This article originally appeared in the June 2001 issue of GammOnLine.
Thank you to Kit Woolsey for his kind permission to reproduce it here.

Backgammon Mathematics

By Kit Woolsey
Most backgammon decisions are not mathematical. The randomness of the dice rolls makes it too difficult to calculate everything out. In a game such as chess, one can look ahead several moves and anticipate how the game will go, so accurate calculation may be necessary. This is not the case in backgammon. There are 21 different possible dice rolls at each turn, so just looking ahead to ones next roll would require analyzing 441 different combinations of dice rolls. This is too difficult to do over the board. Instead, experts rely on their experience and intuition to lead them to the proper conclusion.

There are certain common situations for which some mathematical calculation can be helpful. The calculations involved are seldom complicated and usually require only simple arithmetic. However, doing the simple arithmetic in our heads at the table can be very difficult. Many players are afraid to put forth the effort, and fail to gather information which will be helpful to their checker play and cube decisions.

What I will examine in this article is how one can simplify the mental arithmetic so it can be done at the table. I should emphasize that the techniques I show here are what works for me. They will not be right for everybody. Each person has his own way of approaching things, and it is important for the player to develop techniques which he can use successfully.

The Pip Count

Pip-counting is the most obvious mathematical application. Simple counting will tell you how many pips each side has to go. How simple is it? Easier for some than for others. Let's look at an example:





money game


Yes, the pretty computer-generated board diagram is nice enough to show you the pip count so you don't have to do the work. If you are playing online, the server or a client will also give you the pip count at the click of your mouse. In real live play, however, you don't have these tools available. You have to get the count yourself, and do it in your head.

Let's see what a brute force approach would look like. To get White's pip count, we would say: 2 X 4 = 8, plus 2 X 5 = 10, for 18, plus 4 X 6 = 24 for 42, plus 4 X 8 = 32 for 74, plus 3 X 13 = 39 for a total of 113. Hopefully we didn't make a mental arithmetic error. Now we have to remember that 113 number, and go after Blue's pip count. Here we go: 3, plus 2 X 4 = 8 for 11, plus 2 X 5 = 10 for 21, plus 3 X 6 = 18 for 39, plus 7 for 46, plus 8 for 54, plus 10 for 64, plus 2 X 13 = 26 for 90, plus 2 X 18 = 36 for a total of 126. So we have 126 minus 113 (that was White's count, wasn't it?) so if we haven't made a mistake White is ahead by 13 pips going into Blue's roll.

Whew! That was a lot of work. Plenty of chance for error there, particularly if you aren't gifted in mental arithmetic manipulations. For most of us it would take a long time, and there would be a fair chance that we would make an error.

If you are like me, you look for shortcuts to things. My preference is to take a cancellation count. Instead of getting the total pips, I look at the differences between Blue's position and White's position to get the relative pip count. In most situations, that is all I need. The exact total for each player isn't important. If I know that White is ahead by X pips, that will usually be sufficient for any play or cube decision. If I really need the total I can do the dirty work and get that total, but that is rarely necessary.

Let's see how this is done. Starting at each player's ace point and moving upwards, I start taking the differences. So my thinking would go as follows: Blue has 3 (for the checker on the three point). The checkers on the 4 and 5 points cancel out. White has one more than Blue on the 6 point, so now White has 3. In the outfield, I see that Blue's checkers on the 7, 8, and 10 point are 1 pip more than if they were all on the 8 point. Thus, these checkers cancel 3 of the White checkers on the 8 point so White now has 2. White has that fourth checker on the 8 point, giving him 10. White has an extra checker on the midpoint, now giving him 23. And finally Blue has 2 checkers on the 18 point, for 36 pips, giving him a net extra of 13 pips. Therefore, White is ahead 13 pips.

The above method works for me. I can usually count a position in 5 to 10 seconds and be accurate. It won't work for everybody. It is up to the individual to find methods which work for him. A good article on pip counting is Cluster Counting by Jack Kissane. Jack is probably the world's fastest pip counter, and in this article he describes the tricks he uses to help him count quickly and accurately. It is an excellent method. It doesn't work for me, but I'm sure it does for others.

