I need the equation to the odds of this happening

Question

So there is a group of 9

they are all drawing names for secret santa

what are the odds of:

1 getting 2 and 2 getting 1

3 getting 4 and 4 getting 3

5 getting 6 and 6 getting 5

7 getting 8, 8 getting 9, 9 getting 7

Answer 1

If you are #1 and you get ready to choose, you have 8 possible choices that you can draw: 2, 3, 4, 5, 6, 7, 8, or 9 so the probability of choosing 2 is 1/8. However, you are asking what the odds are. The odds (in favor of choosing 2) are the number of successes out of the number of failures. In this case, there is one possible success (when you pick "2") and 7 possible failures (when you pick either 3, 4, 5, 6, 7, 8, or 9) so the odds in favor of choosing 2 are 1:7.

But you are being asked for the odds of TWO things happening. It's a little funny though because 1 has already selected. So either the odds are zero, or, when 2 selects they get to select from all the options as if 1 had not selected yet. So, let's assume that's the case. When 2 selects then, they have 8 possible choices that they can draw: 1, 3, 4, 5, 6, 7, 8, or 9 so the probability of choosing 1 is 1/8. That means the probability of both events happening are 1/8 * 1/8 = 1/64. That means the odds in favor of this happening are 1:63 (one possible success where 1 picks 2 and 2 picks 1, and 63 other possibilities where other things happen).

This same scenario holds true for each of the other cases except the last one. The probability of 7 picking 8 is 1/8, the probability of 8 picking 9 is 1/8, and the probability of 9 picking 7 is 1/8 so the probability of all three events happening are 1/8 * 1/8 * 1/8 = 1/512 which means the odds in favor of this happening are 1:511