Assume a distribution of an infinite number of tosses of a coin: Hs and Ts. Further assume the incidence of Hs and Ts within this distribution is randomly distributed (never mind worrying about what “random” means; that’s for another thread – axiomatize it).
Somewhere within this infinite distribution will be a gazillion consecutive Hs (same for Ts throughout this discussion). Indeed, given an infinite number of tosses, any finite number of consecutive Hs is not only likely but necessary. That’s the nature of infinity – sort of like Murphy, if it can happen, it will – eventually.
All the conjectures above are based on my admittedly shaky recall of probability theory. If I’m full of shit the question I’m about to ask may be meaningless. The question arises from my understanding of number theory which posits that, among other things, within the positive integers, there is (are?) an infinite number of numbers which are even multiples of a bazillion or a bazillion bazillion.
So…here’s my question: will there be an infinite (not just very large – infinite) string of consecutive Hs lurking somewhere within the infinite distribution of tosses? [A corollary of this suggests (to me, anyway) that if there is an infinite string of consecutive Hs then there are an infinite number of infinite strings of Hs scattered about.]
I suspect that your error is not in your understanding of probability, but in your understanding of infinity.
Or maybe it’s me who is misunderstanding. Anyway…
On the one hand, a six-inch line has twice as many points on it as a three-inch line, despite the fact that they both have an infinite number of points on them. One explanation of that is that “infinity” is not a number, but a concept. Another explanation is that it works because the points are infinitely small.
But none of that applies to coin-tossing. Each toss is a specific event. You can’t have an infinite string of all-heads tosses within the greater infinite string of all tosses.
Compare this to the infinite monkeys who eventually type out Shakespeare. At some point, one of the monkeys will begin Shakespeare’s first work, finish his last work, and then go on to type more gibberish. That’s a finite bunch of typing. Very different from yor idea to have an infinite string of heads and then go back to randomness.
1.) The only ways I can see an infinte number of head tosses in a run of coin tosses that’s not entirely heads is if you have a run of tosses that includes heads and tails (which might itself be infinite if you’ve been tossing since minus infiity) followed by an unbroken string of heads ever after, or the converse of this (you get heads from minus infinity up to a point, followed by a mixture of heads and tails after a point). This isn’t a practical situation, but we knew that going in.
2.) You can calculate the probability of such a run by simply multiplying the probability of heads on each flip times the number of flips – it’s infinitesimal. That’s not terrificalluy helpful – so is the probability of any flip. But you can say that in any large but finite run the probability of all heads is much smaller than any other possibility, and becomes smaller and smaller as the length of the run increases.
Well It’s been a while since I actually did these proofs, so the details are left as an exercise for the reader.
Given any arbitrary number N, where N is the number of heads thrown in a row that you are seeking;you can always find a number of total throws T that makes a streak of N heads in a row a 99% possibility. Now let’s replace 99 with M, an arbitrarily chosen percentage to represent the possibility you want. Over infinity as T increases M can reaches the limit of 100% if you choose.
As N increases that property doesn’t change, so you will have runs of infinte length since M can effectively be 100%. and there is no point at which N grows to a point you can’t find a T.
Well I could have just defined it as M in the first place. I used 99 to avoiding making too many points at the same time, since I can’t explain it very well. M N can be arbitrarily chosen, and there will always be a T.
Here’s the model: the sample space X is the set of all functions from N to {0, 1}. We assume that P(0) = P(1) = 1/2. Because the sample space is uncountably infinite, the probability of any single outcome (i.e., a specific function from N to {0, 1}) is exactly equal to zero.
This isn’t true. As the string becomes arbitrarily long, the probability of getting a run of heads of any finite length approaches 1, but there’s no reason why the coin can’t alternate between heads and tails forever.
Your corollary is true, but by the same reasoning as above, there’s no guarantee that you’ll have an infinitely long run as either symbol.
The most basic truth about infinity is that there is exactly the same number of points in a three-inch line as a six-inch line. There are exactly the same infinite number of points in a three-inch line as in a line the diameter of the universe and in a sphere the size of the universe and in any n-dimensional space. That’s the definition of a certain type of infinity. That’s why infinity is not a number and why standard rules of math don’t work.
