Probability - Equally likely outcomes

[CAF123]

Homework Statement

1) N people, including A and B, are randomly arranged in a line. Compute the probability that A and B sit next to each other.
If instead they sit in a circle, compute the probability that A and B sit next to each other.

The Attempt at a Solution

To the first part, I got 2/N which is correct. My reasoning is:
In the line, there are N! different rearrangements of the N people. Now glue A and B together, then you would be ordering (N-1)! people. Since A can sit to the left of B or B can sit to the left of A there are 2! additional rearrangments which gives a total probability of (since each rearrangment is equally likely as the rest) 2!(N-1)!/N! = 2/N
Now for the circle: same as before, but instead I thought of 'wrapping' the line to make a circle. So A could be at the start of this line and B could be right at the other end. I believe this gives another 2! possiblilties which gives a total of 2!(N-1)! + 2! possibilities.
Divide this by the sample space (N!) gives \frac{2((N-1)! +1)}{N!}. I think the answer is supposed to be 2/(N-1). Can anyone see my error?

 

[Ray Vickson]

Homework Statement

1) N people, including A and B, are randomly arranged in a line. Compute the probability that A and B sit next to each other.
If instead they sit in a circle, compute the probability that A and B sit next to each other.

The Attempt at a Solution

To the first part, I got 2/N which is correct. My reasoning is:
In the line, there are N! different rearrangements of the N people. Now glue A and B together, then you would be ordering (N-1)! people. Since A can sit to the left of B or B can sit to the left of A there are 2! additional rearrangments which gives a total probability of (since each rearrangment is equally likely as the rest) 2!(N-1)!/N! = 2/N
Now for the circle: same as before, but instead I thought of 'wrapping' the line to make a circle. So A could be at the start of this line and B could be right at the other end. I believe this gives another 2! possiblilties which gives a total of 2!(N-1)! + 2! possibilities.
Divide this by the sample space (N!) gives \frac{2((N-1)! +1)}{N!}. I think the answer is supposed to be 2/(N-1). Can anyone see my error?

For any location of A there are N-1 other locations B can occupy, and two of them are next to A.

You might want to worry about why this same argument does NOT apply to the linear case!

RGV

 

[CAF123]

For any location of A there are N-1 other locations B can occupy, and two of them are next to A.

You might want to worry about why this same argument does NOT apply to the linear case!

RGV

Yes, this makes sense. In a row, if A is at the end, then B can only be in one possible place.
What is wrong with my argument? I thought by adding another 2! rearrangements, I had covered this?