# Discrete Random Variable of a die

A icosahedral die has each face marked with different numbers from 1 to 20. Assume that each face is equally likely to occur on a single roll, the outcome is a random variable $X \sim DU(20)$

1) If a die is rolled twice, find the pdf of the smallest value obtained, say $Y$

2) If the die is rolled three finds, find the probability that the largest integer value is 3

Okay so DU just means "Discrete Uniform Distribution"

The answers are apparently $f(y) = (41 - 2y)/400$ for $y=1,2,\dots,20$ and 19/8000 respectively.

I am trying to work backwards and I've noticed that they probably some how multiplied 1/n twice to get that 400 in the bottom.

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1) We calculate the probability that the minimum is $y$. This can happen in three ways: (i) We get a $y$ on both rolls; (ii) We get a $y$ on the first roll and something bigger than $y$ on the second; (iii) the other way around.

For (i), the probability is clearly $(1/20)^2$.

For (ii), the probability first is $y$ is $1/20$. There are $20-y$ numbers greater than $y$, so the probability of being greater than $y$ on the second is $(20-y)/20$. Thus the probability of (ii) is $(20-y)/400$.

The probability of (iii) is the same as that of (ii).

Add up. We get $\dfrac{1}{400}+\dfrac{20-y}{400} +\dfrac{20-y}{400}$.

There is a nicer (I think) way to solve 1). We have $Y=y$ iff (a) both tosses are $\ge y$ and (b) it is not the case that both tosses are $\ge y+1$. The probability of (a) is $\left(\dfrac{21-y}{20}\right)^2$. The probability of (b) is $\left(\dfrac{20-y}{20}\right)^2$. Subtract, and, if desired, simplify.

The second way may seem harder. But note that the cumulative distribution function of $Y$ is most easily arrived at in this way. Similar ideas, for $\max$ as well as $\min$, will continue to be useful.

2) Here is a nice way to solve the problem. The probability they are all $\le 3$ is $\left(\dfrac{3}{20}\right)^3$. The probability they are all $\le 2$ is $\left(\dfrac{2}{20}\right)^3$. Subtract.

Alternately, we can divide into cases: all tosses are $3$; there are two $3$'s; there is one $3$. The relevant probabilities are not hard to calculate. For example, the probability of two $3$'s is $(3)(1/20)^2(19/20)$. Add the three answers we get.

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Could you explain (2)'s first method? I don't understand why it is $(3/20)^3$, where does the 3 inside come in? –  Hawk Dec 4 '12 at 2:07
We are tossing a die $3$ times. The probability that on the first toss we get a number $\le 3$ is $3/20$. Just like probability we get a number $\le 8$ is $8/20$ ($20$ possibilities, of which $8$ are favourable). We have same probability for second toss, third toss, so multiply. (Unfortunately, the $3$'s come from two different places, $\le 3$, which gives $3/20$, and $3$ tosses, which gives the exponent.) I will be away for several hours, so probably cannot answer further questions until roughly $4$ hours from now. –  André Nicolas Dec 4 '12 at 2:14
Wait, so why are you subtracting? –  Hawk Dec 4 '12 at 7:05
We found the probability they are all $\le 3$. But that includes having everybody being $\lt 3$, in which case the largest integer would not be $3$. If $a$ is probability all $\le 3$, and $b$ is probability all $\le 2$, then probability the maximum is $3$ is $a-b$. –  André Nicolas Dec 4 '12 at 7:09