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?
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?