I was playing backgammon and one of my blots was on the bar, I wanted to calculate the chances that I would hit my opponents blot.

For those of you unfamilliar with backgammon, it is a game played by rolling two dice (with 6 faces, which you throw simultaneously) and to simplify (as was the case in my game), throwing the same number (a double) does not create any special cases.

I had to throw $\{1, 2, 4\}$ to enter my opponents board (which you have to do, else you cannot move something else) and I had to throw $\{1, 2, 5\}$ with the remaining die to hit my opponent.

My question: What is the chance that I would hit my opponents blot (after first entring my opponents board) and if there is an easy way to calculate it or at least approximate it (without counting all possibilities)? Also can this be extended to other cases?

My attempt: I did manage to calculate this which I did by enlisting all possible dice rolls that would succeed, namely $\{ 1,1 \}, \{ 2,2 \}$ which account for a $\frac{1}{36}$ probability each and $\{ 1,2 \}, \{ 1,4 \}, \{ 1,5 \}, \{ 2,4 \}, \{ 2,5 \}, \{ 4,5 \}$ accounting for a $\frac{2}{36}$ probability each, for a total chance of $\frac{14}{36}=\frac{7}{18}$.

However, I did not see another way to calculate this and when changing the numbers to $\{1, 2, 5\}$ to enter the board and $\{1, 2, 5\}$ to hit my opponent the chance suddenly decreased to $\frac{1}{4}$, which is quite a significant decrease. I can only understand this very vaguely when looking at the possible dice combinations and have no intuition for this at all.

My question: Is there some sort of intuition that tells you this difference in chance (or what the chances approximately are) or should I just always calculate the probability by enlisting all posibilities?

Any help is welcome!

Edit: To introduce some notation:

If we have sets $A$ and $B$, what is the chance that when I throw two dice that one result is contained in $A$ and the other in $B$? Can we write that as a function of $\#A, \#B$ and $\#(A\cap B)$ or some other set-theoretic property? And what would the answer be when we extend it to multiple dice (which is not the original question, but nonetheless interesting)?


1 Answer 1


In problems like this, it helps to pretend that one die is red and the other one green. This helps with the computations.

For $\{1,2,5\} \times \{1,2,4\}$:

There are $3$ different ways that the red die can be $\{1,2,5\}$ and there are $3$ different ways that the green die can be $\{1,2,4\}$. This makes a total of $9$ ways, so far.

Superficially, you repeat this computation for when the red and green switch so that the green is in $\{1,2,5\}$ and the red is in $\{1,2,4\}.$

So, this is the starter computation : $(18)$.

However, this is wrong, because you have overlaps.

The case of (red,green) of $(1,1), (1,2), (2,1), (2,2)$ was double counted.

So, the correct number of different throws, out of $(36)$ is $\left(2 \times 3^2\right) - 2^2.$

For $\{1,2,5\} \times \{1,2,5\}$:

The analysis is the same, except that there are $(3)$ overlaps, rather than $(2)$.

So, the correct number of different throws, out of $(36)$ is $\left(2 \times 3^2\right) - 3^2.$

  • $\begingroup$ Great answer! Is there a way to make a formula out of the double counted rolls? Or something that does not require listing them (see also my edit of the original problem)? $\endgroup$ May 2, 2022 at 14:01
  • 1
    $\begingroup$ @jorisperrenet The number of double counted rolls is always the square of the number of overlaps. So, for example, the computation of $\{1,2,3,4\} \times \{1,2,5,6\}$ would be $[2 \times (4)^2] - 2^2$. For another example, the computation of $\{1,2,3,4\} \times \{1,2,3,6\}$ would be $[2 \times (4)^2] - 3^2$. $\endgroup$ May 2, 2022 at 14:04
  • $\begingroup$ Is this also the same for 3 dice? For example by now taking $3\times (4)^2 - 3^2 - 2^2 - 2^2$ for $\{ 1,2,3,4 \} \times \{ 1,2,3,5 \} \times \{ 1,2,5,6 \}$? $\endgroup$ May 2, 2022 at 14:05
  • $\begingroup$ @jorisperrenet No, $3$ dice (i.e. one red, one blue, one green) is much more complicated. I suggest writing computer programs, to simulate the rolls, to sanity check such problems. If you have never programmed before, then I recommend Python, as your first language. $\endgroup$ May 2, 2022 at 14:07
  • $\begingroup$ Thank you @user2661923 for those helpful answers! I will try to implement it into python for $n$ dice and come back to it. $\endgroup$ May 2, 2022 at 14:13

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