>>49723779
>>49723780
Duly noted, I will make them even easier for my dumbed down /v/hu friends

But anyway here's how you would solve this:
There are 4 people, let's number them from 1 to 4.
Each one gets to write down the name of a person who is not themselves, so person 1 for instance can write down person 2, 3 and 4's names
So in total we have 3^4 = 81 arrangements for this game
We can represent all these arrangements with a string of four numbers, for instance 3412 is when person 1 wrote person 3's name, person 2 wrote 4's name, person 3 wrote 1's name and person 4 wrote 2's name

For idiots, you could just write down all 81 possible arrangements and count manually, that would be brute force.

But there's a simpler way: recognize that one and only one of the following can occur
(1)there are no pairs
(2)there is exactly one pair
(3)there are exactly two pairs
We want the probability of (2) plus (3)

The probability of having exactly 2 pairs is easy to calculate. The first person gets 3 choices, but whoever he chooses is now locked into choosing him, this leaves two persons who are also locked into choosing eachother, so the probability for (3) is 3/81 = 1/27

The probability of having exactly 1 pairs is more difficult to calculate, but not too difficult. First we choose our pair from the four people, so we have (4 choose 2) = 6 choices, this pair is locked, the other two persons can write down anyone's name except their own, so we have (3*3)-1 = 8 possibilites. So in total we have 6*8=48 arrangements.

Now finally the probability of (3) is 48/81 = 16/27. So the final answer is (1/27) + (16/27) = 17/27
Which is about 63% . . .