Backgammon, and the frequency of dice rolls
I’ve just recently started to get back into playing Backgammon after a spell of a year or so.
One of the rules of backgammon allows your opponent to send a counter to effectively the start of the board. This can only happen if you have a single counter on a location (or spike in BG terminology). It is an obvious strategy to ensure that you move pieces in pairs, so no piece is left vulnerable in this way.
However due to the random nature of dice, it is often the case that a calculated gamble must be made as to what counter to leave open. The strategy is that it is better to leave a piece nearer the start of the board (less to lose if it goes wrong). The other is based on the frequency of certain roles of the dice.
If I make a move that leaves a piece open which is 12 spikes away from my opponents nearest piece I know that there is a small chance that they will role two 6’s is knock me back to the bar (start of the board). Leaving a piece seven spikes away open is not such a good idea as 7 is one of the most common roles with two dice.
Since I thought it would be handy to see the distribution of these rolls, I produced a quick table.
| Roll | Percentage | Frequency |
| 6 | 14.58% | 7 |
| 5 | 12.50% | 6 |
| 7 | 12.50% | 6 |
| 4 | 10.42% | 5 |
| 8 | 10.42% | 5 |
| 3 | 8.33% | 4 |
| 9 | 8.33% | 4 |
| 2 | 6.25% | 3 |
| 10 | 6.25% | 3 |
| 1 | 4.17% | 2 |
| 11 | 4.17% | 2 |
| 12 | 2.08% | 1 |
There is a slight scew to the first 6 numbers, since the table is based on the fact that to roll a combination of 3 the following could take place (where * denotes any value).
| Die 1 | Die 2 |
| 1 | 2 |
| 2 | 1 |
| 3 | * |
| * | 3 |
For a combination below six, we have to be aware that the face value of 1 die and the combination of the two should be taken into account.
Of course this approach is a simple one at best, and there is much more factors in place to determin the correct move.
February 4th, 2007 at 5:33 pm I think your calculations are incorrect. It may be my understanding of what you are trying to show. I think 11 is equally common as 12 and 1 should be much, much more common. The way I am thinking, all the rolls less than 7 would be more common than those requiring two dice, though I can’t immediately prove either completely. A roll of 11 must be a 5-6 or 6-5, or two possibilities of the 36 possible outcomes. A roll of 12 can be 6-6 or 4-4, also 2 of 36. Wouldn’t they have equal probability? Your chart, however shows 10 as more likely than 1. A roll of 10 can be 6-4, 4-6, or 5-5, but a move of 1 can result from any of the 11 outcomes with a 1 showing, 1-* and *-1, minus 1-1 so it isn’t counted twice.
February 5th, 2007 at 6:54 pm Your right of course. Bit of an embarrassment really, but it appears that I forgot about doubles counting twice. Will re-calculate when I get a moment. Oliver