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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 time

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.
    - kareemsabri 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. That's true, but only if you pick ONE specific sequence. The likelihood of any precise sequence of heads/tails is equally unlikely as all heads. However, we are not fixing the outcomes ordering, but simply saying taken to infinity the distribution of outcomes will tend to 50/50. > I still would not really understand what forces my coin to show tails eventually Nothing forces it to show heads either. > but if there were a measure showing that those two sets have measure 1 respectively 0 it would already be easier to swallow Well, we've already proven it's impossible to have all heads infinitely. The same would be true of any non-uniform sequence. Any sequence that favours heads by any margin at all, taken to infinity, would necessarily contain an infinite sub-sequence that contains all heads, which we've proven is impossible by the limit to infinity.
      - danbruc 10y ago
        Well, we've already proven it's impossible to have all heads infinitely. The same would be true of any non-uniform sequence. Any sequence that favours heads by any margin at all, taken to infinity, would necessarily contain an infinite sub-sequence that contains all heads, which we've proven is impossible by the limit to infinity. Have we? We only said that the probability of having an infinite sequence of all heads is zero but not that it is impossible to obtain that result. After all the probability of any infinite sequence is zero and therefore if we would equate probability zero with impossible to obtain, then we could obtain no infinite sequence at all. The second part is obviously wrong, H H T H H T... ad infinitum obviously favors heads over tails but does not contain an infinite subsequence of all heads.
- conjectures 10y ago
  When heads has probability p the probability for any n-length sequence of all heads is p^n. For p between 0 and 1 this vanishes as n goes to infinity. The reason is that the conjunction of some events is always less (than or equal to) probable than the events themselves. This should make intuitive sense as the middle bit of a Venn diagram. Even more so for a conjunction of events with a conjunction of events. When taken to the infinite limit you tend to end up with the conjunction having probability 0 or 1.
  - danbruc 10y ago
    I have no problem with that, all infinite sequences of coin tosses have probability zero. The question is whether a infinite sequence of all heads is an obtainable outcome which would be an counterexample to the statement that the relative frequency converges to 0.5 in the limit.
deleted 10y ago
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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.
  - kareemsabri 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. That's true, but only if you pick ONE specific sequence. The likelihood of any precise sequence of heads/tails is equally unlikely as all heads. However, we are not fixing the outcomes ordering, but simply saying taken to infinity the distribution of outcomes will tend to 50/50. > I still would not really understand what forces my coin to show tails eventually Nothing forces it to show heads either. > but if there were a measure showing that those two sets have measure 1 respectively 0 it would already be easier to swallow Well, we've already proven it's impossible to have all heads infinitely. The same would be true of any non-uniform sequence. Any sequence that favours heads by any margin at all, taken to infinity, would necessarily contain an infinite sub-sequence that contains all heads, which we've proven is impossible by the limit to infinity.
    - danbruc 10y ago
      Well, we've already proven it's impossible to have all heads infinitely. The same would be true of any non-uniform sequence. Any sequence that favours heads by any margin at all, taken to infinity, would necessarily contain an infinite sub-sequence that contains all heads, which we've proven is impossible by the limit to infinity. Have we? We only said that the probability of having an infinite sequence of all heads is zero but not that it is impossible to obtain that result. After all the probability of any infinite sequence is zero and therefore if we would equate probability zero with impossible to obtain, then we could obtain no infinite sequence at all. The second part is obviously wrong, H H T H H T... ad infinitum obviously favors heads over tails but does not contain an infinite subsequence of all heads.
conjectures 10y ago
When heads has probability p the probability for any n-length sequence of all heads is p^n. For p between 0 and 1 this vanishes as n goes to infinity. The reason is that the conjunction of some events is always less (than or equal to) probable than the events themselves. This should make intuitive sense as the middle bit of a Venn diagram. Even more so for a conjunction of events with a conjunction of events. When taken to the infinite limit you tend to end up with the conjunction having probability 0 or 1.
- danbruc 10y ago
  I have no problem with that, all infinite sequences of coin tosses have probability zero. The question is whether a infinite sequence of all heads is an obtainable outcome which would be an counterexample to the statement that the relative frequency converges to 0.5 in the limit.

deleted 10y ago

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