One further point involving pip counting. I strongly urge all players to get used to estimating the pip count by eyeballing the position before they do any actual counting. This will give them a feel for what is going on. This is one area where the online pip counting mechanisms can be of value. Before peeking at the pip count, make a mental estimate as to who is ahead and by how much. With practice, you will get very accurate at this. I can guess the relative (not total) pip count within 5 pips over 75% of the time, which is usually all I need to do. In addition, making a mental estimate is a good double-check on your count. When I am going to take a pip count, I first make an estimate. If my count is 10 pips or more different from my estimate, I know it is time to take a recount. More often than not, it was my count which was wrong, not my estimate.

How does this help in actual play. Let's examine the position we have counted, with Blue having 2-2 to play. We know that after the play Blue will be behind 5 pips with White on roll. Thus Blue should not touch the back checkers -- he should play for maximum contact. If Blue were ahead in the race, he would play 18/14(2), and if the race were just about even he might compromise by shifting up to the 16 point. In practice, I would not need to take a pip count here. The feel of the position tells me that I am behind, therefore I would instinctively keep the back anchor.

What does one do with the pip count in races for cube decisions? The following is a very simple rule which is usually accurate. Provided all other things are equal, if the leader has a 10% lead in a moderate to long race, he is smack in the middle of the doubling window. If he is two pips worse than 10%, it is a borderline double. If he is two pips better than 10%, it is a borderline pass/take. Note the importance of the statement: All other things being equal. If one player has a smoother bearoff structure, that can make a significant difference. Also, it would be necessary to have the total pip count in order to determine where the racing lead is compared to the 10% mark, But it isn't necessary to have it exactly. One can easily estimate to within 10 or 20 pips how many pips the leader has to go, and that is all that is necceary. If it really is a close decision, then the total can be taken.

Match Equity

Match play can lead to unusual situations which can only be resolved with proper use of match equities. It doesn't really matter which match equity table you use. They are all about the same, and provided you use one properly you aren't going to be far off. I use the table I developed 10 years ago because I am used to it. I am aware it has faults, largle due to underestimation of gammons. The table slightly underestimates the trailer's chances at a lopsided match score, and also underestimates the importance of reaching 4 away where a doubled gammon puts you out exactly. I still use the table, and make minor adjustments when necessary to compensate for these deficiencies.

Learning the table can be difficult. For those of you with good memories, simply memorize the table. The rest of us mortals need a crutch. There are several formulas which give some kind of decent approximation. My preference is Neil's numbers. If you can memorize the 1 away and 2 away equities, Neil's numbers will make it easy to get the equity for any other score with little memorization and effort. Click here to see my match equity table and get a description of Neil's numbers.

Using the figures for cube decisions is another matter. Some mathematical wizards do the multiplication of percentages in their heads, and come up with the equity for taking and the equity for passing based on their assessment of wins, losses, and gammons in the position. This is fine if you are a bot. For example: Suppose you are ahead 3-1 in a 5-point match, and are deciding whether or not to take the double. If you pass, you will be ahead 3-2 for 60% equity. You estimate that you will win the game 35% of the time. Of your 65% losses, you estimate that 1/4 of them or 16% will be gammons -- the remaining 49% will be single game losses. If you take, your equity is 100% if you win, 50% if you lose a single game, and 0% if you get gammoned. Thus, your total equity comes to: (35% X 100%), + (49% X 50%) + (16% X 0%) which is .35 + .245 = 59.5%. This is just under 60%, so it is correct to pass.

Whew! That was a lot of work! And this was with nice easy numbers to work with such as 0%, 50%, and 100%. Imagine the mental arithmetic which would have been involved if the possible outcomes weren't the end of the match. The bots can do it, and maybe some mental giants can, but not mere mortals such as you and I.

