<p>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.</p>

<p>Suppose you move the three to the one and leave the six there on the <em>first</em> turn. On the <em>second</em> 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.</p>

<p>On your <em>third</em> 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 <em>fourth</em> turn. If you had both checkers left, you   bear off both 17/18 to finish and require a <em>fourth</em> turn for the remaining 1/18.</p>

<p>Overall, this is an average of <strong>2.604 turns to bear off both</strong>.</p>

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

<p>Note that in every subsequent turn, you are better positioned to finish on that exact turn by moving the six forward on <em>first</em> 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.</p>

<p>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.</p>

<p>In terms of strategy, this means</p>

<ol>
<li><p>Bear off the most checkers possible.</p></li>
<li><p>Move forward the highest remaining checkers that cannot be borne off.</p></li>
</ol>

CC BY-SA 3.0
2018-04-08T10:35:13.777Z
Small correction: when you have 6,1, a double 2 is also good
Dotan
CC BY-SA 3.0
2018-04-08T10:46:11.317Z
Fixed and the ensuing correction also made. Only drops 3 hundredths of a turn to include that 22 roll! @Dotan
Nij
CC BY-SA 3.0
2018-04-08T12:53:08.100Z
`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
CC BY-SA 3.0
2018-04-08T19:08:16.060Z
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
CC BY-SA 4.0
2018-05-28T23:16:40.133Z
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
CC BY-SA 4.0
2018-04-08T10:01:45.913Z
2018-05-29T04:08:55.930Z
2018-05-29T04:08:55.930Z
Nij
<p>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.</p>

<p>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.</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2018-04-08T12:52:23.533Z
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
CC BY-SA 3.0
2018-04-09T05:44:20.700Z
With two checkers left, its the same principle. Counting favourable combinations is just easier over the board.
Skytten
CC BY-SA 3.0
2018-04-08T10:28:02.627Z
2018-04-08T10:34:40.523Z
2018-04-08T10:34:40.523Z
Skytten
<p>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:</p>

<p>Divide the home board into three buckets as follows:</p>

<ul>
<li>Bucket 1 is the 1 &amp; 2 points;</li>
<li>Bucket 2 is the 3 &amp; 4 &amp; 5 points; and</li>
<li>Bucket 3 is the 6 point.</li>
</ul>

<p>Minimize the number of moves that <strong><em>don't</em></strong> reduce a chequer's bucket number; or alternatively, maximize the number of moves that <strong>do reduce</strong> a chequer's bucket number.</p>

<p>This can be recalled more succinctly as:</p>

<blockquote>
  <p>Wherever possible: reach the 5(-point) or reach the 2(-point) with every chequer that cannot be <em>borne off</em>.</p>
</blockquote>

CC BY-SA 4.0
2018-06-03T04:05:11.523Z
2018-06-03T04:05:11.523Z
Forget I was ever here
<p>All my checkers are in my home, all my oponent's checkers are not in my home.</p>

<p>I roll the dice.</p>

<p>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?</p>

<pre><code>#| | | | | | |#
#|X| | |X| | |#
</code></pre>

<p>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?</p>

<hr>

<h2>example</h2>

<p>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:</p>

<p><strong>option 1</strong>: move the 3</p>

<p>result:</p>

<pre><code>#|X| | | | |X|#
</code></pre>

<p>chance of winning:</p>

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

<pre><code># 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
</code></pre>

<p><strong>option 2</strong>: move the 6</p>

<p>result:</p>

<pre><code>#| | |X|X| | |#
</code></pre>

<p>chance of winning:</p>

<p>I have to get at least a 3,4:</p>

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

<p>So close call, but it's better to move the 6.</p>

<p>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?</p>

CC BY-SA 4.0
2018-05-29T16:17:47.340Z
"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
CC BY-SA 4.0
2018-05-30T12:22:30.880Z
@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
<backgammon>
Backgammon best strategy for clearing my home
Dotan
# Backgammon best strategy for clearing my home

(Asked by: username_0 on 2018-04-08)

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 username_5. 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?

**Tags:** backgammon

> **Comments:**
> **username_1**: "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.  
> **username_0**: @username_1 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..  

---

## 3 answers

### Answer by username_2 on 2018-04-08. Score: 3

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.

> **Comments:**
> **username_0**: Small correction: when you have 6,1, a double 2 is also good  
> **username_2**: Fixed and the ensuing correction also made. Only drops 3 hundredths of a turn to include that 22 roll! @username_0  
> **username_0**: `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 username_4'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  
> **username_2**: 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.  
> **username_3**: 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.  

---

### Answer by username_4 on 2018-04-08. Score: 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.

> **Comments:**
> **username_0**: It's true, but I think the better measurement is the one suggested by @username_2 - you should count the average number of turns left and not the chances of finishing in the next step (well, depends on the game)  
> **username_4**: With two checkers left, its the same principle. Counting favourable combinations is just easier over the board.  

---

### Answer by username_5 on 2018-06-03. Score: 2

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*.


---

