How to calculate the probability of dice outcomes of a D$30$?

Question

One D$30$ dice is rolled three times, will the difference between the maximum and minimum value be greater than $15$?

So I guess this is the same as just picking three random numbers between $1-30$, and looking at the biggest and smallest number to see if the difference is greater than $15$. But what I am wondering is; How do I calculate the different probabilities here?

Answer 1

First choose which two dice will be min and max. For each combination of min and max on the two chosen dice, the third dice will be between min and max.

There are $3 \cdot 2 = 6$ ways to choose two dice and which will be max and which will be min.

If dice A has min and dice B has max,

For example, if A shows $1$ then B must show between $17$ and $30$ for the difference to be greater than $15$. If B is $17$, C can be any number between $1$ and $17$. So you can count the number of permissible outcomes as,

$\displaystyle 6 \cdot \sum_{i=1}^{14} \sum_{j=16+i}^{30} (j - i - 1)$

This does not include cases when the third number is equal to min or max. There are $ \ 3 \cdot (15-i) \ $ possibilities for third number to be equal to min and same for third number to be equal to max. So the final expression is,

$\displaystyle 6 \cdot \sum_{i=1}^{14} \sum_{j=16+i}^{30} (j - i) = 12810$

Now divide by the total of $30^3$ unrestricted outcomes for desired probability.

Answer 2

First we will calculate the probability such that the difference of maximum and minimum will be $d$.

There are $30-d$ different possible extreme cases, say $(1,d+1),\, (2,d+2),\, \ldots ,\, (30-d,30)$. Now, if all three dice have different values then there would be $3!=6$ ways to order them, and each pair could have the middle value different from the extremes of $d-1$ different ways, so for each pair there are $6\cdot (d-1)$ different possible triplets of values.

If the middle value repeat then the resulting triplet can be ordered of three different ways, so if it repeat for each pair will be $2\cdot 3=6$ possible different triplets with a repetition (the $2$ there comes because it can repeat the minimum or the maximum value).

Therefore in total you have $(30-d)(6(d-1)+6)=6(30-d)d$ valid cases where the difference of the maximum and the minimum is exactly $d$. Then the probability that you want to calculate is the sum

$$ \frac6{30^3}\sum_{d=16}^{29}(30-d)d=\frac{6}{30^3}\left(29\frac{30\cdot 29-16\cdot 15}{2}-\frac{30\cdot 29\cdot 28-16\cdot 15\cdot 14}{3}\right)\\ =\frac{12810}{30^3}\approx 0.47 $$