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> Is there any (uncontroversial) theory that rigorously defines what a 50 % probability for heads and tails means? It certainly doesn't mean that in the long ru

> Is there any (uncontroversial) theory that rigorously defines what a 50 % probability for heads and tails means? It certainly doesn't mean that in the long run you will obtain the same number of heads and tails because there is a (vanishing) chance that you will always get heads even though the coin is actually fair. It doesn't really require a theory, outside of probability theory itself. According to the law of large numbers, P(x) = 0.5 does imply that, for N trials, the probability of N/2 outcomes being x (heads) approaches 1 as N approaches infinity. And as you pointed out, the probability of all heads is vanishing, as N approaches infinity. P(all heads) = 1 - P(not all heads), and P(all heads) approaches 0 as N approaches infinity, so P(not all heads) must approach 1. The same exercise could be done with the relative distributions of heads and tails, in a large number of trials. Each coin flip is random, but the results will converge to the mean. You will obtain the same number of heads and tails with certainty, not high probability, if you can make N big enough.

danbruc 10y ago
That means it is totally impossible to get H H H H... or H T T H T T... ad infinitum? I really have no well-founded opinion on that, it just seems very counterintuitive that the coin is not allowed to yield any such sequence, they same as good as any other sequence to me.
- kareemsabri 10y ago
  Not at all! It yields that sequence all the time, in small numbers. But, it is impossible to get only heads, if you could truly flip a coin infinitely many times, which of course you can't. The larger you make the sequence, the smaller you make the probability of all heads. In your example: P(H,H,H,H) = 0.5 ^ 4 = 0.0625 P(H,H,H,H,H,H,H,H) = 0.5 ^ 8 = 0.00390625 As you can see it's getting pretty small already. However, the probably will never actually reach 0. It's the half step concept, if you keep halving the distance between yourself and an object, you will never actually reach it. But the distance between yourself and the object will approach 0 as the number of steps approaches infinity. 0.5 ^ infinity = 0 Note that P(H|H,H,H) = probability of heads, given 3 previous heads, is still just 0.5. Any independent coin flip is 0.5 chance of being heads. But, when you've done 0 coin flips, and you ask, what are the odds of all heads for the next, say, million flips, they are nearly 0.
  - danbruc 10y ago
    But now you are contradicting yourself, aren't you? We agree that for any finite sequence of tosses it is unlikely but possible to get all heads and therefore it is possible to not converge to 0.5. The question is what difference it makes to go from a large but finite to an infinite number of tosses. Either it is impossible to get only heads an infinite number of times, then I have a problem understanding why that is, or all heads is still a possible outcome even in the infinite case, then the process does not necessarily yield 0.5 even in the limit of an infinite number of tosses.
    - kareemsabri 10y ago
      > The question is what difference it makes going from a large but finite to an infinite number of tosses. Convergence is only guaranteed as N -> ∞. The difference between large but finite and infinite is.. well, infinite :) So that's a pretty significant difference. > Either is impossible to get only heads an infinite number of times, then I have a problem understanding why that is, or all heads is still a possible outcome even in the infinite case, then the process does not necessarily yield 0.5 even in the limit of an infinite number of tosses. It's the former. It is impossible to get only heads an infinite number of times. It is possible, but increasingly unlikely, to get only heads a REALLY LARGE, but finite, number of times.
      - danbruc 10y ago
        Okay, assuming that is true, is there an intuitive way to understand that? And the fact that lim[n-> ∞] 0.5^n = 0 unfortunately won't do for me, that is true for any specific infinite sequence, even those containing 50/50 heads and tails. I think most of the sequences - hand-waving, most of an infinite set - are 50/50 heads and tails just because there are more possibilities - hand-waving again - to arrange 50/50 heads and tails versus 100/0 or 40/60 heads and tails. So can I sum over all the 50/50 sequences and get 1 and sum over the rest and get 0? I still would not really understand what forces my coin to show tails eventually, but if there were a measure showing that those two sets have measure 1 respectively 0 it would already be easier to swallow.

danbruc 10y ago

That means it is totally impossible to get H H H H... or H T T H T T... ad infinitum? I really have no well-founded opinion on that, it just seems very counterintuitive that the coin is not allowed to yield any such sequence, they same as good as any other sequence to me.

kareemsabri 10y ago
Not at all! It yields that sequence all the time, in small numbers. But, it is impossible to get only heads, if you could truly flip a coin infinitely many times, which of course you can't. The larger you make the sequence, the smaller you make the probability of all heads. In your example: P(H,H,H,H) = 0.5 ^ 4 = 0.0625 P(H,H,H,H,H,H,H,H) = 0.5 ^ 8 = 0.00390625 As you can see it's getting pretty small already. However, the probably will never actually reach 0. It's the half step concept, if you keep halving the distance between yourself and an object, you will never actually reach it. But the distance between yourself and the object will approach 0 as the number of steps approaches infinity. 0.5 ^ infinity = 0 Note that P(H|H,H,H) = probability of heads, given 3 previous heads, is still just 0.5. Any independent coin flip is 0.5 chance of being heads. But, when you've done 0 coin flips, and you ask, what are the odds of all heads for the next, say, million flips, they are nearly 0.
- danbruc 10y ago
  But now you are contradicting yourself, aren't you? We agree that for any finite sequence of tosses it is unlikely but possible to get all heads and therefore it is possible to not converge to 0.5. The question is what difference it makes to go from a large but finite to an infinite number of tosses. Either it is impossible to get only heads an infinite number of times, then I have a problem understanding why that is, or all heads is still a possible outcome even in the infinite case, then the process does not necessarily yield 0.5 even in the limit of an infinite number of tosses.
  - kareemsabri 10y ago
    > The question is what difference it makes going from a large but finite to an infinite number of tosses. Convergence is only guaranteed as N -> ∞. The difference between large but finite and infinite is.. well, infinite :) So that's a pretty significant difference. > Either is impossible to get only heads an infinite number of times, then I have a problem understanding why that is, or all heads is still a possible outcome even in the infinite case, then the process does not necessarily yield 0.5 even in the limit of an infinite number of tosses. It's the former. It is impossible to get only heads an infinite number of times. It is possible, but increasingly unlikely, to get only heads a REALLY LARGE, but finite, number of times.
    - danbruc 10y ago
      Okay, assuming that is true, is there an intuitive way to understand that? And the fact that lim[n-> ∞] 0.5^n = 0 unfortunately won't do for me, that is true for any specific infinite sequence, even those containing 50/50 heads and tails. I think most of the sequences - hand-waving, most of an infinite set - are 50/50 heads and tails just because there are more possibilities - hand-waving again - to arrange 50/50 heads and tails versus 100/0 or 40/60 heads and tails. So can I sum over all the 50/50 sequences and get 1 and sum over the rest and get 0? I still would not really understand what forces my coin to show tails eventually, but if there were a measure showing that those two sets have measure 1 respectively 0 it would already be easier to swallow.