# Cumulative distribution and probability mass functions.

I was faced on my probability course with the following problem:

A die is rolled four times. Let $U$ be the smallest value. Calculate the probability density function and probability mass function of the random variable $U$.

How can this be solved? Now, I started solving this, but apparently I calculated the probability mass function for U being the biggest value. For example I got $P(U = 1) = 1/1296$ and $P(U = 6) = 671/1296$, which is obviously wrong.

• if you think you solved the case of the max value, why dont you just "flip things over"? i.e. by symmetry, $P(max = 1) = P(min = 6) = 1/6^4$. – antkam Apr 18 '18 at 13:27
• It seems to be working. After using the method you provided below, I got the same results in reverse. Thanks! – HaydenSpetting Apr 18 '18 at 18:19
• You're welcome, but my point is: If you haven't already solved the max case, then finding e.g. $6^4 - 5^4 = 671$ would indeed have been helpful. But since you have already solved the max case (by whatever method) and found $P(max = 6) = 671/1296$, then that's also $P(min=1)$. – antkam Apr 18 '18 at 18:27

If you roll a dice four times you obtain a four digit number, so there are $6^4$ different outcomes. How many outcomes will result in having $6$ as the minimum value? Clearly only one, namely $6666$. So $P(U=6)=\frac{1}{6^4}$.
How many outcomes are there where $5$ is the minimum value? Well one posibility is that $5$ is rolled on the first roll, i.e. the outcome is of the form $5xyz$, where $5\leq x,y,z \leq 6$. There are clearly $2^3$ outcomes of this form. Another possibility is that the minimum $5$ is rolled not on the first but on the second roll, i.e. the outcome is of the form $65xy$, where $5\leq x,y \leq 6$. There are clearly $2^2$ outcomes of this form. Similarly there are $2$ outcomes where the minimum $5$ is rolled first on the third roll and only one where it is rolled first on the last roll. So there are $2^3 + 2^2 + 2 + 1=15$ outcomes where $5$ is the minimum value. So $P(U=5)=\frac{15}{6^4}$.
• An easier way: $P(U \ge 5) = \big(\frac{2}{6}\big)^4$ and $P(U = 5) = P(U \ge 5) - P(U = 6)$. Similarly $P(U = 4) = P(U \ge 4) - P(U \ge 5)$ etc. – antkam Apr 18 '18 at 13:30