# What is three different way to solve this probability / combinatorics problem of dice

I wanted to obtain as many way as possible for this problem so i decided to asked for three different approach for this problem:

A standard cubical die is thrown four times, with respective outcome a,b,c,d. what is the probability that $a\leq{b}\leq{c}\leq{d}$

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Here's one way. When throwing the die $n$ times with results $X_1, \ldots, X_n$, let $f_n(x)$ be the probability that $x \le X_1 \le X_2 \le \ldots X_n$. By conditioning on $X_1$, this satisfies the recurrence $$f_{n}(x) = \dfrac{1}{6} \sum_{y=x}^6 f_{n-1}(y)$$ with initial conditions $f_1(x) = P(X_1 \ge x) = \dfrac{7-x}{6}$ for $1 \le x \le 6$.

It is easy to calculate $f_4(1) = \dfrac{7}{72}$.

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Let $X = \{1,2,3,4,5,6,l,m,b\}$ be a set.

Sample space is $\Omega = \{1,2,3,4,5,6\}^4$

$F = \{A \subset X | |A| = 4\}$ is the favourable event. Here,

1) Drawing $l$ with numbers represents the smallest number came twice. For example: drawing $\{1,2,3,l\}$ represents throwing $1,1,2,3$.

2) Drawing $m$ with numbers represents the middle number came twice. For example: drawing $\{1,2,3,m\}$ represents throwing $1,2,2,3$.

3) Drawing $b$ with numbers represents the biggest number came twice. For example: drawing $\{1,2,3,b\}$ represents throwing $1,2,3,3$.

4) Drawing $l,m$ with numbers represents the event that the smaller number repeats thrice. For example: drawing $\{1,2,l,m\}$ represents throwing $1,1,1,2$.

5) Drawing $b,m$ with numbers represents the event that the bigger number repeats thrice. For example: drawing $\{1,2,b,m\}$ represents throwing $1,2,2,2$.

6) Drawing $l,b$ with numbers represents the event that both numbers repeats twice. For example: drawing $\{1,2,l,b\}$ represents throwing $1,1,2,2$.

7) Drawing $l,b,m$ with a number represents the event where the number appears all the time. For example: drawing $\{1,m,l,b\}$ represents throwing $1,1,1,1$.

A little thought shows that this establishes a bijective map between this sample space and the actual events and thus $|F| = \binom{9}{4}$. Thus

$\Pr \{F\} = \dfrac{|F|}{|\Omega|} = \dfrac{126}{6^4} = \dfrac{7}{72}$

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Whenever you fix the 3rd at $k$, you have $1+2+...+k$ valid sequences at dice first 3 dice. When the 3rd dice is $k$, the 4th can have one of the $6-k+1$ choices.

So it is

$\sum^6_{i=1}(7-i)i(i+1)/2$, adding up to 126 valid sequences in total. And the probability of a sequence is $(1/6)^4$.

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