Backgammon doesn’t just depend on the player’s decisions, but also on the randomness of the dice.

This is different from Chess. In Chess, some games end as a draw because of the threefold repetition because no player wants to back out of the repetition cycle. Imagine if you have a choice between either continuing the repetition cycle, or breaking the cycle by putting yourself at a disadvantage on the board. Then your opponent has the same choice. Neither you nor your opponent wants to be the one to break the cycle, and therefore the game is declared a draw. If one of the two players instead thought that breaking the cycle would be advantageous to them, then they would do it and avoid the draw.

But Backgammon uses dice. Which means that even if both players want to continue the cycle, the cycle might be broken by a roll of the dice.

For instance, imagine you and I each only have one checker left, both starting in our furthest point, so that we have to run around the board. The game would be an infinite loop only if our checkers never passed eachother without hitting. But even if we’re both trying to hit the opponent because we love hitting, then the probability of our checkers passing through without being able to hit each-other is about 59%. The probability of hitting each-other is only 41%, even if we absolutely want to hit each-other. And when we hit each-other, it starts again, which means that the probability that we will hit each-other n times before finally passing through follows a geometric distribution: the probability of hitting each-other two times is only 0.41^2 = 16%; the probability of hitting each-other 3 times before passing through is only 0.41^3 = 7%; the probability of hitting each-other 4 times before passing through is only 0.41^4 = 3%; 5 times is only 0.41^5 = 1%; and the probability of an infinite cycle where we always hit each-other without passing through is exactly 0%.

Small note about the 59% / 41% probabilities: These probabilities can be calculated easily. The proba that our checkers will be able to hit each-other depends on their distance. This allows us to calculate the six probabilities P(hit | dist=k) for k=1..6 recursively:

P(hit | dist=1) = 11/36
P(hit | dist=2) = 12/36
P(hit | dist=3) = 13/36 + 1/36 * P(hit | dist=1)
P(hit | dist=4) = 14/36 + 1/36 * P(hit | dist=2) + 2/36 * P(hit | dist=1)
P(hit | dist=5) = 15/36 + 1/36 * P(hit | dist=3) + 2/36 * P(hit | dist=2) + 3/36 * P(hit | dist=1)
P(hit | dist=6) = 16/36 + 1/36 * P(hit | dist=4) + 2/36 * P(hit | dist=3) + 3/36 * P(hit | dist=2) + 4/36 * P(hit | dist=1)

and finally, when the checkers start from the furthest point on the board, they are guaranteed to eventually reach one of the six distances 1..6, and the probability for each distance is about 1/6, so that:

P(hit | dist=furthest) ≈ 1/6 * (P(hit, dist=6) + P(hit, dist=5) + P(hit, dist=4) + P(hit, dist=3) + P(hit, dist=1) + P(hit, dist=1))

Other interesting values can be calculated by writing the whole transition matrix for this Markov chain.