Here's a problem in my textbook:
"From a set of 2n + 1 consecutively numbered tickets, three are selected at random without replacement. Find the probability that the numbers of the tickets form an arithmetic progression. [The order in which the tickets are selected does not matter.]"
I tried to solve it by first arbitrarily choosing 2 numbers from the first n+1 tickets. This is because in any 3 tickets in an arithmetic progression there must be at least 2 tickets from the first half of the pile. Once we've chosen our two numbers there is only one possible third ticket. Thus the probability must be 2C3/[(2n+1)C3]. However, the textbook gives n^2/[(2n+1)C3].
What went wrong?