<s>[INST] What is the probability of strictly increasing digits in a randomly generated 4 figure number? Suppose I make a $4$ digit number with each digit chosen at random from $\{0,1,...,9\}$, with repetition allowed. How do I find the probability that the digits in the number are in strictly increasing order?
I don't quite see how to proceed (unless naively writing out all possible combinations)?
 [/INST] The first digit cannot be greater than 6, because with 7, 8, or 9 as the first digit it is not possible to create a 4 digit number with strictly increasing digits.
Let's denote the possibility of the first digit as A, then as the first digit is restricted with the set {0, 1, ..., 6} we have $$P(A)=\frac{7}{10}$$
The second digit is restricted with the set {1, 2, 3, 4, 5, 6, 7} by the same requirement of strictly increasing digits. Furthermore, the probability of the second digit is conditioned by the first digit. Denoting the probability of the second digit as B and the value of the first digit as x we have this




1st digit
0
1
2
3
4
5
6




P(B|x)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$




Further, the third digit is also restricted with the relative set {2, 3, 4, 5, 6, 7, 8} and its probability is conditioned by the second digit. Let's denote the probability of the third digit as C and the value of the second digit as y.




2nd digit
1
2
3
4
5
6
7




P(C|y)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$




The fourth digit is restricted with the set {3, 4, 5, 6, 7, 8, 9} and its probability is conditioned by the third digit. Let's denote the probability of the fourth digit as D and the value of the third digit as z.




3d digit
2
3
4
5
6
7
8




P(D|z)
$\frac{7}{10}$
$\frac{6}{10}$
$\frac{5}{10}$
$\frac{4}{10}$
$\frac{3}{10}$
$\frac{2}{10}$
$\frac{1}{10}$




So actually the probability of the number with strictly increasing order is the sum of all the possible products of the 4 digits probabilities like $\frac{7}{10}*\frac{7}{10}*\frac{7}{10}*\frac{7}{10}$.
Eventually we can define the final formula as $$\sum\frac{7*a*b*c}{10^4}$$ for each a, b, and c $\in \{1, 2, 3, 4, 5, 6, 7\}$ where $a\geq b \geq c$
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