4 ms·

> This algorithm works because no matter what the bias is, the odds of getting a Heads then Tails will always be exactly the same as the odds of getting a Tails

> This algorithm works because no matter what the bias is, the odds of getting a Heads then Tails will always be exactly the same as the odds of getting a Tails then Heads. That may be easy to state, but I don't see it at all as a self-evident Truth. Hence the need for jargon, in order to actually come up with proofs in the language of mathematics.

sp332 14y ago
The odds of getting Heads on any given flip are always (x)%, regardless of the previous result. Likewise, the odds of getting Tails is always (100-x)%. So you will get HT with probability (x)% * (100-x)% which might be as high as 25% with a fair coin but might be much lower. Regardless of how small the probability, it is exactly the same as getting TH which is (100-x)% * (x)%. edit: typos
- klibertp 14y ago
  Is something like this sufficient as a proof? It's so natural and easy, yet convincing and accurate, that I have hard time believing it. :) Even if it's just an explanation it's still awesome. How I wish I could find materials teaching math ideas without relying heavily on math language! (And no, because I'm not going to contribute to the field I don't really need to know that language... Except I need, because it's the only way to obtain what is useful to me, however uncomfortable it is for me...)
  - gizmo686 14y ago
    That is a completely valid and rigorous proof. The difficulty with reading more "formal" proofs is that they try to make each statement as consise as possible, because it makes it easier to manipulate the results, and hold more of the proof in your head at a time. For perspective, consider the quadratic equation. This equation was first discovered in 628 AD, India. An English translation (from 1817) reads: "To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value" Or, as modern mathematicians would right, x=(sqrt(4ac+b^2)-b)/2a. The equation is slightly different for two reasons. First, it applied to equations of the from ax^2+bx=c, as apposed to ax^2+bx+c=0, so the sign of 4ac is inverted. It also misses the second answer provided by the '±'. The original paragraph form is probably easier for someone unfamiliar with anything beyond arithmetic, however the modern form is far easier to reason about, and use to those who invest the time to learn the notation.
    - klibertp 14y ago
      > The original paragraph form is probably easier for someone unfamiliar with anything beyond arithmetic Isn't this because of the fact that at the time (628 AD as you say) there wasn't that much to be familiar with beyond arithmetic? (I don't really know the history of maths, just curious.) > and hold more of the proof in your head at a time Is it really the case? I mean that no matter with which notation you start with, you need to parse and evaluate it mentally to convert it into terms you actually think with. I can see that for maths 'natives' it is helpful, because their mental models are very closely tied to the formal notation. But for me mathematics is a foreign language, it seems that I think in slightly different terms and this succinct notation makes it harder for me to actually understand what it is really about. (I don't mean your example specifically - it's still well inside my comfort zone). As a programmer I do struggle for "math literacy" because I have to. I need to be able to read proofs, understand some concepts to be able to use them and to know when I need them. The problem is it's not natural for me. Yesterday I found here on HN a book that just may be exactly what I'm looking for: http://greenteapress.com/thinkstats/html/index.html http://greenteapress.com/thinkstats/html/index.html (didn't have time to read it yet, but it looks promising). > who invest the time to learn the notation I think I said it before but maybe it's worth noting explicitly. I am somewhat familiar with the notation, I consider myself 'literate'. But I'm far from fluent and I probably never will be. Similarly, I can read and write cyrillic script (with pen and paper). But it takes me more than twice the time to read russian in cyrillic than reading the same text written with latin letters. I don't need someone to translate russian to english for me. It's just the character set that is a problem. I think it's very similar to what I experience with math.
michaelcampbell 14y ago
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 :)

sp332 14y ago

The odds of getting Heads on any given flip are always (x)%, regardless of the previous result. Likewise, the odds of getting Tails is always (100-x)%. So you will get HT with probability (x)% * (100-x)% which might be as high as 25% with a fair coin but might be much lower. Regardless of how small the probability, it is exactly the same as getting TH which is (100-x)% * (x)%. edit: typos

klibertp 14y ago
Is something like this sufficient as a proof? It's so natural and easy, yet convincing and accurate, that I have hard time believing it. :) Even if it's just an explanation it's still awesome. How I wish I could find materials teaching math ideas without relying heavily on math language! (And no, because I'm not going to contribute to the field I don't really need to know that language... Except I need, because it's the only way to obtain what is useful to me, however uncomfortable it is for me...)
- gizmo686 14y ago
  That is a completely valid and rigorous proof. The difficulty with reading more "formal" proofs is that they try to make each statement as consise as possible, because it makes it easier to manipulate the results, and hold more of the proof in your head at a time. For perspective, consider the quadratic equation. This equation was first discovered in 628 AD, India. An English translation (from 1817) reads: "To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value" Or, as modern mathematicians would right, x=(sqrt(4ac+b^2)-b)/2a. The equation is slightly different for two reasons. First, it applied to equations of the from ax^2+bx=c, as apposed to ax^2+bx+c=0, so the sign of 4ac is inverted. It also misses the second answer provided by the '±'. The original paragraph form is probably easier for someone unfamiliar with anything beyond arithmetic, however the modern form is far easier to reason about, and use to those who invest the time to learn the notation.
  - klibertp 14y ago
    > The original paragraph form is probably easier for someone unfamiliar with anything beyond arithmetic Isn't this because of the fact that at the time (628 AD as you say) there wasn't that much to be familiar with beyond arithmetic? (I don't really know the history of maths, just curious.) > and hold more of the proof in your head at a time Is it really the case? I mean that no matter with which notation you start with, you need to parse and evaluate it mentally to convert it into terms you actually think with. I can see that for maths 'natives' it is helpful, because their mental models are very closely tied to the formal notation. But for me mathematics is a foreign language, it seems that I think in slightly different terms and this succinct notation makes it harder for me to actually understand what it is really about. (I don't mean your example specifically - it's still well inside my comfort zone). As a programmer I do struggle for "math literacy" because I have to. I need to be able to read proofs, understand some concepts to be able to use them and to know when I need them. The problem is it's not natural for me. Yesterday I found here on HN a book that just may be exactly what I'm looking for: http://greenteapress.com/thinkstats/html/index.html http://greenteapress.com/thinkstats/html/index.html (didn't have time to read it yet, but it looks promising). > who invest the time to learn the notation I think I said it before but maybe it's worth noting explicitly. I am somewhat familiar with the notation, I consider myself 'literate'. But I'm far from fluent and I probably never will be. Similarly, I can read and write cyrillic script (with pen and paper). But it takes me more than twice the time to read russian in cyrillic than reading the same text written with latin letters. I don't need someone to translate russian to english for me. It's just the character set that is a problem. I think it's very similar to what I experience with math.

michaelcampbell 14y ago

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 :)