Monthly Archives: December 2011

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

Betting is often a good way of settling arguments: “I bet you so much that my sports team will beat yours”, can quickly lead to a resolution. But some of the most heated arguments are between people who completely agree with each other. So what happens when you want to gamble with someone else, but both of you want to bet on the same outcome?

You could bet on a particular score, or the timing of an event. But how about 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 this party will win the election?

Here, the bet can go ahead as long as one of you is more sure than the other that they’re correct.  Continue reading

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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 a, b, c such that \|a\|_1 > \|b\|_1 > \|c\|_1, \|b\|_{2} > \|c\|_{2} > \|a\|_{2} and \|c\|_{\infty}>\|a\|_{\infty}>\|b\|_{\infty}. 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 c=(10,0,0,0) and a=(6,6,0,0) or a=(4,4,4,0), and then I’d got stuck choosing b. So I decided to try to prove the opposite.

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

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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.  Continue reading

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