# Die Rolled 3 times probabilites

So I have a practice exam with this question:

Suppose that a balanced die is rolled 3 times, and let $X_i$ denote the number that appears on the $i^{th}$ roll (i=1,2,3). Evaluate:

a. $P(X_1>X_2)=\frac{{6 \choose 2}}{6^2}$

I don't understand how they got this answer?I think I understand the denominator, because each die has 6 numbers, so there are $6^2$ total possibilities, but I don't understand how they got the answer for the numerator.

Similarly,

b.$P(X_1>X_2>X_3)=\frac{{6 \choose 3}}{6^3}$

c.$P(X_1+X_2+X_3\le4)=\frac{4}{216}$

for this one I just wrote out the possibilities and counted them, is there a more sophisticated way of doing this?

• What's this: $P(X_1+X_2+X_3+X_4)$??? The probability that the sum of these $4$ variable is what??? Feb 18, 2015 at 20:27
• Oh typo, less than or equal to 4 Feb 18, 2015 at 20:41
• Write down the ways this can happen. E.g., for the X1>X2>X3 case, something like {{3, 2, 1}, {4, 2, 1}... {6,5,4}. You'll see there's 20 of them, corresponding to the ways you can choose three numbers from six, hence C(6,3)... Feb 18, 2015 at 23:24

For a.: You want to count how many ways there is to pick $X_1$ and $X_2$, ordered (permutation), so that $X_1 > X_2$.
This is equivalent to asking how many ways your can pick $Y_1$ and $Y_2$, both different. Whichever choice you make, one will be larger than the other. Just set $X_1$ to be the least of the two and $X_2$ the greatest. This is a combination, ${6 \choose 2}$.
Similarly, for b. and c. You can just count the numbers of way to pick $n$ different values and see that you can then order them from least to greatest. There'll be just one order you can put them in for each combination of numbers you picked.
It is actually counter-intuitive, but it works. In a nutshell: For each combination of $n$ different numbers you can take from a set, there's one and only one permutation of those same numbers that happens to be ordered correctly.