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If an infinite number of coins are tossed infinitely often, is it true that there will be infinite subsets of those coins that repeat any finite sequence of heads/tails infinitely often? I.e., infinitely many coins will always produce heads, infinitely many always produce tails, infinitely many produce HTHTHT..., THTHTH..., HHTHHTHHT..., TTHTTHTTH..., etc.

And on each toss of all the coins, would some infinite subset of coins begin reproducing a finite sequence infinitely often? I.e, infinitely many coins that had previously produced irregular sequences of HT would begin producing HTHTHT..., THTHTH..., et

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  • $\begingroup$ If one fair coin is tossed repeatedly, then, with probability one, it will produce any finite sequence of heads and tails infinitely often. In fact, it doesn't have to be a fair coin, just as long as it can do both heads and tails, and has no memory of what it has already done. $\endgroup$
    – Gerry Myerson
    Commented Jul 15, 2013 at 8:50
  • $\begingroup$ Already 29 answers over there. $\endgroup$
    – Did
    Commented Jul 15, 2013 at 9:16

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The answer to the question in your first sentence is yes, with probability $1$. However, the assertion of your second sentence, beginning with "I.e.", is false, also with probability $1$. The same applies to the question and assertion of your second paragraph.

For any coin, the probability that a given finite sequence will occur infinitely often is $1$, while the probability that the finite sequence will occur sequentially infinitely often is $0$.

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  • $\begingroup$ Thank you for reply. However, I don't understand why a finite sequence cannot occur sequentially infinitely often. If an infinite number of coins are tossed once, an infinite number, H1, will land heads. Discard all coins that land tails and toss the remainder again. This time, an infinite number, H2, will land heads again. Why can this not occur infinitely often, given that discarding coins that land tails will always leave an infinite number of coins that have previously always landed heads? $\endgroup$
    – Mathamator
    Commented Jul 15, 2013 at 9:47
  • $\begingroup$ @Mathamator: Let's consider your example with all heads; the argument is essentially the same for any other defined sequence. The probability of H on the first toss is a half. For HH on the first two tosses, it is a quarter. For HHH, it is an eighth, and so on: For H...H (with $n$ occurrences of H) the probability is $2^{-n}$. The probability of an infinite H-sequence is less than $2^{-n}$ for all n; namely it is zero. For large $n$, to get a sizeable probability of an all-head result, you would need to throw the order of $2^n$ coins each $n$ times. (Continued in the next comment) $\endgroup$
    – John Bentin
    Commented Jul 15, 2013 at 13:10
  • $\begingroup$ It is perhaps arguable that, if we toss continuum-many independent coins (whatever that means), we might get a sizeable probability of an all-H result coming up. But here we are in the deep waters of measure theory if we are to try to make a mathematically meaningful interpretation of ordinary language. $\endgroup$
    – John Bentin
    Commented Jul 15, 2013 at 13:25

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