6

All my checkers are in my home, all my oponent's checkers are not in my home.

I roll the dice.

Assuming I can't take out checkers, what's my best strategy of advancing them? e.g. when this is my home and I roll a 2, should I move the 6 or the 3?

#| | | | | | |#
#|X| | |X| | |#

Since I'm not dependent on my opponent, I assume there's an optimal strategy I can use here. Is it "always move the highest numbers"?, "always move the lowest"?, something more complicated?


example

In the simple case I described above I know it's best to move the 6, since my chances of clearing out in the next round are higher:

option 1: move the 3

result:

#|X| | | | |X|#

chance of winning:

I have to get a 6 at least in one dice, so it's

# get a 6 in the first dice
1/6 +
# get 2-5 in the first dice, 6 in the second or double
4/6 * (1/6 + 1/6) +
# get 1 in the first dice, 6 in the second
1/6 * 1/6 =
# result:
0.416

option 2: move the 6

result:

#| | |X|X| | |#

chance of winning:

I have to get at least a 3,4:

# both dice get 3-6:
4/6 * 4/6 +
# double 2:
1/6 * 1/6
# result:
0.472

So close call, but it's better to move the 6.

But this is a simplified case, assuming I will either finish in the next round or the one after. What about when I have more than 2 checkers?

  • "Since I'm not dependent on my opponent" Not quite true. What is relevant is neither, as answers have suggested, probability of winning on next turn nor expected number of turns to win, but probability of bearing off before your opponent does. How far your opponent is from bearing off could theoretically affect what optimal play is. – Acccumulation May 29 '18 at 16:17
  • @Acccumulation good point. I was expecting a simple answer but it seems to only get more complex the more you look at it. For a complete answer I guess you would also need to know the opponent's strategy, and assume it remains constant.. – Dotan May 30 '18 at 12:22
3

The logic of finding the “general rule” in these situations is flawed and will lead to potential mistakes. There are positions where you have to move the highest or the lowest number and need to be able to quickly calculate the numner of dice combinations that are favourable or not for your next roll.

In your example, if you leave the checkers at 6,1 points: you need all sixes (11 combs) plus 55,44,33,22 (4 combs), total of 15.

If you however leave the checkers on 4,3 points, you need 65,64,63,54,53,43 (12 combs) plus 66,55,44,33,22 (5 combs) for a total of 17 rolls. The latter play is superior.

Just in case one wonders why not always move the highest point, consider the example of having two checkers on 5,3 with a need to play a deuce. if your move is 5-3, leaving checkers on 3,3 you will clearly lose more games than if you have moved 3-1, receiving the position of 5,1. I will let you calculate the combs and compare.

  • It's true, but I think the better measurement is the one suggested by @Nij - you should count the average number of turns left and not the chances of finishing in the next step (well, depends on the game) – Dotan Apr 8 '18 at 12:52
  • With two checkers left, its the same principle. Counting favourable combinations is just easier over the board. – Skytten Apr 9 '18 at 5:44
2

Mathematically speaking, you should minimise the average number of additional dice needed to bear off all checkers. This implies a minimisation of the number of dropped rolls - rolls that do not remove a checker from your home, but only advance it further.

Suppose you move the three to the one and leave the six there on the first turn. On the second turn, you now need to roll a 6 and a {1,2,3,4,5} or a double {2,3,4,5,6} to clear both checkers with probability 15/36. Anything less will bear off one checker with probability 19/36 or nothing with probability 2/36.

On your third turn, you bear off the last checker with probability depending on the previous roll, but an average of 96.6% of the time, you're now finished. The remaining 3.4% require a fourth turn. If you had both checkers left, you bear off both 17/18 to finish and require a fourth turn for the remaining 1/18.

Overall, this is an average of 2.604 turns to bear off both.

A similar analysis finds that by moving the six to the four and leaving the three, you are 47% to finish on the second turn, and 50% to finish on the third turn. The average from this position is 2.553 turns to bear off both.

Note that in every subsequent turn, you are better positioned to finish on that exact turn by moving the six forward on first turn, not the three. The reason is simple: the single die roll has a uniform distribution and the pair dice roll has a triangular distribution. This means you are more likely to hit values in the middle of the range, either 3 or 4, and less likely to hit those at the ends, 1 or 6, while 2 and 5 fall in the middle. Since bearing off requires at least a particular number, and small rolls only bear off small numbers, it is more likely that a checker in the middle can bear off than a checker at the six. By moving checkers from the high end to the middle, you gain better chances of bearing off much quicker than by moving from middle to the low end.

You can test this yourself: set up a board with both colours in their own homes, white split between the one and six points, black between the three and four points. The vast majority of games will end with a black victory: white struggles to use middle rolls to move the high checkers forward, while black uses almost every turn to remove at least one checker. To really exemplify the point, use identical turns for both sides by rolling the dice once and playing them both with those numbers.

In terms of strategy, this means

  1. Bear off the most checkers possible.

  2. Move forward the highest remaining checkers that cannot be borne off.

  • Small correction: when you have 6,1, a double 2 is also good – Dotan Apr 8 '18 at 10:35
  • Fixed and the ensuing correction also made. Only drops 3 hundredths of a turn to include that 22 roll! @Dotan – Nij Apr 8 '18 at 10:46
  • the single die roll has a uniform distribution and the pair dice roll has a triangular distribution I didn't follow you're reasoning on this. Two dice distribute uniformly over {1-6}x{1-6}. also, see Skytten's example, if you have 5,3 and rolled a 2, it's better to move the 3 for average turns of 1.3 rather than 1.52 if you move the 5 – Dotan Apr 8 '18 at 12:53
  • True, it ends up being corner-cased with two checkers. When dealing with more though, is when it becomes apparent. You're getting the "most of the most" by being in the middle, rather than hoping for extreme rolls by being at the ends. – Nij Apr 8 '18 at 19:08
  • 1
    When you changed denominators in the second paragraph you generated a total possibility that is greater than 1 (37/36). The 1/18 is correct (2 ways to get 0 checkers off, with the 2-3 or 3-2 rolls). But the probability for removing only 1 checker is not 10/18 (20/36); it is 19/36. I am not sure how this affects the remainder of the answer. – Doug Edmunds May 28 '18 at 23:16
1

For small numbers of chequers, the relative odds can be calculated exactly as described by other answers. For cases of too many chequers to perform calculations in your head, the following heuristic has long been recommended by championship players:

Divide the home board into three buckets as follows:

  • Bucket 1 is the 1 & 2 points;
  • Bucket 2 is the 3 & 4 & 5 points; and
  • Bucket 3 is the 6 point.

Minimize the number of moves that don't reduce a chequer's bucket number; or alternatively, maximize the number of moves that do reduce a chequer's bucket number.

This can be recalled more succinctly as:

Wherever possible: reach the 5(-point) or reach the 2(-point) with every chequer that cannot be borne off.

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