Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
    
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
1  
@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.

share|cite|improve this answer

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: $$P(z\le k) = \left( \frac{k}{6}\right)^2$$

and

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

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:

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.