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Backgammon probability of last roll being doubles estimation
Posted By: Bob Koca In Response To: Backgammon probability of last roll being doubles estimation (Jason Lee)
Date: Thursday, 12 June 2025, at 6:22 p.m.
I think you got 2/7 from the adding up 36 dice rolls and getting, including extra moves from doubles a total of 294 pips. From the doubles there are 84 pips. 84/294 = 2/7.
Using the branch of probability called Renewal Theory, I think this is an extremely good (probably within 0.1%) estimate for the following situation: Start a single player with 100 pips. Roll until the player is down to 0 (or negative pips). As the initial pipcount goes to infinity the limiting value is indeed 2/7. If anyone wants to write a Python code to check it should be quick to do so.
There are a couple ways (other than pipcount of 18 being less than infinity) that the result is altered for the given situation.
i) The pipcount is spread over several checkers. Doubles give more flexibility. For example 22 is always at least as good as 62 but often is better. For example with the position 000200 or 010110. I consider that a small factor.
ii) There are two players both racing towards the end. This gives a bias towards a double being the last roll since the side that ends first is more likely to have been lucky and that includes getting doubles on the last roll.
For the aforementioned single checker model it could also be checked. My guess is that as the initial pipcount for each player N goes to infinity that this effect disappears and the limiting probability is again 2/7. The longer the game the less likely it is to be close at the end.
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