Probability and arithmetic progression Suppose $4$ numbers are chosen at random from $1, 2, ..., 20$. What is the probability so that the $4$ numbers chosen can be somehow arranged to make them an arithmetic sequence? I have no clue how to start with this problem. I can do it with 3 numbers by counting all the arithmetic sequence formed by taking $i^{th}$ number (where $i = 1, 2, ..., 20$) as the middle term.
 A: As I understand it from the wording of the question, you need to count the number of unordered sets of four numbers which can be firmed into an arithmetic progression, so we count the progressions $1234$ and $4321$ as the same set (combination).
You can enumerate the possible sets as:
Common difference $1$, $n=17$
Common difference $2$, $n=14$
....and so on...
Common difference $6$, $n=2$
This makes a total of $57$ possible sets. The total possible sets is $\binom {20}4$.
So the required probability is $$\frac{57}{\binom{20}4}=\frac{1}{85}$$
This could give you some idea as to how to figure out the general case.
A: $\underline{Simplified formula}$
You already understand that $1 \le d \le 6$ for ascending sequences.
I have taken the numbers to be chosen without replacement, and as subsequently clarified by OP, only ascending sequences.
The number of such sequences for any given $d$ can easily be seen to correspond to $20 - 3d$, since this automatically gives the highest possible starting number for the sequence.
$$Thus\quad Pr = \dfrac{\sum_{d=1}^6 (20-3d)}{\binom{20}4}$$
PS
The numerator can be further simplified to $6*20 - 3(1+2+...+6)$
Taking $D$ to be the maximum value of $d$ possible, the numerator reduces to
$20D - 3D(D+1)/2$
and the formula becomes $\dfrac{20D - 3D(D+1)/2}{\dbinom{20}4}$
A: I don't see any general way to solve this in few seconds. However there are facts that can help you. For example $d=a_2-a_1$ can be maximum $6$ in this case and the sequence can be $1,7,12,17$ and minimum $-6 (17,12,7,1)$. When $d=1$ you have numbers from $1$ to $4$ in any order, that is $4!=24$ ways to arrange them, then $2$ to $5$ similarly and so on up until $17$ to $20$. So a total of $(17-4)*4!$ when $d=1$. For $d=2$ you will have less possible ways and so on up until $d=6$ where you have $2*4!$ ways. For the cases when $d<0$ It looks its the same numbers as when its positive so you can count the cases when $d>0$ and double it. Don't forget $d=0$ case it's also considered arithmetic sequence.
