Reply #21: Friendly amendment [View All]

[eomer]

"There are 374 distinct precincts, each precinct (except the
first and last) has 2 adjacent precincts in the sequence, --
one above and one below. In total there are (n-1) x 2 = 746
possible adjacent pairs."

I submit that the number of adjacent pairs is (n-1).

In the following list how many adjacent pairs are there?
6.5
7.8
7.8
3.7
9.0

I count 4. If you count (n-1) x 2 = 8 then you are counting some of them twice.

The way I think of the probability is that once you have the first number in the sequence then you have (n-1) tries and the question each try is whether you match the previous number in the list.

If you try to think of the probability in terms of each number in the list possibly matching either the one before it or the one after it then I think you will get it wrong.

The rest of your approach makes total sense to me.