5 ms·

Perhaps not self evident, but easily provable. Now, having to COME UP WITH this when all you know is p(a) = x; p(b) = 1 - x, and the fact that p(a)p(b) == p(b)

Perhaps not self evident, but easily provable. Now, having to COME UP WITH this when all you know is p(a) = x; p(b) = 1 - x, and the fact that p(a)p(b) == p(b)p(a) is some awesome insight.

sp332 14y ago
You don't even need the first part. As long as p(a) and p(b) are both constant, they can add to any number, not just 1. They could even be negative, which is impossible for a physical coin of course but my point is: you only need to know that there are two constants, and multiplication is commutative, and you're done :) An example: if you have a die and you don't know exactly how many sides it has, or whether it's fair, you can still get a perfectly fair 50/50 outcome. Just count "1 followed by 2" vs. "2 followed by 1" and ignore every other result.
- michaelcampbell 14y ago
  You don't need it for arbitrary problems, no, but this problem was stated as a coin toss.
  - sp332 14y ago
    Ah true, you are given it. I just meant that you don't need it :)
- kbolino 14y ago
  They could even be negative Not under any standard formulation of probability. Probability is a measure, and measures are nonnegative by definition.
  - sp332 14y ago
    That's an interesting idea. I know negative probabilities are used in quantum mechanics http://news.ycombinator.com/item?id=4319276 http://news.ycombinator.com/item?id=4319276 and they come up in some math problems http://en.wikipedia.org/wiki/Negative_probability http://en.wikipedia.org/wiki/Negative_probability so, even if you can't think of a case where they would occur, it's nice to know we have the math in case it happens :)
    - kbolino 14y ago
      Quantum mechanics: frustrating formal mathematicians for over a century. You'd have to exercise caution when using (quasi-)probabilities less than zero or greater than one, because many of the classical properties of probability theory depend upon those conditions.

Perhaps not self evident, but easily provable. Now, having to COME UP WITH this when all you know is p(a) = x; p(b) = 1 - x, and the fact that p(a)p(b) == p(b)p(a) is some awesome insight.

sp332 14y ago

You don't even need the first part. As long as p(a) and p(b) are both constant, they can add to any number, not just 1. They could even be negative, which is impossible for a physical coin of course but my point is: you only need to know that there are two constants, and multiplication is commutative, and you're done :) An example: if you have a die and you don't know exactly how many sides it has, or whether it's fair, you can still get a perfectly fair 50/50 outcome. Just count "1 followed by 2" vs. "2 followed by 1" and ignore every other result.

michaelcampbell 14y ago
You don't need it for arbitrary problems, no, but this problem was stated as a coin toss.
- sp332 14y ago
  Ah true, you are given it. I just meant that you don't need it :)
kbolino 14y ago
They could even be negative Not under any standard formulation of probability. Probability is a measure, and measures are nonnegative by definition.
- sp332 14y ago
  That's an interesting idea. I know negative probabilities are used in quantum mechanics http://news.ycombinator.com/item?id=4319276 http://news.ycombinator.com/item?id=4319276 and they come up in some math problems http://en.wikipedia.org/wiki/Negative_probability http://en.wikipedia.org/wiki/Negative_probability so, even if you can't think of a case where they would occur, it's nice to know we have the math in case it happens :)
  - kbolino 14y ago
    Quantum mechanics: frustrating formal mathematicians for over a century. You'd have to exercise caution when using (quasi-)probabilities less than zero or greater than one, because many of the classical properties of probability theory depend upon those conditions.