If I put nine pieces of paper, with a digit from 1-9 written on each one, into a hat, and pull 3 out without replacement, what is the likelihood that I will select three numbers that result in a winning tic-tac-toe?

1 | 2 | 3
---+---+---
4 | 5 | 6
---+---+---
7 | 8 | 9

(In reply to

Answer by K Sengupta)

Since 3 numbers are drawn without replacement from the 9 available numbers, the total number of ways this is possible is comb(9, 3) = perm(9, 3)/3! = 504/6 = 84

Now, one can choose three numbers, disregarding order, in the following manner to produce a winning combination,:

** Along the row: (1,2,3), (4,5,6), (7,8,9), so the number of ways is 3

** Along the columns: (1,4,7), (2,5,8), (3,6,9), so the number of ways is 3

** Along the main diagonals: (1,5,9), (3,5,7), so the number of ways is 2

Thus, we have a total of 3+3+2 = 8 distinct winning combinations.

Consequently, the required probabiblity is 8/84 = 0.095238(approx.)