anaxagoras
King
According to my calculations the odds of all four civs getting the same tech is around 3.7% and all 3 techs being available between the four civs 44.4%.
The first calculation is easy, and Othniel had it almost right. It is 1/3^4 = 1/81 = 1.2% for all four civs to have one tech, but then you need to multiply by three, because there are three ways this could happen. So 1.2% is the odds of all 4 civs having a particular one of the three, and 3.7% is correct for all four civs to have exactly one of the three, but not any particular one.
For the second question, it is a bit tougher math. The first choice is irrelevant. You will get at least one of the three. On the second pick, you have a 1 in 3 chance of selecting the same tech, and a 2 in 3 chance of selecting something different.
First, take the case where the first two choices match. Now you must pick two different, non-matching choices with your next two picks. That's a 2/3 chance of a new tech on choice three, then a 1/3 chance on choice four of picking the last remaining tech. Therefore, total odds of this case = 1/3 * 2/3 * 1/3 = 2/27.
In the case where your first two picks do NOT match, you have a 1/3 chance to pick the last tech right off. In the 2 of 3 times when you fail, you then have a second 1/3 chance to pick the last tech. So that's 1/3 + 2/3*1/3 = 5/9. Times the 2/3 chance that we're in this situation to begin with, or 10/27.
The total answer is the combined totals of the two cases, or 12/27. That's a 44.4% chance, just as Thar concluded.
Ain't math great?