There is nothing the gambler, investor, forecaster, or Match.com dater likes as much as the sure thing. Don't we all? Back in April I stated what I claimed to be a sure thing forecast:
In any group of 2 or more people, there is at least one pair of people who have the same number of friends in the group.
This forecast is based on the definition of friendship as a mutual relationship between two people -- either they are friends with each other, or they are not. You can't be friends with yourself.
Readers were challenged to come up with a counterexample (which would have won them a signed copy of The BFD (the book)), or offer a proof of my claim. Since I received neither counterexamples nor proofs, I must conclude that The BFD's followers were either too engrossed watching Scotty win American Idol, or else too busy deleting objectionable tweets from Congressman Weiner. So here is the proof:
Consider any group with n members (where n > 1, which is why we call it a group and not just some guy).
Suppose (for contradiction) that each of the n members has a different number of friends within the group.
Since you can't be a friend of yourself, the most friends anyone can have in the group is n-1.
Since you can be friendless, the least number of friends anyone can have in the group is 0.
So for the n people to each have a different number of friends within they group, they must be spread over the n possibilities: 0 friends, 1 friend, ..., n-1 friends.
But this results in a contradiction: if someone is friends with everyone else in the group (has n-1 friends), then there can't be someone in the group with 0 friends.
Therefore, the supposition that each of the n members of the group has a different number of friends within the group is impossible, so there must be at least one pair of people in the group with the same number of friends in the group.
It's too bad that all our real-life forecasting problems aren't this easy to solve.
Goodnight and good luck.