Are you 100% sure? Betting for the same team.

Christmas is often a time for getting together and getting in arguments. But some of the most heated arguments are between people who completely agree with each other.

Betting is often a good way of settling arguments: I bet you so much that my sports team will win against yours, can quickly lead to a resolution.

But what happens when you want to gamble with someone else, but both of you want to bet on the same outcome occurring? You could bet on a particular score being achieved, or the timing of an event. But what happens if the options are limited: is the correct direction to turn left or right ? Or if you have identical views: the game will end in a nil-nil draw, or a mutual friend will give up their resolution on a certain day because of some specific event?

The bet can go ahead if one of you is more sure than the other that they’re correct. 

Getting into Norms: a technical postscript

For mathematicians’ eyes only. The post doesn’t require much theoretical knowledge to understand, but I haven’t given many definitions.

In the last two posts I’ve been talking about an example I made with four-dimensional vectors such that , and . Finding it was more difficult than I at first expected, so I thought I would write the investigation up, which happily gives me an excuse to introduce a useful inequality.

My first thoughts were to choose something like and or , and then I’d got stuck choosing . So I decided to try to prove the opposite.

First of all, it’s not possible to create such an example in two dimensions.

Intransitive votes

In the previous post I gave an example of students deciding who best guessed some lecturers’ ages.

I chose the numbers carefully so that under three reasonable methods of measuring:

Method	First place	Second place	Third place
Method a	Adam	Beth	Charlie
Method b	Beth	Charlie	Adam
Method c	Charlie	Adam	Beth

This is actually almost identical to Condorcet’s voting paradox:

Voter	First preference	Second preference	Third preference
Voter 1	A	B	C
Voter 2	B	C	A
Voter 3	C	A	B

If three people in an election vote for candidates A, B, and C this way, then even using a method that takes account of all the preferences in one of the Condorcet voting system leads to a deadlock.