5 and a double 6 are winning rolls. It is clear, then, that double 4 and double 3 are also winning rolls, so the odds of winning in the above scenario are 4 in 36 or an 11% chance or odds of 1 in 9.

The table of probabilities then showed that the odds of winning if the player had TWO rolls instead of one was 78% and in another part of the book, that this represented 169/47, odds of around 3.5 to 1. I have been wracking my brains over this for some weeks and have finally sucumbed to the maths forum. Any smart guys or guyesses out there? How did they calculate the odds?
Yes, this is a tricky one! I got the answer of 75% and want to know if I am wrong or the book is. I suspect it is me! The author is a former World Champion! I will show my workings and perhaps someone can spot a mistake! As mentioned, the problem occurs in the "bearing-off" stage of a backgammon game.

the table of possible outcomes with two dice, corrected for backgammon doubles if one starts with two men on position 6, will reduce a 6-6 after a single roll as follows: in other words, taking the first entry, if I roll 1-1, in backgammon this is 1-1 followed by another 1-1, so I can move my two men from position 6 such that both go to position 4. With the next entry in the table 1-2, I would move one man from 6 to 5 and the other from 6 to 4, leaving a result after the first throw of 5-4, and so on. "win" signifies that a 3-3 for example, wins outright.

4-4 5-4 5-3 5-2 5-1 5-0
4-5 2-2 4-3 4-2 4-1 4-0
3-5 3-4 win. 3-2 3-1 3-0
2-5 2-4 2-3 win 2-1 2-0
1-5 1-4 1-3 1-2 win 1-0
0-5 0-4 0-3 0-2 0-1 win

Now, after the second dice throw, the table of possibilities will be reduced to the following:

2-2 4-2 4-0 1-1 win win
2-4 win 2-0 win win win
0-4 0-2 win win win win
1-1 win win win win win
win win win win win win
win win win win win win

this shows that there are 27/36 possible wins after 2 rolls i.e 75%.