# Probability of multiple dice rolls with constraints

I am having problems understanding how to tackle part b of the following question.

A fair die is rolled three times. What is the probability that
A) her second and third rolls are both larger than her first roll?
B) the result of her second roll is greater than that of her first roll and the result of her third roll is greater than the second.

In order to solve part A of this question I did the following:

$p_r(A) = \frac{{5\choose1}^2+{4\choose1}^2+{3\choose1}^2+{2\choose1}+{1\choose1}^2}{216}$

And for part B I attempted solving it the following way:

$p_r(A) = \frac{{5\choose1}{4\choose1}+{4\choose1}{3\choose1}+{3\choose1}{2\choose1}+{2\choose1}{1\choose1}}{216}$

but that is clearly not the answer. The answer to part b is $5/54$. Although I do not understand how to get there.

I am assumer the answer (un-simplified) is $\frac{{5\choose1}{4\choose1}}{216}$ although I am having problems understanding the logic of obtaining this.

Any help would be much appreciated.

The problem here is that for your first case when you fix your first roll to $1$, you assume that after you choose ${5\choose 1}$, you always have four things left. However this is not true. When you choose say $3$, for example, you have only $3$ choices left instead of $4$.
You need to think in a different way, which is ${5\choose 2}$ instead of ${5\choose 1}{4\choose 1}$.
• thank you for your answer. Is this what you meant by thinking in a different way ${5\choose2}+{4\choose2}+{3\choose2}+{2\choose2}$? I apologize if my response seems uneducated, we are all here to learn. Thank you. Cheers – Miroslav Glamuzina Oct 9 '15 at 5:21
• Yes this is correct. You need to choose the second and the third at once instead of picking the second and then picking the third. If you insist on your way you can only break into further cases like $(4+3+2+1)+(3+2+1)+(2+1)+(1)$ which is essentially listing all the probabilities. – cr001 Oct 9 '15 at 5:40