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You roll 2 ordinary dice. Let X denote the maximum of the two numbers you get. What is the distribution of X?

I did the problem as follows:

$$\begin{array}\\ X &= 1: (1, 1) \\ X &= 2: (1, 2), (2, 1), (2, 2) \\ X &= 3: (1, 3), (3, 1), (2, 3), (3, 2), (3, 3) \\ X &= 4: (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4) \\ X &= 5: [9 \; \text {possibilities}] \\ X &= 6: [11 \; \text {possibilities}] \end{array}$$ and $$\begin{array}\\ P(X = 1) &= 1/36\\ P(X = 2) &= 3/36\\ P(X = 3) &= 5/36\\ P(X = 4) &= 7/36\\ P(X = 5) &= 9/36\\ P(X = 6) &= 11/36 \end{array}$$

Is there a counting formula I can use to do this problem without enumerating all the possibilities? It was a bit tedious.

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up vote 2 down vote accepted

Hint: Look for a pattern in the numerators of your probabilities! See if you can derive the counting formula for yourself.

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I notice that they are consecutive odd numbers, but I wouldn't know that if I didn't enumerate the possibilities... – Frank Epps Aug 1 '13 at 18:30
@FrankEpps: Exactly right. I would not have noticed this either. However, a lot of mathematics is noticing patterns and trying to extrapolate. Now that we've noticed a pattern, we should try to see the general reasoning. Suppose we want our largest roll to be $n$. If it appears on the first die, there are $n$ choices for the second die. Similarly for the case where $n$ appears on the second die. Now we've counted the case $(n,n)$ twice, so subtracting this off we see there are $2n-1$ possibilities. This gives a probability of $\frac{2n-1}{n^2}$. – Jared Aug 1 '13 at 18:33

Break this down into cases:

If the dice both roll the same value that is 1 combination.

Otherwise, one of the die has to be lower than X and can have values from 1 to (X-1) and there are 2 dice which means there are $2*(X-1)=2X-2$ combinations in this case

Now, add these together and we get: $2X-1$ which is what you have there.

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An approach that does not involve any counting follows from taking into account that the $z=\max(x_1,x_2)$ translates into the following cumulative distribution functions:

$$P(z\le k) = P(x_1\le k,x_2\le k) = P(x_1\le k)P(x_2\le k)$$

The last equality follows from the independence of the random variables $x_1$ and $x_2$. In order to obtain the distribution function of the random variable $z$, $P(z=k)$, and given that $z,x_1,$ and $x_2$ are discrete random variables we use:

$$P(z=k) = P(z\le k) - P(z\le k-1).$$

For the particular case of a two fair die with 6 sides: $$P(z\le k) = \left( \frac{k}{6}\right)^2$$


$$P(z=k)= \left( \frac{k}{6}\right)^2-\left( \frac{k-1}{6}\right)^2.$$

Which is a different way of obtaining the above probabilities without counting.

I think this approach is better suited for generalisation to unfair die and die with different number of sides.

By the way, if there are $n$ die all with $p$ sides the solution is:

$$P(z=k)= \left( \frac{k}{p}\right)^n-\left( \frac{k-1}{p}\right)^n.$$

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