My personal approach is to use what I call a gain/loss table. I determine the equity gain from taking and winning the game (as opposed to passing), and the equity loss from taking and losing the game (as opposed to passing), and compare these numbers. I like to think in terms of odds rather than percentages for these comparisons. Perhaps that comes from my youthful days at the race track. I can easily convert these odds back to percentages when necessary. It is only necessary to know a few basic numbers, and you can interpolate from there. For example, 2 to 1 odds is the same as 33 1/3%, 3 to 2 odds is 40%, 3 to 1 odds is 25%, and so on. Let's see how this same problem would be handled:

We pass the double: Ahead 3-2 (2 away, 3 away), 60% match equity.
We take and win: Win match, 100% equity.
We take and lose. If we lose a single game, the score is tied at 3-3 for 50% equity. If we lose a gammon, we lose the match for 0% equity. From our estimate that 1/4 of our losses will be gammons our average equity if we lose the game will, therefore, be 1/4 of the way down from 50%, or 37 1/2%.

Thus, we gain 40% equity if we take and win, but cost 22 1/2% equity if we take and lose. We can see that this is worse than 2 to 1 odds (40 to 20 would be 2 to 1), but better than 3 to 2 odds, being closer to 2 to 1 than 3 to 2. Thus we are closer to 33 1/3% than 40% -- probably a little above 35%.

Therefore, we would have to win around 36% of the time. to justify a take. If our estimate of the position is that we win 35% of the time, we would have a close pass.

Note that this is exactly the same conclusion we came to when we did all the multiplication of percentages. Here we never had to do anything more than some simple subtraction and an interpolation estimate. Thus, this method is (for me, at least), much easier to use at the table. However, other players do things differently, and if the calculations are correct they will come to the same conclusions.

Shot Counting

Sometimes it is important to know exactly how many shot numbers there are. Perhaps safety is the ultimate priority, perhaps the number of shots will determine whether to make a risky play with potential gains if it works, or perhaps the number of shots will determine a cube decision. At any rate, it must be valuable to be able to tell at a glance how many shot numbers there are for a position.

For starters, let's examine the basics of dice arithmetic. Each die has six sides. For simplicity, let's call one die the green die, and the other die the red die. Each die is independent of the other, and for each die is equally likely to land with a 1, 2, 3, 4, 5, or 6 showing. Thus, there are 36 possible results from our dice roll, all equally likely.

                  Red die

        1     2     3     4     5     6

G  1   1-1   1-2   1-3   1-4   1-5   1-6
E  2   2-1   2-2   2-3   2-4   2-5   2-6
N  3   3-1   3-2   3-3   3-4   3-5   3-6

D  4   4-1   4-2   4-3   4-4   4-5   4-6
E  5   5-1   5-2   5-3   5-4   5-5   5-6

   6   6-1   6-2   6-3   6-4   6-5   6-6

It should be noted that for a given non-doubles, there are two ways it can be thrown. For example, a 3-1 can be produced with a 3 on the red die and a 1 on the green die, or with a 1 on the red die and a three on the green die. Thus, any non-doubles has 2/36 chances of occurring. A specific doubles can occur only one way, so that has 1/36 chance of occurring. We can form the following list:
Roll   # ways

1-1      1
1-2      2
1-3      2
1-4      2
1-5      2
1-6      2
2-2      1
2-3      2
2-4      2
2-5      2
2-6      2
3-3      1
3-4      2
3-5      2
3-6      2
4-4      1
4-5      2
4-6      2
5-5      1
5-6      2
6-6      1


For starters, let's count how many ways a simple ace shot can be hit. It might seem as though there were 6 ways to have an ace on the red die (1-1, 1-2, 1-3, 1-4, 1-5, and 1-6), and 6 ways to have an ace on the green die (1-1, 2-1, 3-1, 4-1, 5-1, and 6-1), so there would be 12 ways to roll an ace. However, if you look carefully you will see that we counted 1-1 twice. Therefore, there are 11 ways out of 36 to roll an ace.