I may be reading the OP in a different and simpler way than some of the math guys. Infinity means without end. A designated subset of infinity the way the OP defined it, however long, must have a starting point and an ending point. Therefore it must be finite.
If you talk about an infinite number of tosses of a coin, that just means without any last toss, i.e., you can keep on going forever. It’s a different kind of infinity from the number of points on a line segment, and a gentleman called Cantor discovered a theorem to prove that.
So, yes, if the sequence is random you can have an infinite number of any finite sequence of results, including an infinite number of sequences of a bajillion heads (where a bajillion is any arbitrarily large number).
But that doesn’t mean that you have an infinite number of infinite sequences. If the sequence contains one infinite sequence of heads, that’s it, because (remember) here “inifinite” means without any last toss. So once you’ve started that infinite sequence of heads, it’s never going to end, i.e., you never get another tail. So an inifinte sequence can only contain one such subsequence – and because of that, it’s no longer random, i.e., it can be completely specified in a finite description, e.g., “Up to the nth position it’s TH…HT, then it’s all H from then on.”
Not necessarily true. A subset of the positive integers is the positive even integers. While they have a starting point, they don’t have an ending point and
the subset is infinite.
As Gile points out, if this happens the sequence isn’t random. That is, it can be algorithmized (if that’s not a word it should be) as “T, then H, then T, …, then H followed by T forever”.
Yes, but they’re not consecutive, and you specified consecutive strings in the OP. The only consecutive subsequences of an infinite sequence are the kind where you take the whole sequence and leave off finitely many terms at the beginning.
In other words, if the Hilbert Hotel is full, and infinitely many people show up wanting a place to stay, you can accommodate them, but they can’t all have adjoining rooms.
This is true only if you consider the sequence to have a start but no end. If you start in the middle and say (pointing to the left) “It’s all heads that way” and (pointing to the right) “It’s all tails that way,” then you have two infinite sequences within the infinite sequence.
The other way to approach the problem is to get a little more specific on “infinity.” Are you starting with an infinite number of coin tosses or an infinite number of sequences of coin tosses, where “sequence” is defined as “tossing the coin until you get a result different than the first toss in this sequence?” If the latter, then you’re guaranteed to get an infinite string of heads and of tails.
Are we basically looking for an aleph-null sequence of heads in an aleph-one string of tosses?
The way these things seem usually conceptualised is that you start a string of coin tosses now, and throw coins indefinitely into the future. You could hypothesise a sequence of coin tosses that stretches out both ways, i.e., one that was already in progress at the Big Bang, before there were any coins, or even any metals to make them out of. But that just means you can have two infinite sequences of heads, i.e., one with no start and another with no end. You still can’t have an infinite number of such sequences.
Since in the latter case each such sequence has a finite number of tosses, it’s the same thing.
I don’t think aleph-one has been adequately identified. It’s the infinite cardinal number next greater than aleph-one, but may or may not equal 2 raised to the power aleph-null, i.e., the number of points on a line segment, depending on what axioms you add to the usual axioms for set theory.
It is entirely possible, if you toss a coin an infinite number of times, to end up with (T, H, T, H, T, H …) The probability of this happening, though, is exactly 0. However, since we are working in an uncountable probability space, this doesn’t mean that it is an invalid result. It can happen, it just “almost surely” (technical term) won’t happen.
As you pointed out, in an infinite sequence of coin tosses, you will almost surely (i.e. with probability 1) find arbitrarily long finite sequences of consecutive Ts and Hs. Will you find an infinitely long sequence of consecutive Hs? Well, unless I’m missing something, the only way to get this is to start getting only Hs after N coin tosses. The result will therefore be something like (H, T, T, H, …, T, H, H, H, H, H, …) with only Hs from that point on. The probability of this happening is 0. It may happen anyway, but it almost surely won’t happen. Therefore, the answer is no, in all likelihood you won’t find infinitely long sequences of consecutive Hs or Ts in your sequence.
Right. The probability of an infinite sequence of heads is zero, just as the probability of any sequence is zero. Similarly, if you randomly choose a point on a line, with all points having the same probability, the probability of any specific point is zero.
But in your infinite sequence of coin tosses, the probability that there is an infinite number of sequences of heads only each a googolplex long is 1. But don’t hold your breath waiting for the first one to turn up: infinity is a very long time indeed.