Single event probability equivalent to that of its permutations?

[Loren Booda]

Is the probability for a particular event, out of a set of events, equal to the total [normalized?] probability for all permutations of events from the set, including the particular event?

Say you have a 2 x 2 square with cells numbered 1 to 4. I am asking if the probability for square 1, p(1) is equal to the total probability for its permutations, p(1,2)+p(1,3)+p(1,4)+p(1,2,3)+p(1,2,4)+p(1,3,4)+p(1,2,3,4). Or should I divide this by the number of permutations?

 

[matt grime]

Is the probability for a particular event, out of a set of events, equal to the total [normalized?] probability for all permutations of events from the set, including the particular event?

The answer is almost surely no, whatever question you meant to ask. Just suppose that something has zero probability.

Say you have a 2 x 2 square with cells numbered 1 to 4. I am asking if the probability for square 1,

This doesn't make sense. I have a square with 4 numbers. This is nothing to do with probabilities.

p(1) is equal to the total probability for its permutations, p(1,2)+p(1,3)+p(1,4)+p(1,2,3)+p(1,2,4)+p(1,3,4)+p(1,2,3,4). Or should I divide this by the number of permutations?

How is P(1,2) a permution of P(1)?

 

[Loren Booda]

Consider this, demonstrable no doubt by the binomial theorem:

Given a set of events occurring with equal probability, the total number of permutations including a given event is one more than the total number of permutations without that event.

 

[matt grime]

I'll say it again: the total number of permutations of what?

There are 6 permutations of the numbers 1,2,3 and 4 permutations of 1,2. You know, n!, right? And again, probability has nothing to do with this fact.

 

[Loren Booda]

matt,

It turns out, upon further inspection, that what I was trying to describe is a redundancy. Sorry for the wild goose chase.