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July 3[edit]

Probabilities being between 0 and 1 - an arbitrary scale or a necessary one?[edit]

So I think I read in a textbook somewhere something like this: Since probability is defined as the number of desirable outcomes ( $r$ , say) over the total number of possible outcomes ( $n$ ), $0\leq r\leq n$ and so $0\leq {\frac {r}{n}}\leq 1$ . However, Probability_theory#Discrete_probability_distributions implies that the 0 to 1 scale is part of the definition of probability (and, as such, appears to be an arbitrary scale - like degrees Celsius or something). Thus, is it arbitrary or does it emerge necessarily out of the classical definition of probability? Thanks. It Is Me Here ^{t / c} 19:37, 3 July 2011 (UTC)[reply]

It's not really defined as the ratio you speak of. That ratio is correct if the outcomes are all equally probable. As a definition, though, that's circular, because it relies on knowing what probability means to start with.

I think the values 0 and 1 are pretty natural, so not "arbitrary", but there's no reason you couldn't use different endpoints if you wanted to. It would just complicate the formulas a bit. Not sure what the point would be. --Trovatore (talk) 19:44, 3 July 2011 (UTC)[reply]

There are other methods of representing probabilities, such as percentages (although there technically 100% = 1) and odds. However, the math is simplest when you use the direct 0-1 scale:

0.5 × 0.5 = 0.25

50% × 50% = 25%

1:2 × 1:2 = 1:4

So, I'd say that the probabilities either need to be in a 0-1 scale or be in some form that can be converted to that scale. StuRat (talk) 19:52, 3 July 2011 (UTC)[reply]

Note that for the last one, odds ratios are frequently stated as (outcome one):(outcome two) - e.g. chance for winning:chance for losing. So a 50% chance would be 1:1, and a 40% chance would be 2:3. This only makes your point better, as the odds of two 1:1 events both happening is 1:3 (or 1:1 × 1:1 = 1:3). The way you have it stated is more commonly represented with fraction notation - e.g. 1/2 × 1/2 = 1/4, which of course is simply 0.5 × 0.5 = 0.25 re-written. -- 174.31.196.47 (talk) 15:45, 4 July 2011 (UTC)[reply]

The classical definition is in terms of relative frequency. If you define probabilities in terms of relative frequency, then probabilities are between 0 and 1, and the probabilities of all possible events add up to 1. The problem with the definition in terms of relative frequency is that it requires being able to sample an event an unlimited number of times. To avoid all the difficulties that restriction causes, mathematicians prefer to simply define a probability distribution as a set of numbers between 0 and 1 that add up to 1. So it isn't really arbitrary, it is more like an axiom. Looie496 (talk) 18:44, 4 July 2011 (UTC)[reply]

Even if you solve the problem of being able to sample arbitrarily many times, it still doesn't make this a non-circular definition. The reason is that it is not certain that the limit of the relative frequency will equal the probability. It is almost certain, but that simply means the probability is 1. Again, this is circular. --Trovatore (talk) 05:40, 5 July 2011 (UTC)[reply]

You might to look at likelihood, for another way of looking at the whole concept of probability/likelihood/belief values. -- The Anome (talk) 18:54, 4 July 2011 (UTC)[reply]

Zero to one is a very sensible scale to use; motivated by the first example the OP gave. But you could use any scale you choose to define the range of probability. You could use the range α ≤ P(X) ≤ β, provided α < β, but you'd have to change all of the equations to reflect the new scale, and it wouldn't have a very natural explanation. That's one of the beauties of mathematics: there is a great freedom to define things as you choose. However, over time, what you end up with is the most natural, and most favoured way of doing things. Much like Wikipedia itself. — Fly by Night (talk) 19:48, 4 July 2011 (UTC)[reply]

In one sense it's a "necessary" scale; see Richard T. Cox's book Algebra of Probable Inference. But here we'd have to regard it as "the same" because of an isomorphism) as the odds scale:

{\text{odds}}={\frac {\text{probability}}{1-{\text{probability}}}}

and also the same as other scales such as the logarithm of the probability (which by some accounts, is how "surprised" one should be) and also the same as any of a number of other such scales. Michael Hardy (talk) 21:08, 4 July 2011 (UTC)[reply]

Set theory proof help[edit]

My question regards the last two pages (182 and 183) of this paper. It is stated in the paper that Axiom A implies Axiom B (these two axioms are stated on the last 2 pages of the paper), and that this implication "can be derived from Zermelo's axioms quite easily." Well, I could not figure out the derivation, regardless of how easy it is. Can someone please help me with this proof?? The paper states that Replacement is not required, but if it makes the derivation easier, I have no problem with it being used. On the other hand, the axiom of choice should not be used, because the point of this derivation is to show that Axiom A implies the axiom of choice, which is done by noting that A implies B, and that B implies the axiom of choice (and this second implication in the syllogism is already proven earlier in the paper). Thank you in advance to anyone who wants to help me with this proof! JamesMazur22 (talk) 20:06, 3 July 2011 (UTC)[reply]

I suspect you can do better by proceeding directly. You can derive AC from the proposition "any two sets have comparable cardinality", using the Hartogs number. Axiom A implies that a lot of sets have comparable cardinality, because for a given N, any set having incomparable cardinality to a subset of N must be an element of the M whose existence is guaranteed by the axiom. (Oh, not quite any set, but any subset of M having incomparable cardinality to a subset of N.)

Once you have that Axiom A => AC, you can probably also show that Axiom A <=> there are arbitrarily large inaccessible cardinals. Then you can use that proposition (together with AC) to derive B. --Trovatore (talk) 20:22, 3 July 2011 (UTC)[reply]

I would just take the author's word for it, rather than having to actually work out a proof for yourself. After all, it wouldn't have gotten past his editors otherwise. --188.28.150.187 (talk) 00:25, 4 July 2011 (UTC)[reply]

Dear 188.28.150.187,

You are a moron.

That is all.

Sincerely,

--COVIZAPIBETEFOKY (talk) 00:56, 4 July 2011 (UTC)[reply]

Not that I disagree, but this kind of language doesn't inspire confidence for outside observers. -- Meni Rosenfeld (talk) 09:07, 6 July 2011 (UTC)[reply]

What an interesting way to put it! Only beat your kids in the dark, folks... --188.29.111.138 (talk) 14:38, 6 July 2011 (UTC)[reply]

That's right! That way, they learn nothing, and you get to continue being superior to them! --COVIZAPIBETEFOKY (talk) 15:08, 6 July 2011 (UTC)[reply]