Blogs

Blogs

A Certain Forecast is Proven

2
By Topics | Advanced Analytics

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.

Tags
Share

About Author

Mike Gilliland

Product Marketing Manager

Michael Gilliland is a longtime business forecasting practitioner and formerly a Product Marketing Manager for SAS Forecasting. He is on the Board of Directors of the International Institute of Forecasters, and is Associate Editor of their practitioner journal Foresight: The International Journal of Applied Forecasting. Mike is author of The Business Forecasting Deal (Wiley, 2010) and former editor of the free e-book Forecasting with SAS: Special Collection (SAS Press, 2020). He is principal editor of Business Forecasting: Practical Problems and Solutions (Wiley, 2015) and Business Forecasting: The Emerging Role of Artificial Intelligence and Machine Learning (Wiley, 2021). In 2017 Mike received the Institute of Business Forecasting's Lifetime Achievement Award. In 2021 his paper "FVA: A Reality Check on Forecasting Practices" was inducted into the Foresight Hall of Fame. Mike initiated The Business Forecasting Deal blog in 2009 to help expose the seamy underbelly of forecasting practice, and to provide practical solutions to its most vexing problems.

2 Comments

  1. Marge on

    Interesting proof, Mike. I can't agree with how your logic works though. There's something off about the way you've shown the contradiction. In order to contradict your statement, other statements must also be true. This is where I have a problem. Can being friendless be considered as having "the least number of friends"?

  2. Mike Gilliland on

    Marge -- thanks for reading my blog and your comment, but not sure I'm following. I understand "friendless" to mean having zero friends. Are you suggesting someone can have a negative number of friends?
    The proof is achieved via the Pigeonhole Principle, that if you stuff p pigeons into p-1 pigeonholes, at least one of the pigeonholes will have more than one pigeon. I just pray that the pigeonholes are roomy enough so that no pigeons are harmed during the aforesaid stuffing.

Back to Top