Cool Trick w/ Numbers

Originally Posted by abelardx

The Collatz conjecture is an Erdos problem. They're all deceptively easy to state.

Ramsey numbers are a perfect example. How many people do you have to invite to a party so that there's always a subset of n people who either all have met before or all haven't met before? For n = 3, the answer is 6. For n = 4, the answer is 18 (hard). For n = 5, no one knows. Finding Ramsey(5) is also an Erdos problem.

I don't get it, shouldn't there be some bayesian probability in there or is it just assumed that every person has either met or not met another with probability 0.5?

10-02-2013 03:21 PM #1
Penneywize View Profile View Forum Posts Private Message View Blog Entries Join Date Jan 2005 Posts 2,122	Originally Posted by abelardx The Collatz conjecture is an Erdos problem. They're all deceptively easy to state. Ramsey numbers are a perfect example. How many people do you have to invite to a party so that there's always a subset of n people who either all have met before or all haven't met before? For n = 3, the answer is 6. For n = 4, the answer is 18 (hard). For n = 5, no one knows. Finding Ramsey(5) is also an Erdos problem. I don't get it, shouldn't there be some bayesian probability in there or is it just assumed that every person has either met or not met another with probability 0.5?

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10-02-2013 04:03 PM #2
abelardx View Profile View Forum Posts Private Message View Blog Entries Join Date Jul 2013 Posts 646 Location navy yard	Originally Posted by Penneywize I don't get it, shouldn't there be some bayesian probability in there or is it just assumed that every person has either met or not met another with probability 0.5? IMO the first person to use the word 'Bayesian' in a conversation wins.

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Cool Trick w/ Numbers

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