You roll a d6, d8, d12, and d20 all at the same time. What is the probability the d6 results in the highest number? d8? d12? d20?First, let's clarify by what we mean by "highest number''. Does that mean that the d6 is

*greater*the other dice, or just that no

*other*dice are greater? In other words, how do we count ties? Let's not count ties for now, and deal with them last. With four dice, they will be the most tricky. (This is due to the number of cases involved. There could be 2, 3, or 4 dice tied. Or two sets of 2 for that matter!)

The faces of each of these dice are numbered 1 through the number of sides that it has. Each side has equal probability of occurring. If we label the results of the four dice as random variables $X_6$, $X_8$, $X_{12}$, and $X_{20}$, where the subscript refers to the number of sides of the die, then we can write the statement of equal probability in the following equation.

\begin{align}

P(X_N = x) = \begin{cases}

\frac{1}{N} &\qquad x \in \mathbb{W}, 1 \leq x \leq N \\

0 &\qquad \text{else}

\end{cases}

\end{align}

Here $\mathbb{W}$ is the set of all whole numbers (i.e. 0, 1, 2, 3, etc.). Recall that the probability of rolling less than a number on such a die is proportional to said number,

\begin{align}

P(X_N < x) = \frac{x-1}{N} &\qquad x \in \mathbb{W}, 1 \leq x \leq N .

\end{align}

But perhaps I'm putting the cart before the horse. How do we approach this problem? We can write the question as,

\begin{align}

P(X_6 > \max(X_8, X_{12}, X_{20})) = ?

\end{align}

This is not a great way to write it though. It tends to make us think that we should compute the distribution of the maximum term $\max(X_8, X_{12}, X_{20})$. We can write them separately also,

\begin{multline}

P(X_6 > \max(X_8, X_{12}, X_{20})) \\= P((X_6 > X_8) \cap (X_6 > X_{12}) \cap (X_6 > X_{20}))

\end{multline}

This means we are looking for the probability that $X_6 > X_8$

*and*$X_6 > X_{12}$

*and*$X_6 > X_{20}$. Unfortunately, these events are not independent, so we can not break up the intersection easily. How do we know that they are not independent? The requirement for independence is that knowing one event does not affect the probability of the other. It holds that $P(A | B) = P(A)$ if $A$ and $B$ are independent. Consider the case when $X_6 > X_{20}$. If true, that makes it more likely that $X_6 > X_8$ as then $X_6$ is more likely to be high.

Instead of using independence, let's consider that we roll the d6 first. All the rolls are independent, so this doesn't affect the probabilities. We can then consider the conditional probability,

\begin{multline}

P(( (X_6 > X_8) \cap (X_6 > X_{12}) \cap (X_6 > X_{20}) ) | X_6 = x) \\= P((x > X_8) \cap (x > X_{12}) \cap (x > X_{20})).

\end{multline}

Here, the three events are independent, since they depend on independent random variables, thus,

\begin{multline}

P((x > X_8) \cap (x > X_{12}) \cap (x > X_{20})) \\= \underbrace{P(x > X_8)}_{\frac{x-1}{8}} \cdot \underbrace{P(x > X_{12})}_{\frac{x-1}{12}} \cdot \underbrace{P(x > X_{20})}_{\frac{x-1}{20}}.

\end{multline}

We can then fill in the individual terms using a previous equation as shown above.

\begin{align}

P((x > X_8) \cap (x > X_{12}) \cap (x > X_{20})) = \frac{(x-1)^3}{8 \cdot 12 \cdot 20}.

\end{align}

We can then use the conditional probability to compute the total probability, using the following form,

\begin{align}

P(A) = \sum_{x \in \Omega} P(A | X=x) \cdot P(X=x),

\end{align}

where $A = (X_6 > X_8) \cap (X_6 > X_{12}) \cap (X_6 > X_{20})$ is an event and $X = X_6$ is a random variable. Note that this is an extension of the basic statement of conditional probability, namely that,

\begin{align}

P(A \cap B) = P(A|B) \cdot P(B),

\end{align}

where $A$ and $B$ are events. Consider a set of mutually exclusive events $B_i$, the union of which encompasses the entire sample space. That is,

\begin{align}

&B_i \cap B_j = \varnothing, \quad i \neq j \\

&\bigcup_i B_i = \Omega .

\end{align}

Suppose we take the intersection of our event $A$ with the whole sample space. The resulting event is equivalent to the event $A$,

\begin{align}

A = A \cap \Omega ,

\end{align}

and thus the probabilities are the same,

\begin{align}

P(A \cap \Omega) = P(A | \Omega) \cdot P(\Omega) = P(A) .

\end{align}

Now, but substituting $\Omega$ for the union of our events $B_i$, we find a new way to write the probability of $A$.

\begin{align}

P(A) & = P(A \cap \Omega) \\

&= P\left (A \cap \left (\bigcup_i B_i \right ) \right ) \\

&=P\left( \bigcup_i \left( A \cap B_i \right) \right)

\end{align}

The last line above distributes the intersection across all the terms in the union. This is analogous to the distributive property of multiplication. That is, treat intersection like multiplication and union like addition when rearranging. A simple statement of the idea is that

\begin{align}

A \cap (B \cup C) = (A \cap B) \cup (A \cap C).

\end{align}

We know that all that sets $A \cap B_i$ are disjoint since $B_i$ are disjoint by construction. The probability of disjoint events add when considering their union, thus,

\begin{align}

P(A) &=P\left( \bigcup_i \left( A \cap B_i \right) \right) \\

&=\sum_i P(A\cap B_i).