For other direct shots there will be more, since not only will we have the single number on one die but the two die may combine to hit the shot. For example, let's see how many ways there are to hit a blot 4 away, assuming no point in between:

We start with the 11 ways we can get a 4 on at least one of the die. Note that this is exactly the same as the 11 ways we can get an ace on at least one of the die. Now we have to tack on the combination shots. 3-1 adds up to 4, and there are two of these (3-1 and 1-3). 2-2 adds up to 4, and there is one of these. Also, since doubles allows us to play the number four times, 1-1 also hits the four shot. Thus, there are a total of 15 ways to hit a 4 shot.

Continuing in the same manner, and being careful to count the doubles properly (for example, 2-2 and 3-3 hit a 6-shot), we get the following results:

  1     2     3     4     5     6

  11    12    14    15    15    17
Once you commit the above table to memory, you won't have to do this calculation every time you get a single direct shot. Of course, you now know how to do the calculation if you forget the table.

For the indirect shots, only combinations will count. Craps players will recognize that there are 6 ways to roll a 7 (6-1, 5-2, 4-3, 3-4, 2-5, and 1-6). This goes down by one as we go up each number -- 5 ways for an 8, 4 for a 9, etc. Of course we have to watch those sneaky doubles. 2-2 hits an 8 shot, 3-3 hits a 9 shot, and 3-3 and 4-4 hit a 12 shot. The results are:

  7     8     9    10    11    12

  6     6     5    3     2     3
Once again, committing these to memory will make life easier.

When we get to multiple shots, things get complicated. The key is to avoid double-counting numbers.

For a simple example, let's examine how many ways we can hit a shot if we have aces and twos to hit. We know there are 11 ways we can roll an ace, and we know there are also 11 ways we can roll a two. Does that mean that there are 22 numbers which hit? No. 1-2 and 2-1 contain both an ace and a two, and we can't can't count these rolls twice. Therefore, there are 20 ways to hit a direct double shot not counting combinations.

For other double shots, it can get more complicated when we have to take combinations into account. For example, suppose we wanted to count the ways we can hit a blot when we have fours and fives to hit and there are no points in our way. There are two ways we could do our counting.

One is to add the number of hits with fives and fours, and then subtract out duplications. There are 15 hits with a four, and 15 hits with a five. 5-4 and 4-5 are obviously double-counted. 4-1 and 1-4 are also double-counted. Therefore the answer is 30 - 4 = 26 hitting numbers.

The other approach is to look at the direct shots and then add in the extra combinations. We know we have 20 ways to find a 5 or a 4 on one of the dice -- that is the same as the double-shot with aces and twos. Looking for combinations which don't contain a five or a four, we have 1-1, 2-2, 3-1 (and 1-3), and 3-2 (and 2-3). Six combination shots plus the initial 20 gets us up to the same 26 hitting numbers.

In this manner, it is fairly easy to calculate any double-shot number. However, doing it at the table can be tedious, and there is always a chance of making a mistake either by forgetting a combination or by accidentally double-counting something. I found it easier to memorize what I call the "multiplication table" for double shots. There are only 15 numbers to remember, and once the table is memorized you won't have to worry about counting double shot numbers again. The table is:

        2       3       4       5       6

1       20      20      21      22      24
2               21      23      23      24
3                       24      25      28
4                               26      27
5                                       28
To see how easy it is, let's try calculating the number of shots when we have 2's, 5's, and 9's which hit. From the table, we have 22 hits from the 2's and 5's. From the 9's, we have 6-3 and 3-3 which are new hitting numbers -- we already hit with the five from the 5-4. Thus, we have 25 hitting numbers.

It should be noted that the direct shot numbers (without combinations) are excatly the same as the number of ways to enter from the bar against various boards. Against a 5-point board, there are 11 entering numbers. Against a 4-point board, there are 20 entering numbers. Continuing the calculations will show that there are 27 entering numbers agains a 3-point board, 32 entering numbers against a 2-point board, and 35 entering numbers against a 5-point board. I know it doesn't seem that way -- our opponent never seems to miss a triple shot, while we flunk against a three-point board all the time. But that is just our selective memory -- the numbers are the same.

