1. ## Hypotheses and Probability

For Evan,

Note: c(e) refers to the logical content of e (not including tautologies).

If p(h|e) is greater than p(h), then p(e|h) = 1.

If p(e|h) = 1, then c(e) is a subset of c(h).

If c(h1) = c(h) - c(e), then c(h1) + c(e) = c(h)

Therefore, h = h1 + e

We can now replace any instance of h with h1 + e.

If p(h1 + e|e) is greater than p(h1 + e), then p(e|h1 + e) = 1.

However, if we break apart h1 + e, we can isolate h1: that part of h which is not equal to e. We can then ask whether h1 is supported by e,

Is p(e|h1) = 1?

Since h1 is derived from c(h) minus c(e), then the answer is no.

In other words, that portion of h which goes beyond e is not made more probable by e. The apparent increase in probability of all of h arises from treating hypotheses as indivisible elements -- a fallacy of composition. What probabilities you get, and what supports what, just depends on how you say it.

Edit: I haven't actually tried writing this out in this form before, so I probably (heh) made some mistakes, knowing me.

2. Why didn't you just PM this?

3. Originally Posted by SmileyMan
Why didn't you just PM this?
Because I felt like it?

4. Originally Posted by reason
For Evan,

Note: c(e) refers to the logical content of e (not including tautologies).

If p(h|e) is greater than p(h), then p(e|h) = 1.

If p(e|h) = 1, then c(e) is a subset of c(h).

If c(h1) = c(h) - c(e), then c(h1) + c(e) = c(h)

Therefore, h = h1 + e

We can now replace any instance of h with h1 + e.

If p(h1 + e|e) is greater than p(h1 + e), then p(e|h1 + e) = 1.

However, if we break apart h1 + e, we can isolate h1: that part of h which is not equal to e. We can then ask whether h1 is supported by e,

Is p(e|h1) = 1?

Since h1 is derived from c(h) minus c(e), then the answer is no.

In other words, that portion of h which goes beyond e is not made more probable by e. The apparent increase in probability of all of h arises from treating hypotheses as indivisible elements -- a fallacy of composition.
Ah, I see what you were trying to say yesterday. That's true.

The cool part of Bayes' rule is when you have mutually exclusive hypotheses, because then you can add all of them to 1. When evidence makes the probability of one of those hypotheses go to zero, the probability of others go up.

Also, if you think about coin flips, you could have 3 hypotheses before you see an outcome: a head/head coin, a head/tail coin, and a tail/tail coin.

You see another head. Now the fair coin only has a 25% chance of having produced that but the head/head has a 100%.

So say you see 10 heads in a row and no tails. A fair coin would've had a 1/1024 chance of producing it. Whereas a heads/heads would have a 100% chance. Factor in how likely heads/heads coins are in general, and you have a sweet decision making engine.

Abstract from that, and you can at least produce a descriptive account of human reasoning.

Edit: some math --

h2 = heads/tails coin produced the sequence

p(h1 |d) / p(h2 | d) = (p(d | h1) * p(h1)) / (p(d | h2) * p(h2))

if you assume that before you see any evidence, a heads/heads coin is just less likely, you could start with something like this:
p(h1) = .01
p(h2) = .99

So you see 5 heads in a row (the sequence HHHHH), and you're trying to figure out which of your two hypotheses is more likely given the assumptions.

p(h1 |d) / p(h2 | d) = (1*.01) / (.03125 * .99)
= .323

Say you see 10 heads in a row:

p(h1 |d) / p(h2 | d) = (1*.01) / (0.0009765625 * .99)
= 10.34

At about 7 heads in a row, you should favor the heads/heads hypothesis. Before that, you should favor the fair coin hypothesis. The further you go from the point where the ratio is 1, the more you should weigh those beliefs.

Say you saw someone cough. You could think they had a random cold. Or you could think they had some chronic long lasting cough. Since colds are more likely, you'd probably think they had a cold. But if you see them cough every day for some months, you'd probably reassess your original belief.

This logic works for like everything, bro.

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