\end{align}

Now, we can use our earlier statement of conditional probability to rewrite this as,

\begin{align}

P(A) &=\sum_i P(A\cap B_i) \\

&=\sum_i P(A | B_i) \cdot P(B_i).

\end{align}

By defining the event $B_i$ as when $X = x$, where $x \in \mathbb{Z}$, for some mapping of $i$ onto $x$ (or should it be vice-versa?), we get the equation we are looking to prove. QED.

Now we can move on with the problem at hand, and compute the probability we are interested in.

\begin{align}

P((X_6 > X_8) \cap (X_6 > X_{12}) \cap (X_6 > X_{20})) &= \sum_{x=1}^6 \frac{(x-1)^3}{8 \cdot 12 \cdot 20}\cdot \frac{1}{6} \\

&\approx 0.0195

\end{align}

We can similarly compute the other probabilities, but there is a slight wrinkle here. We tacitly benefited from the fact that the d6 has the lowest maximum. If we instead look at the probability that the d20 is highest, we have to consider that the d20 can roll a number not possible on the other dice. This shows up when we use the earlier equation for $P(X_N < x)$ outside the range we stated. We can extend it as follows.

\begin{align}

P(X_N < x) = \begin{cases}

\hfill 0 \hfill & \, x < 0 \\

\hfill \cfrac{x-1}{N} \hfill &\, x \in \mathbb{N}, x \leq N \\

\hfill 1 \hfill & \, x > N

\end{cases}

\end{align}

Focusing on the probability that the d8 is highest, there are thus two cases two consider: when the d8 rolls up to 6, and when it rolls above 6.

\begin{align}

P((x > X_6) \cap (x > X_{12}) \cap (x > X_{20})) = \begin{cases}

\cfrac{(x-1)^3}{6 \cdot 12 \cdot 20}&\, x \leq 6 \\ & \\

\cfrac{(x-1)^2}{12 \cdot 20} &\, x > 6

\end{cases}

\end{align}

From this we can compute the probability that the d8 is higher than all other dice using two summations.

\begin{multline}

P((X_8 > X_6) \cap (X_8 > X_{12}) \cap (X_8 > X_{20})) \\= \left (

\sum_{x=1}^6 \frac{(x-1)^3}{6 \cdot 12 \cdot 20}

+ \sum_{x=7}^8 \frac{(x-1)^2}{12 \cdot 20}

\right ) \cdot \frac{1}{8}

\end{multline}

\begin{align}

P((X_8 > X_6) \cap (X_8 > X_{12}) \cap (X_8 > X_{20}))&\approx 0.0638

\end{align}

Computing the probability for the d12 and d20 are similar, except that there are even more ranges.

\begin{multline}

P((X_{12} > X_6) \cap (X_{12} > X_{8}) \cap (X_{12} > X_{20})) \\= \left (

\sum_{x=1}^6 \frac{(x-1)^3}{6 \cdot 8 \cdot 20}

+ \sum_{x=7}^8 \frac{(x-1)^2}{8 \cdot 20}

+ \sum_{x=9}^{12} \frac{(x-1)}{20}

\right ) \cdot \frac{1}{12}

\end{multline}

\begin{align}

P((X_{12} > X_6) \cap (X_{12} > X_{8}) \cap (X_{12} > X_{20}))&\approx 0.222

\end{align}

\begin{multline}

P((X_{20} > X_6) \cap (X_{20} > X_{8}) \cap (X_{20} > X_{12})) \\= \left (

\sum_{x=1}^6 \frac{(x-1)^3}{6 \cdot 8 \cdot 12}

+ \sum_{x=7}^8 \frac{(x-1)^2}{8 \cdot 12}

+ \sum_{x=9}^{12} \frac{(x-1)}{12}

+ \sum_{x=13}^{20} 1

\right ) \cdot \frac{1}{20}

\end{multline}

\begin{align}

P((X_{20} > X_6) \cap (X_{20} > X_{8}) \cap (X_{20} > X_{12}))&\approx 0.622

\end{align}

Now, let's consider the probability of a tie. If we just lump them all together, then we compute the probability relatively easily, by looking at it in terms of there

*not*being a tie.

\begin{align}

P(\text{tie}) = 1 - P(\text{no tie})

\end{align}

We can construct the case when there is no tie. Consider rolling the d6 first. Any face is allowed. Now consider rolling the d8. All but one of the faces are allowed; if we roll the same as on the d6 we have a tie. Similarly for the d12, all but 2 are allowed. For the d20, all but 3 are allowed.

\begin{align}

P(\text{no tie}) &= \frac{6}{6} \cdot \frac{7}{8} \cdot \frac{10}{12} \cdot \frac{17}{20}\\

&\approx 0.620\\

P(\text{tie}) &\approx 1 - 0.620 \\

&\approx 0.380

\end{align}

Note that when we compare this to what's left when none of the four dice are higher than all the others, the values are not equal.

\begin{align}

P(\text{tie for highest}) &= 1 - P(\text{d6 strictly higher}) \\

&\qquad -P(\text{d8 strictly higher}) \\

&\qquad -P(\text{d12 strictly higher}) \\

&\qquad- P(\text{d20 strictly higher}) \\

&\approx 1 - 0.0195 - 0.0638 - 0.222 - 0.622 \\

&\approx 0.0727

\end{align}

This is because many, most actually, of the ties are between dice which are not amongst the highest. For example, the dice may roll 6, 6, 10, 15.

(I originally found this question here on BGG.)