Racing Plays

Finding the best racing play can often be a matter of counting. You know what you need to do, so for each possible play you count the number of rolls for which the play succeeds and for which it fails. That can be tedious, and there is some chance of missing a roll and making an error. For example:





money game


I'm sure that just about all readers know this position, and know that the correct play is 6/0, 3/2 rather than 6/0, 4/3. But let's suppose that we didn't know the correct play and had to work it out.

The brute force approach would be to go through all 36 rolls and see which ones succeeded with each play. After playing 6/0, 3/2, we would get off on all 6's except 6-1 -- that's 9 rolls. We would also get off on all 5's except 5-1 -- 9 more rolls, but we have to avoid double-counting 6-5, so that's really only 7 more rolls. 16 so far. No non-doubles which are any lower will work, but we have some doubles. 4-4, 3-3, and 2-2 all get off, getting us up to a total of 19 rolls. Now, let's look at 6/0, 5/4. We get off on 6's except 6-1 and 6-2 -- so that's 7 rolls. For the fives we don't make it on 5-1 and 5-2, and 6-5 is already counted, so we have 5 more rolls getting us up to 12. We also get off on 4-3, but that's the only non-double 4 which makes it -- 14 rolls. We still get off on 4-4, 3-3, and 2-2, totalling 17 rolls.

Whew! That was a lot of work. The results are correct (19 winners after 6/0, 3/2 and 17 after 6/0, 5/4), but there was a lot of counting to be done and several chances for a mental error. Is there an easier way?

If all we want to do is find the better play, there is an easier way. The key is that rolls such as 6-6 and 5-5 will get off next turn with both plays, while rolls such as 1-1 and 2-1 won't get off with either play. What we want to concentrate on are the rolls which make one play better than the other. Clearly a 5-2 roll works only with 6/0, 3/2 and a 4-3 roll works only with 6/0, 5/4. We can see at a glance that a 6-2 roll works only with 6/0, 3/2. A quick runthough of the possibilities shows us that for every other roll, it either works with both plays or with neither play. Therefore 6/0, 3/2 is superior by 2 rolls. This approach won't tell us how many rolls we get off from each play, but it does tell us which play is superior.

The same kind of comparison technique is handy when racing to avoid a gammon. Consider the following series of positions:





money game


If Blue plays 12/6, 2/1, he will get off next roll if he rolls a 6-1 -- he would not have gotten off on a 6-1 had he played 12/11, 12/6. However, this is the only roll for which 12/6, 2/1 is superior. After 12/11, 12/6, Blue gets off with 5-2, 5-3, or 5-4, while he does not after 12/6, 2/1. Thus 12/11, 12/6 is superior for more rolls. Note that we never counted exactly how many rolls get off for each play. We only looked at rolls for which one play succeeds and the other fails. It is obvious that for all doubles 2 or higher both succeed, for 6-2 or higher sixes both succeed, and for 5-1 or all non-doubles with the highest number smaller than five both fail. Thus, we don't need to count these rolls, because they give the same result for both plays.





money game


12/6, 2/1 is superior if we roll 6-1 or 5-1 next turn. 12/6, 11/10 is superior if we roll 4-3 or 4-2 next turn. Otherwise, it doesn't make any difference. The plays are equally good, and it is a guess.





money game


12/6, 2/1 is superior if we roll 6-1, 5-1, or 4-1 next turn. 12/6, 10/9 is superior only if we roll 3-2 next turn. Therefore, 12/6, 2/1 is the correct play.

Note that for none of these problems did we ever count all the rolls which got off. A simple comparison of the rolls for which the play makes a difference is sufficient to reach the correct conclusion. This approach not only saves a lot of mental wear and tear, but minimizes the risk of a miscalculation from miscounting rolls.

Case Counting

Probably the most difficult kind of mathematical calculation occurs when a player tries to look ahead at all the next 36 possible rolls and break them down. In essence he is doing the equivilent of a Snowie 2-ply evaluation. This is very difficult, but sometimes it can be done and some positions lend themselves to a case-by-case analysis. For example:





money game


Blue doubles. Does White have a take? The answer is not immediately obvious. How can one attempt to figure this out. Most of us would just go on our gut feeling and hope to be right, but some players might try to calculate it out by breaking down Blue's rolls.

