Because “approach infinity” is a technical term that does not require an actual number infinity. It means that a sequence can be as large as you like. The formal definition is that for any number, there is a point in the sequence such that, past that point, the sequence is greater than the number.
An interesting counterintuitive result is that if your coin keeps coming up heads, you’re probably a childhood friend of the prince of Denmark.
Anyway, John Baez has a good discussion on the problems inherent in the definition of probability (and how those problems express themselves in the problems with interpreting quantum mechanics) here.
Yeah, I’d lose it after about 12 flips.
Right. The whole modern standard formulation of limits (and of Calculus, which is based on limits) is designed to allow you to avoid assuming that “infinity” (or things like “infinitely close to zero”) actually exists.
So instead of talking about doing something (like flipping a coin) an infinte number of times (whatever that would mean), you can talk about doing it an arbitrary-but-finite real number of times n, and then see what happens as n approaches infinity in the technical sense described above.
As the number of flips approaches infinity, the chance that you lose it approaches 100%
Ok, instead of inifintely deconstructing the problem, is the common mathematical definition of probability actually appropriate for this question? It sounds like you end up with 1/infinity, which looking over the related threads, has no common definition.
Worse than that, could the question be how many times does the set of infinite flips which are all heads, occur in the set of infinite sets of flips? Which sounds to me like infinity/infinity, or infinity1/infinity2, where infinity2>infinity1. But I need to take off my pants to count to 21, so I have no clue.
It’s like this, TriPolar.
What’s the probability that you flip a coin once and get heads? 1/2
What’s the probability that you flip a coin twice and get heads each time? 1/2 * 1/2
What’s the probability that you flip a coin three times and get heads each time? 1/2 * 1/2 * 1/2
What’s the probability that you flip a coin a bajillion times and get heads each time? 1/2[sup]bajillion[/sup].
Now, 1/2[sup]bajillion[/sup] is still greater than zero, but it’s tiny. However small you want to make this positive probability, I can pick a high enough value of “bajillion” to do so. The probability of you being able to get all heads hurtles towards zero as the number of coin flips increases. In that sense, the probability of getting all heads over an infinite number of coin flips is taken to be zero.
Another way of putting it: the probability that you flip heads each of infinitely many times = the probability that you flip heads the first time * the probability that you flip heads each of the remaining infinitely many times = 1/2 * the probability that you flip heads each of the remaining infinitely many times. So if we call the probability of flipping heads on each of infinitely many trials P, then we have that P = 1/2 * P. Ergo, P = 0.
So if we call the probability of flipping heads on each of infinitely many trials P, then we have that P = 1/2 * P. Ergo, P = 0.
At last. Thank you. Ignorance fought. Please edit the Wiki page and every other source that fails to put it in succinct terms. Not necessarily you, but somebody.
Indistinguishable is always kind of awesome when mathy threads come along (I’m sure Indistinguishable is awesome more generally, but this is the only time I notice).
Even if s/he makes me feel dumb for forgetting all the math I took in college.
I have nothing constructive to add.
An event with probability 0 will occur with probability 1.
Here is the way I would formulate the problem. Consider heads=1 tails =0, and use the infinite sequence of binary flips to indicate the binary representation of a real number on the interval [0,1]. Infinite heads with no tails =1, infinite tails with no heads =0. If the coin is fair, then the probability distribution of the resulting numbers will exactly equal the uniform distribution on [0,1].
No need to worry about limits or anything of that sort. For a fair coin determine the probability of an event, simply work out the Lebesgue measure of that set on the [0,1] interval. So the probability of any one particular sequence of coin flips is zero, yet you are guaranteed with probability 1, do get some sequence. For an unfair coin the distribution will be more a more complicated discontinuous distribution on [0,1], but still we be a legitimate measure, and unless P(heads)=0 or P(heads)=1 you will still get that the probability of any given sequence is 0, yet some sequence will occur with probability 1.
For an infinite sequence it doesn’t make sense to talk about the ratio of heads to tails (you get infinity/infinity). But if you take a particular sequence of heads and tails, and after each flip calculate #heads/#total flips, this will be a sequence that may or may not be convergent. However you will find that the set of those sequences for which it converges to P(head) will have measure (probability) 1.
Ya. That’s what I was gonna say.
Well, I, for one, consider that a constructive comment. 
Thanks for the compliment; it makes my day.