3

So I was playing a lot of backgammon lately, and I noticed something. It seemed like I was "going out" on doubles more often than not. And my initial reaction was "Shouldn't that be, at best, a 1 in 6 chance of getting doubles on that final roll?"

But then my math/statistics side kicked in and I realized it's probably slightly more complicated than that. So being out of practice, I figured I'd drop this hear to get a "proper" answer, and correct my math :) I also wasn't sure whether I should ask here, or over in : https://math.stackexchange.com/questions But figured I'd start here. If it get's too "mathy" perhaps I'll re-ask over there :)

Here's my thoughts so far:

our end states are:

5+ checkers
4 checkers
3 checkers
2 checkers
1 checker

Now with 5+ checkers left, we cannot go out, so whether we roll doubles or not is irrelevant, it will leave us with 1, 2, 3 or 4 checkers, at which point we could go out if:

1) 4 checkers:
   a) roll doubles of a value equal to or larger than the Runner (the checker farthest in)
         (ie if we have a checker on each of 2, 3, 4 and 5 point in your home, we need to roll double 5's or double 6's to go out)
2) 3 checkers:
   a) roll doubles of a value equal to or larger than the Runner (the checker farthest in)
         (ie if we have a checker on each of 2, 3 and 4 point in your home, we need to roll double 4's, 5's or 6's to go out)
   b) roll doubles of a value equal to half or more of the Runner, and equal or larger than the other 2.
         (ie if we have a checker on each of 2, 3 and 4 point in your home, we can go out with double 3's: half of 4 is 2. next largest is 3. so double 3's puts us out)
3) 2 checkers:
   a) roll doubles of a value equal to half or more of the runner.
         (ie if we have a checker on 2 and 4 point. We can go out with double 2's)
4) 1 checker:
   a) roll doubles of a value equal to a quarter or more of the runner.
         (ie if we have a checker on the 4 point, we can go out with double 1's)

I'm not sure these are all the cases .. but it's showing that there's a lot of variations .. and I'm not sure how to go about calculating all the variations accurately :)

So back to my question :

How to calculate the odds of going out by rolling doubles ? (let's ignore the chance your opponent goes out first at any time ... or interferes in any way ... )

  • As in all math problems, it is often best to start with concrete/simple examples, then start to generalize. I would approach this problem, by looking at the case where all the checkers are on the 1. (My back of the envelope gives a probability of going out with doubles of 11/42 in that case.) Then, I would start generalizing, and making sure my results were consistent with the above. My gut is that the probability in general is closer to 1/3 than it is to 1/6. It's definitely above 1/6. – John Apr 27 '15 at 16:24
  • 1
    If you like the math, here's an actual mathematical paper on the subject. – ghoppe Apr 27 '15 at 16:54
  • @ghoppe: Neat ... thanks. I browsed through it, I'll have to ponder it more deeply with alcohol this weekend ;) – Ditto Apr 27 '15 at 17:50
  • John's figure sounds close to what I'd intuitively expect - it's near 1/3 because in most games you'll have two attempts at finishing with a double, once when you have 3 or 4 men remaining, and once when you have 1 or 2 remaining. – Julia Hayward Apr 28 '15 at 8:22
4

Let's set a lower bound on the likelihood of winning with doubles, by simply ignoring all cases where it is impossible to win without doubles.

Assumption: All board positions considered are equally likely. This is probably not true, but will approach truth in longer games.

Consider the case of two men only left on the board, both in the home court, and not both on the 6-point (so that it is possible to win without doubles). There are 20 possible board positions for the two men, of weight 2 when on separate points and of weight 1 when on the same point. Likewise there are 30 possible non-double dice rolls and 6 possible double rolls. Averaged over all (the assumed equally likely) board positions under consideration, a simple counting exercise gives a probability of winning without doubles of 550 out of 1050, and a likelihood of winning with doubles of 168 / 210. (Note that the total possibilities = 1050 + 210 = 1260 = 35 * 36 as required.)

These are probabilities of 53.9% and 80% respectively, with a ratio of double-win to total wins of 23.4%.

This ignores the relatively small case-set of single blot board positions and the significantly larger case-set of board positions with 3 or 4 pieces or one or more pieces out of the home board. The net affect of these will clearly increase the likelihood of winning with a double roll.

This analysis can be extended over the additional case-sets, but it is clear that the likelihood of winning with a double roll far exceeds the 16.7% chance of rolling doubles.

                   **Wins**       **Wght * Wins**
Wght  Position  Single  Double   Single    Double
 1      1 - 1      30      6       30         6
 2      1 - 2      30      6       60        12
 2      1 - 3      28      6       56        12
 2      1 - 4      26      5       52        10
 2      1 - 5      24      5       48        10

 2      1 - 6      22      5       44        10 
 1      2 - 2      20      5       20         5
 2      2 - 3      20      5       40        10 
 2      2 - 4      18      5       36        10 
 2      2 - 5      16      5       32        10 

 2      2 - 6      14      5       28        10 
 1      3 - 3      12      5       12         5
 2      3 - 4      12      5       24        10 
 2      3 - 5      10      5       20         8
 2      3 - 6       8      4       16         8

 1      4 - 4       6      4        6         4
 2      4 - 5       6      4       12         8
 2      4 - 6       4      4        8         8
 1      5 - 5       2      4        2         4
 2      5 - 6       2      4        4         8
----------------------------------------------------------------
35                                550/1050   168/210
  • guess it's been long enough .. I should accept the answer :) hehe sorry, lost track of this and had forgotten about it ... was a nice re-visit and re-read though. I think it actually made more sense to me today on re-reading it than it did when I first read your answer ... shrug Thanks! – Ditto Mar 10 '17 at 14:18

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