Solutions to the variations on the birthday problem:

1. In a group of 8 people, what is the probability that two people's birthdays are in the same month?

P(all different) = (12/12)(11/12)(10/12)(9/12)(8/12)(7/12)(6/12)(5/12) = .05

P(all different) = 5%

P(2 in the same month) = 1 - (12/12)(11/12)(10/12)(9/12)(8/12)(7/12)(6/12)(5/12) = .95

P(2 in the same month) = 95%

2. In a group of 6 people, what is the probability that two people's birthdays are in the same month?

P(all different) = (12/12)(11/12)(10/12)(9/12)(8/12)(7/12) = .58

P(all different) = 58%

P(2 in the same month) = 1 - (12/12)(11/12)(10/12)(9/12)(8/12)(7/12)(6/12)(5/12) = .42

P(2 in the same month) = 42%

3. If everyone in a group of 4 people rolls a 6 sided die, what is the probability that (at least) two people roll the same number?

P(all different) = (6/6)(5/6)*(4/6)*(3/6) = .28

P(all different) = 28%

P(2 the same number) = 1 - (6/6)(5/6)*(4/6)*(3/6)= .72

P(2 the same number)= 72%

4. If everyone in a group of 5 people rolls a 20 sided die, what is the probability that (at least) two people roll the same number?

P(all different) = (20/20)(19/20)*(18/20)*(17/20)*(16/20) = .58

P(2 the same) = 1 - (20/20)(19/20)*(18/20)*(17/20)*(16/20) = .42

P(2 the same) = 42%

5. If I roll 3 8-sided dice, what is the probability that (at least) two of the dice will have roll same number?

P(all different) = (7/8)(6/8) = .66

P(2 the same) = 1 - (8/8)(7/8)(6/8)

P(2 the same) = 34%

Notice that for all of these, the first fraction in the product can be omited, because it is 1: 1=366/366=12/12=6/6=20/20=8/8. That's because the first person is guaranteed to not overlap with any previous person (because there isn't a previous person).