Blue's rolls can be split reasonably enough into 4 categories:

1) The great rolls. These are obviously any roll with a six. There are 11 of these.
2) Rolls which hold the board. Blue has several small numbers which allow him to keep the closed board another roll. 4-2, 4-1, 3-2, 3-1, 2-1 and 1-1 fit into this category -- 11 rolls.
3) Rolls which break the six point. These include the following: 5-5, 5-4, 5-3, 5-2, 5-1, 4-3, and 2-2 -- a total of 12 rolls.
4) Rolls which break two points. There are 2 such horror rolls: 3-3 and 4-4.

A quick check: 11 + 11 + 12 + 2 = 36. Everything has been accounted for.

Let's suppose we play the game out 36 times, with Blue starting with each roll once. If White passes the double, he will be -36 for the 36 games. If he takes the double, the game will be played with the cube on 2. Now we have to estimate how White will do for each category of rolls, add up the results, and see if White does better or worse than -36.

Category 1: White is virtually always lost when Blue rolls a six. In addition, White will sometimes have the outfield checker hit and be gammoned. Of the 11 games where Blue rolls the six, I estimate that on 4 of them White will lose a gammon, and on 7 of them White will lose a single game. This gives White a score of 4 X -4 for the gammons and 7 X -2 for the single losses, for a total of -30 so far.

Category 2: Blue holds the fort. Blue will still be the favorite because of his chance of rolling a six next turn, but he won't be too much of a favorite. Of the 11 rolls in this category, I estimate Blue will win 7 of the games (including 1 gammon) and lose 4 of the games. This gives White a net loss of -4 for the gammon and -4 for the extra 2 single game losses, for an additional -8. So far, White is -38.

Category 3: Blue breaks the six point. I believe this makes White a small favorite, particularly since he owns the cube. Of the 12 games I estimate that White will win 7 of them, and lose 5, for a net of +4 points. This brings White's total to -34.

Category 4: Blue rolls a horror. White is now a clear favorite. Of the two games I expect White to win 1 1/2 of them, for a net plus of +2 points. White's total is now -32.

If my estimates are correct, White has a take, because he is -32 for taking and -36 for passing over the 36 games. However, there is no guarantee that my estimates are correct, and if there is a sufficient bias in them I may come to an erroneous conclusion. This is why the case-by-case approach is very risky unless you are very good at this sort of estimation. In addition there was quite a bit of mental arithmetic involved.

This kind of case-by-case analysis can in theory be applied to play decisions also. The bots do it routinely, but they think a lot faster than we do and their mental arithmetic is faultless. If they could always assess the resulting positions accurately their play would be perfect, but since their assessments can be off the bots make mistakes also. It is almost impossible for a human to do this. I once saw a top player playing in an important match at double match point have a choice of 3 candidate plays which were conceptually quite different. He took about 20 minutes attempting to make a case-by-case analysis of each of the three plays. He examined all 36 of his opponent's rolls after each play, and tried to assess the equities and add them together so he could choose the play with the highest equity. He finally came to a conclusion, but it was far from clear that his play was correct. A simple over-the-board judgment after looking at all three candidates would probably have been just as effective.

The truth of the matter is that there simply isn't that much calculation to be done in backgammon. I would estimate that in a normal 15 point match I would find occasion to actually calculate anything fewer than 6 times. Almost all plays are done by feel, based on experience, understanding of the priorities of the position, and general principles which have proven to be effective for the type of position involved. An exact pip count is seldom necessary. A match equity calculation comes up rarely. Most of the time counting shots is unnecessary -- simple general principles of putting your checkers where they do the most good combined with proper use of duplication and diversification will lead to the correct play. End-game racing positions may need a quick calculation, but if the player is familiar with the principles involved even these can be accurately played by feel and experience. Players who try to calculate everything usually wind up losing the forest through the trees. However, there are times when it is handy to be able to make a simple calculation and get it right. The techniques illustrated in this article will give you a good basis for making these calculations painlessly and accurately.

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