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 ... )