We have a finite arithmetic progression $ a_n $, where $ n \geq 3 $ and its $r\neq 0 $. We draw three different numbers. We have to calculate the probability, that these three numbers in the order of drawing will create another arithmetic progression.
My proposition :
$ \Omega={n!\over (n-3)!}$
$ \mathbf{A}= {n \choose 3} \cdot 2$
But I think that my way of thinking about way of counting $ \mathbf{A}$ is incorrect.
Any suggestions how can I count it? Thanks in advice, Kuba!
On
Let the sequence be $a_1,a_2,a_3,...a_n$. Then $|\Omega | = {n\choose 3} = {n(n-1)(n-2)\over 6}$
and each good 3-sequence is uniquely defined by it's middle term and the difference $d$. So if middle term is $a_m$ then the difference can be $d,2d,...,(m-1)d$.
So if we have $n=2k$, then we have $$1+2+...+(k-1)+(k-1)+...+2+1 = k(k-1)$$ good 3-sequences. So the probability is $$P = {6k(k-1)\over 4k(2k-1)(k-1)} = {3\over 2(n-1)}$$
If $n=2k+1$ then we have $$1+2+...+(k-1)+k+(k-1)+...+2+1 = k(k-1)+k=k^2$$ good 3-sequences. So the probability is $$P = {6k^2\over 2(2k+1)k(2k-1)} = {3(n-1)\over 2n(n-2)}$$
$\Omega = n(n-1)(n-2)$
We're interested in triples whose elements differ by $r$, $2r$ ... up to $\lceil n/3\rceil r$
Notice, that if a triple $a_i,a_k,a_l$ is OK, then so is $a_l,a_k,a_i$.
My idea was to "stick together" those desired elements that form each triple and calculate their quantity. To be precise: there are $2(n-2)$ triples whose elements differ by r, $2(n-4)$ triples whose elements differ by $2r$ .... and $2(n-2\lceil n/3\rceil)$ triples whose elements differ by $\lceil n/3\rceil r$. Then:
$P(A) = \frac{2(n-2+n-4...+n-2 \lceil n/3\rceil)}{n(n-1)(n-2)}$