We have $2n+1$ numbered balls in a box. Three balls are chosen randomly and simultaneously. Find the probability that the numbers are in arithmetic progression.
My attempt to solve it:
I have defined $$S=\{\{x_1,x_2,x_3\} | x_1,x_2,x_3 \in \{1,...,2n+1\}, x_1 \not = x_2 \not = x_3\}. $$ Then, I have defined the mass function $p: S \rightarrow [0,1] $ so that for $ \overline x \in S$, $p(\overline x)=k=\frac{1}{Card(S)}$. Correction: $Card(S)=\binom{2n+1}{3}$.
Note: $Card()$ means cardinal.
In consequence, $\forall A \in \mathcal{P}(A)=\sum_{\overline x \in A} k =...=\frac{Card(A)}{Card(S)}.$
Now, we define $A_d=\{\{x_1,x_2,x_3\}\in S | x_2=x_1+d, x_3=x_2+d\}$ for $d>0$. Thus, the probability of the numbers being in arithmetic progression would be $ P(\cup_{d=1}^{2n+1} A_d)$. However, I am having trouble delimitating the range of $d$ so that $A\subseteq S$ and calculating the intersection between different '$A_d$'s.
I feel like there should be an easier way to calculate the probability but I really can't come up with it. Any help would be appreciated. Thank you!
I will try to follow your notation, as you are almost there.
As you stated, let's write
$$A_d = \{(x_1, x_2, x_3) \in S : x_2 = x_1 + d, x_3 = x_2 + d\}$$
We want to remove the "$\in S$ " restriction, so we replace it with the definition of $S$:
$$A_d = \{(x_1, x_1 + d, x_1 + 2d) : x_1, x_1 + d, x_1 + 2d \in [1, 2n + 1]\}$$
Of course, since $d\geq 1$, we only need the final term in the A.P. to be in the interval:
$$A_d = \{(x_1, x_1 + d, x_1 + 2d) : x_1 \geq 1, x_1 + 2d \leq 2n + 1\}$$
Finally, let's rearrange the final condition:
$$A_d = \{(x_1, x_1 + d, x_1 + 2d) : x_1 \in [1, 2n - 2d + 1]\}$$
From this, clearly we have $|A_d| = 2n - 2d + 1$ (iff it's positive of course). Now since the sets $A_d$ are disjoint, We have
$$|A| = \left|\bigcup_{d=1}^n A_d\right| = \sum_{d=1}^n |A_d| = \sum_{d=1}^n 2n + 1 - 2d = n^2$$
Finally, simply divide by $Card(S)$ to get your answer!