Dice rolling / Smallest value probability

Question

If $X$ is the smallest number when rolling a die $n$ times. Then how can we obtain the PDF/CDF? I thought as for $1\leq{i}\leq6$ how to find $P(X=i)$.

Example: Throws$=(3,2,4,4,5)\implies X=2$. If we want to generalize this into a PDF/CDF, I was trying to use the formula $X=\min{(X_1,X_2,\dots,X_n)}$.

I don't know how to apply the whole formula and get to the general PDF/CDF. Any tips would be welcome.

Answer 1

The idea is to find $P(X\ge i)$, it's not too difficult : the event $X\ge i$ is realized when all the throws gave a result greater than $i$, so it's $\left(\frac{7-i}{6}\right)^n$.

After that, $$P(X=i)=P(X\ge i)-P(X\ge i+1)=\left(\frac{7-i}{6}\right)^n-\left(\frac{7-(i+1)}{6}\right)^n$$ for $1\le i\le 5$, and $$P(X=6)=P(X\ge6)=\left(\frac{1}{6}\right)^n$$

Answer 2

Hint: Since different rolls are independent and have the same distribution, you have that \begin{align}P(X> x)&=P(\min{\{X_1,\dots,X_n\}}> x)]=P(X_1> x, X_2> x,\dots,X_n>x)=^{\text{independce}}\\[0.2cm]&=P(X_1>x)P(X_2>x)\dots P(X_n>x)=^{\text{identically distributed}}\\[0.2cm]&=P(X_1>x)^n=(1-P(X_1\le x))^n\end{align} Hence, the CDF $F_X$ of $X$ is given by $$F_X(x)=P(X\le x)=1-P(X>x)=1-(1-P(X_1\le x))^n=1-(1-F_1(x))^n$$ where $F_1$ denotes the CDF of $X_1$. So, all reduces in finding the CDF of one roll.

Answer 3

Here is a verification of the results in the very nice Answer of @NicholasFRANCOIS, for the case $n = 8$, based on simulating the 8-roll experiment a million times and finding the minimum face number $Y$ for each experiment.

m = 10^6;  n = 8;  die=1:6
x = sample(die, m*n, rep=T)
DTA = matrix(x, nrow=m)      # m x n matrix; each row has results of one n-roll expt
y = apply(DTA, 1, min)
table(y)/m
#     y
#        1        2        3        4        5        6  # simulated dist'n
# 0.767262 0.193576 0.035219 0.003779 0.000163 0.000001 

hist(y, prob=T, br=(0:6)+.5, col="wheat")
   pdf=numeric(6)
   i = 1:5;  pdf[1:5] = ((7-i)/6)^n - ((7-i-1)/6)^n
   pdf[6] = (1/6)^n
points(1:6, pdf, col="blue", pch=19)

round(pdf, 6)
# 0.767432 0.193550 0.035112 0.003754 0.000152 0.000001  # exact dist'n

The histogram shows simulated results, and the blue dots atop histogram bars show exact results. The resolution of the graph shows only about two decimal places, simulated results are accurate to about three places.

Note: I started by simulating the case $n = 10,$ but then minimums are almost entirely 1's and 2's.