Hmmm–I guess that stretches the idea of reciprocal a bit further than I am used to. I find myself reaching for the type of quibble that ultrafilter just introduced. It doesn’t make sense (to me) to speak of a reciprocal for a transfinite number when transfinite numbers have no unity, but it would make sense to speak of the reciprocal of a supernatural number whose “value” was equivalent to the “value” of a transfinite number.
Math is fun!
It gets better. The supernatural numbers arise when you take the negation of an undecidable sentence as an axiom for arithmetic. Those crazy mathematicians!
There are also surreal numbers, but I don’t remember exactly what those are.
for infinite coin flips you are GUARANTEED to get heads at least once. Imagine keeping on throwing a coin until you got heads, here are the (non-cumalitive) probabilties for when your game of chance will end (with the cumalitive probabiltiy that your game would of ended by that time in brackets in brackets:
1 flips = 1/2 (1/2)
2 flips = 1/4 (3/4)
3 flips = 1/8 (7/8)
4 flips = 1/16 (15/16)
after infinity flips you will find the cumalitive probabilty that your game has ended (i.e. you have thrown a heads) to equal 1 (as the series tends to 1 as it approaches infinity)
Going back to my original statement regarding the countability of all possible outcomes of the coin toss thingy, I was referring to finite outcomes - ie, a finite string of coin tosses, as a subset of the infinite number of tosses. And I fail to see why this would not be countable. It can be arranged as such:
First, list all possible outcomes of 1 flip (H, T). Then, all possible outcomes of 2 flips (HH, HT, TH, TT). Then 3 flips, 4 flips, and so on. The sequence {1 flip, 2 flips, 3 flips, … } is countable, and any ordering of the possbile permutations of a given number of flips is obviously countable, so why wouldn’t the whole shebang be countable?
Or were you specifically referring to the series of all possible infinite sequences of coin flips, which I can see as being uncountable, but which I didn’t think we were referring to?
Jeff, who hasn’t done this much math in a good 5 years.
Yes, that’s what we had in mind. But you’re right about that first set being countable.
MC, go back and reread the entire thread, and the one I linked to. “happens with probability 1” and “always happens” are not the same in this case.
I stick by what I said, when your talking about an infinite number of events you cannot specify an exact chain of events only make generalizations.
And one of the generalizations we can make is that if you have an uncountable number of possible events, events with probability zero can happen.
btw the logic that because you can’t throow infinite tails in succession that every chain of events is flawed. The only thing that this tells us is that the chain of events can’t be ‘rationalized’ (i.e. put into a form which can be exactly descibed using less than infinte space).
I should make myself clearer:
imagine that every different event was written as a binary fraction (i.e 0.10001…, etc. where 1=H and 0=T ) what I am saying is those events that give an irrational binary fraction are possible chain of events.
This is logical as there are infintely more irrational numbers than rational numbers between 0 and 1.
Every possible outcome has probability zero.
No, you’ve only proved that every ‘rationalizied’ outcome has a probability of zero. It is impossible to prove that an ‘irrational’ outcome has a probility of zero.
no, really, go back and read the whole thread and the one he linked to.
there are probability functions that satisfy the axioms for probability, and having p(s) = 0 does not guarantee s will not occur. it’s counterintuitive, but it arises because “p(s)=0 guarantees s will not occur” is not an axiom of probability.
the example of picking a random real between 0 and 1 was particularly convincing for me. each number has 0 probability of being picked, but you can pick one by the axiom of choice.
That is to say, you can prove that /T/TTTTTTTT… has a probaility of zero, you can prove H/T/TTTTT… has an outcome of zero and you can prove /HTTTHHT/HTTTHHTHTTTHHT… has an outcome of zero (wher the two /'s signifify a repeating unit) but you can’t prove that an outcome which cannot be rationalized in this way has a probabilty of zero.
Not only is it possible, but it’s really easy. Since we’re dealing with a uniform distribution on [0, 1], the probability of drawing an element from a set S is [symbol]m/symbol, the Lebesgue measure of S.
The Lebesgue measure of a single point is always 0, no matter what the point is. So the probability of picking an irrational point is 0, just like a rational point.
If you don’t believe me, go pick up any measure-theoretic text on probability.
(Shouting) Every individual outcome has probability 0!!! Whether or not the corresponding binary number is rational or irrational has no bearing on that fact.
Allow me to repeat myself, MC. Why is the outcome “TTTTTT…” (i.e. all tails) somehow impossible, when an outcome like “HTHTHTHTHTHT…” is possible? These are just two different possible outcomes to the experiment. By symmetry, not only do they have the same probability (zero) but they are also either both possible or both impossible. If we claim that “TTTTTT…” is impossible, then by symmetry every outcome is impossible, which is absurd.
I have yet to see a convincing rebuttal to this argument in this or any other thread on this topic.
I’m still throughly unconvinced. When picking a random real number between 1 and zero you do not have the whole set of numbers to choose from in reality as you are only aware of a finite number of them.
As I said before all you have done is proved rational outcomes are impossible, the symmetry does not apply as you cannot prove irrational outcomes have a probabilty of zero.
just show me a proof that specifically excludes an irrational outcome.
What numbers that you are aware of has nothing to do with choosing a number from the interval [0,1] at random. We’re not picking numbers out of a hat. “Choosing a number at random from the interval [0,1]” has a specific mathatical meaning, precisely because of questions like this, and a consequence of that meaning is that any number in the interval is just as likely to be chosen as any other. If the choice wasn’t symmetric like that, it wouldn’t be random.
Regarding your “rational/irrational” objection, what does the rationality of the corresponding binary number have to do with anything? In fact, why focus on rational vs. irrational? There are plenty of other ways to divide the interval [0,1] into a countable set and an uncountable set. Here’s one: define R to the be the rational numbers in [0,1] together with sqrt(2)/2, and let S be all the other elements of [0,1].
By your logic, therefore, it’s somehow impossible to randomly choose the number “0.1011010100…” (sqrt(2)/2 in binary) from the interval [0,1]. But it somehow is possible if we divide [0,1] the “normal” way, into rationals and irrationals. So I repeat: what does rationality/irrationality have to do with anything here?
The point is you can prove that HTHTHTHT… is an impossible event without resorting to symmetry, but you can’t do the same for an irrational series.