# Maximum of Two Dice

I stumbled over a puzzle, namely figuring out the probability mass function of the maximum of two dice throws. Never afraid of practising my enumerative combinatorics, I couldn’t resist giving it a shot.

Call the highest side shown \(m\). This outcome can be split into two non-interfering cases:

- Either both dice show \(m\). This can happen just one way.
^{1}If \(m = 4\), then there is only one outcome where \((x, y) = (4, 4)\). - Only one die shows \(m\). This die can be selected in two ways
^{2}It’s either one die or the other., and the outcome on the other die can happen in \(m-1\) ways^{3}If the highest die shows 4, the only valid outcomes for the lowest die is 1, 2, 3..

Adding both non-interfering cases up, there are \(1 + 2(m-1)\) ways for a specific side \(m\) to be highest of two dice thrown, and there are \(6^2 = 36\) total outcomes possible. Thus, the probability mass function is

\[\frac{1 + 2(m-1)}{36}\]

for \(m = 1..6\).

# More Dice

It’s the same reasoning for three dice:

- Only one die shows \(m\). There are three ways to select this one, and there
are \((m-1)^2\) ways for the other two dice to land
^{4}If \(m=4\), they can both independently take any of the values \(1..3\).. - Two dice show \(m\). There are three ways to select these
^{5}Think of it the other way around – we’re selecting the one dice that does not show \(m\), out of three, and that can obviously happen in three ways., and there are \((m-1)\) ways for the last die to land. - All three dice show \(m\), and this can happen only in one way.

Summing up the non-interfering cases and dividing by \(6^3\) gives us

\[\frac{3(m-1)^2 + 3(m-1) + 1}{216}\]

again for \(m=1..6\).

For four dice we can benefit from throwing binomial coefficients into the mix. Three cases are similar to before:

- Only one die shows \(m\), which can be selected in four ways, and the other three dice have \((m-1)^3\) outcomes among them.
- Two dice show \(m\). Now these can be selected in \(4 \choose 2\) ways, and the other two dice can have any of \((m-1)^2\) outcomes.
- Three dice show \(m\). These can be selected also in four ways, and the last die has \(m-1\) possible outcomes.
- All four dice show \(m\), which again happens in just one way.

The total probability is

\[\frac{1 + 4(m-1) + {4 \choose 2} (m-1)^2 + 4(m-1)^3}{6^4}.\]

# General Formula

If we aim for notational consistency and expand the numbers in the previous formula into powers and binomial coefficients, we are lead to believe there may be a general formula:

\[\frac{\sum_{i=1}^{D} {D \choose i} (m-1)^{D-i}}{N^D}\]

where \(D\) is the number of dice thrown and \(N\) is how many sides they have. These two are parameters of the probability mass function. The other free variable, \(m\), is the domain and ranges, as before, \(1..N\).

As we increase the number of dice, both the numerator and denominator become very large numbers (thanks, combinatorics) but the ratio between them grows fairly slowly. Even with 10 dice there’s less than a 90 % chance of getting a six. On the other hand, the risk of getting a one is very, very low.