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I am a backgammon beginner and I love this game so much! I tried playing a full backgammon match with computer on Hard difficulty (3rd hardest of 4 difficulties available) and I managed to beat it with a score of 15:2, having just basic knowledge!

Can I ask a question, please? I sometimes feel very unhappy realizing that quite a lot of board games tend to last forever or to come to an infinite loop. Chess and Monopoly are two very good examples. Chess has threefold repetition and fifty-move rule which was revised in 2014 only to become 75-move rule. And a game of Monopoly can in theory last indefinitely long; in practice, the longest game lasted for seventy days. You know, it's because they always play clockwise; passing "Go" gives you your $200 wage, and then if you are lucky enough you may land on your own properties only which prevents you from being bancrupt. An infinite circle!

So what about backgammon? Constantly hitting opponent's pieces is fun, but I wonder so much if there is a proof that a backgammon game can never last forever.

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    To me it seems pretty obvious a game can theoretically last forever; just imagine a situation where each side has one man left and they just keep hitting each other back to the bar.
    – Philip Kendall
    Jun 12 at 15:00
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    @Zibelas The question isn't about practical, in the same way that two or more people in Monopoly won't always land on their own properties.
    – Philip Kendall
    Jun 12 at 16:56
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    Monopoly only has long playtimes if you don't follow the rules and add in a bunch of house rules. Otherwise it will end in a reasonable time due to the low probability of all players only landing on their own property.
    – Joe W
    Jun 12 at 17:54
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    @JoeW The question is about theoretical possibilities; why Monopoly often ends up going for too long isn't really relevant.
    – GendoIkari
    Jun 12 at 17:56
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    @GendoIkari Maybe, but monopoly is only known as a long game because of the house rules that people add to the game which makes it drag on. There are many games that you can use "theoretical possibilities" to show that they go on for a very long time or never end. Also the fact that the OP questions the practically of a situation for backgammon makes me question that it is about "theoretical possibilities"
    – Joe W
    Jun 12 at 18:00

3 Answers 3

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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.

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    Perhaps worth mentioning the concept of "almost never" - there is indeed a 0% chance of a never-ending game, yet it is still possible for it to occur. Somewhat unintuitively, finding a 0% chance of a never-ending game doesn't prove it can't happen.
    – Nuclear Hoagie
    Jun 13 at 13:24
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    "Almost never" is a statistical concept to describe something that has 0% probability but is not technically impossible. If I select a random real number in the range [0, 1], there is a 0% chance you'll guess it exactly right, but it's not categorically impossible you'll guess correctly, since whatever number I pick is available to you as well. A backgammon game will "almost never" go on forever, but it is possible that no matter how many times you roll, the game still hasn't ended. The probability goes to 0%, but there is a sequence of dice rolls that results in no end.
    – Nuclear Hoagie
    Jun 13 at 14:23
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    In the random number example, it's possible I pick the number 0.5194, and it's possible that you do, too. There is nothing that prohibits you from picking my number other than the sheer unlikelihood of it, but it is possible. In talking about the possibility of an infinite game I'm not addressing practical issues like time required or heat death of the universe, etc. What I mean is that I can define a series of dice rolls and moves for each player that can be repeated forever without ever finishing the game.
    – Nuclear Hoagie
    Jun 13 at 14:36
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    To put it another way, there is no value of X where you can claim "all backgammon games end by turn X". The length of a backgammon game is unbounded. It's possible, just vanishingly unlikely, that we could get the exact right series of rolls for an arbitrarily long time. It is not impossible to play a never-ending backgammon game, one could write down a loop of rolls and moves needed to do it.
    – Nuclear Hoagie
    Jun 13 at 14:38
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    Precisely - whatever number either of us picked a priori had exactly zero probability of being picked. And yet, we still picked it despite having exactly zero prior chance of doing so! Events with zero probability aren't necessarily impossible. By "it is possible that the game doesn't end", I mean I can write down a legal and attainable sequence of rolls and moves which can be repeated forever without the game ending.
    – Nuclear Hoagie
    Jun 13 at 14:59
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It has been proven that backgammon terminates with probability 1:

https://www.bkgm.com/rgb/rgb.cgi?view+1423

Essentially, in any legal position there is at least one finite sequence of dice rolls in which the players cannot prevent the game from ending within that sequence of rolls, no matter how they play.

C. T. McMullen writes: Here is a proof that backgammon terminates with probability 1, or equivalently that the dicemaster has a winning strategy from any legal position. This proof is inspired by Koca's observation above.

The strategy is very simple:

    MAKE EVERY ROLL 2-4.

Then the game will terminate.

It works like this. Say a checker has the right parity (R) if it is on a point of the same color it would be if entered by an even number from the bar. Otherwise it has the wrong parity (W). Since all the rolls in this game are even, any time a checker is hit, it has the right parity from that point onward.

The only possible hits are R hits W and W hits R. Every time R hits W the number of W-parity checkers decreases, so that can only happen a finite number of times. Every time W hits R, a W checker advances, so that can only happen a finite number of times (the only way a W can get sent back is to turn in to an R).

Thus in any game where only even rolls occur, there are only a finite number of hits. (An explicit upper bound is 30 + 24x30/2 = 390, the maximal number of W checkers plus the maximal combined pip-count for the W checkers, divided by 2 since all moves are even.)

Now we just have to verify that we can keep both sides moving using only even dice, indeed using only 2's and 4's. But this is obvious. If a checker X cannot move 2 or 4 because both moves are blocked by O, then O can move 2.

QED.

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  • The proof show that a game will terminate "almost certainly" (probability 1). It is trivial to construct the series of throws that would result in a never-ending game, though.
    – L. Scott Johnson
    Oct 20 at 19:37
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It is theoretically possible for a game of backgammon to last forever. Imagine a scenario where both players have a checker on the bar, and they both dance on their turn. While this scenario could repeat for a discrete number of rolls, as the number of rolls approaches infinity, the probability of both players staying on the bar forever tends towards zero.

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    The chances of that are so low it isn't worth mentioning.
    – Joe W
    Oct 20 at 13:47
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    Wrong. as the number of rolls approaches infinity, the probability of both players staying on the bar is exactly 0.
    – Cohensius
    Oct 20 at 15:01
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    Joe has it correct. That the probability "approaches 0" is not the same as "cannot happen". There's a disconnect in people's understanding of what a set of measure zero is. The odds of flipping heads every time for X tosses approaches zero, but it can happen. Even if you flip the coin an infinite number of times, it could come up heads every time.
    – L. Scott Johnson
    Oct 20 at 19:34
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    @Cohensius, I think you need to go reread your freshman Calculus book, paying particular attention to the section on limits, before you just start blindly saying that I am wrong.
    – Joe Guichebarou
    Oct 22 at 15:29
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    Hi @JoeGuichebarou, you right, my previous comment is wrong. However, I still thinks that no backgammon game can last forever. Because from any situation there is a positive probability to end the game. That is a sequence of dice rolls that will end the game no matter how the players will play them.
    – Cohensius
    Oct 23 at 6:24

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