ok, Probability. First of all if something has a probability of 1 it will occur, no maybe about it. second the whole coin thing, if you toss a coin 100 times and get all heads the probability of getting 101 heads is very very very small (but not 0 cause that would imply it can’t happen) but if you look at the probability of getting a head on the 101st toss you are back to .5 (assuming a fair coin). the answer differs depending how you phrase the question.
An event having probability one of occuring is not the same as saying the event must occur. It’s entirely possible for something to have probability one of occuring, but not occur.
For a simple example, suppose we have this:
We are going to pick a real number at random from the interval [0,1]; no individual number is preferred over any other number. (Such a probability model does exist; Lebesgue measure, for example).
Let’s look at some arbitrary number x in [0,1]. It should be clear that the probability that x will be picked is zero; consequently, the probability that x will not be picked is one.
OK, now suppose the random number (call it b) is picked.
What is the probability that b would have been picked? Zero, but it still happened.
What is the probability that b would not have been picked? One, but it didn’t happen.
jayjay
You’re misunderstading what erislover was saying. erislover meant that the more you flip the coin, the more likely a heads will come up at some point (not on the “next flip” necessarily, just somewhere down the line).
In and of itself, your interpretation is correct, i.e., if you’ve flipped the (fair) coin a bunch of times and heads hasn’t come up yet, the probability of heads on the next flip is still 1/2. But that’s definitely different from saying that, if you flip a coin n times, the probability of (at least) one heads occuring increases as n increases (which is what erislover was saying.)
jayjay, consider two coin tosses. Each individual toss has a 0.5 probability of coming up heads. Our four possibilities are
HH
HT
TH
TT
each with a probability of 0.25
But notice that the probability of one of the two tosses yielding at least one head is 0.75!
See, this is why I consider myself dysnumeric. I have trouble wrapping my head around this because it seems to me that each coin flip is a totally separate event. Being totally separate, each coin flip in no way affects or is affected by any previous or subsequent coin flips. So the probability should be .5 down the line.
How does the probability of a heads go up if no coin flip affects or is affected by any other? It shouldn’t matter if I flip the coin 100 times and get 99 tails, the probability of the next flip seems to have to be .5. If that’s not the case, where does the increase in probability come from? What’s causing the .5 to turn into .75 (for example)?
jayjay (still confuzzled)
Guys, I had heard of Canter and set theory before today. Don’t think of my post as an actual mathematical proposition; I intended it more as the equivalent of the bellboy’s missing dollar -a mathematical sleight-of-hand if you will.
Phlosphr states that the universe must be infinite, because, were it finite, it would have a boundary at the end, which is impossible. This reasoning goes back to the Roman poet Lucretius, who imagined standing at the edge of a finite universe and throwing a spear, implying there must always be something beyond and the universe was thus infinite. But doesn`t this confuse the ideas of space and nothing, the one being part of the fabric of the universe, having dimensions through which to move, the other being literally nothing, no dimensions, no space, no distance, just plain nothing. I find it perfectly feasible to envisage a finite universe lying within the bounds of such nothingness.
aldiboronti: Umm… I don’t even know how to begin to answer that one. What do you mean by lying ‘within’ nothingness? For that matter, ‘nothing’ doesn’t seem to be a well defined term if you want to think of it as something bounding hte universe. Anyway, as has been said (I think in this thread, certainly in many others) there are plenty of ways of space being finite without a boundary.
Cabbage: Ok, ya got me there. 
I concede the point. Though when you say lebesgue measure, do you mean P(X is an element of A) = Measure(A). And if so, what’s the probability of it lying in an unmeasurable set?
I acknowledge there are many ways that finite universes may have no boundaries, and also my solecism of nothing being something bounding the universe. All I was trying to convey, as someone admittedly not well-grounded in science, is the difference between nothing and empty space, which is often confused in popular terminology. Do I have it wrong?
Yeah, that’s what I mean.
The probability of it lying in an unmeasurable set would be left undefined. (I don’t think (but offhand I’m not certain) that there is a probability model for randomly picking an element of [0,1] such that no number is preferred over any other number AND is also defined for every subset of [0,1]).
In general, a probability function doesn’t have to be defined for every possible subset of the sample space, it just has to be defined on a sigma algebra of those subsets. A sigma algebra of subsets is one which satisfies the following:
- The empty set has to be included.
- It has to be closed under complements.
- It has to be closed under countable unions.
<< because it seems to me that each coin flip is a totally separate event. Being totally separate, each coin flip in no way affects or is affected by any previous or subsequent coin flips. So the probability should be .5 down the line. >>
Try thinking this way:
The chance of any ONE flip of the coin being heads is 1/2
The chance of TWO flips in a row being heads is 1/4
The chance of THREE flips in a row being heads is 1/8
etc.
Thus, the chance of 10 flips in a row being heads is very small (roughly 1/1000) but not zero. If you take, as a single trial, ten flips in a row… and if you have enough time to do (say) 10,000 trials of 10 flips each, you may well get several strings of 10 heads in a row.
If, in one of those trials, you get 9 heads in a row, the chance of the tenth flip being a head is still 1/2.
The chance of 100 flips in a row being heads is very very very small, but not zero. If you take, as a single trial, a consecutive string of 100 flips … and if you you have enough time to do several million trials, you may well get a string of 100 heads amongst them.
If, in one of those trials, you get 99 heads in a row, the chance of the 100th flip being a head is still 1/2. (Imagine your frustration if, after a few million trials, you get 99 heads in a row and the 100th flip is a tail!)
I’m reminded of the beginning of the Tom Stoppard play Rosencrantz and Guilderstern are dead, in which one of the confusingly interchangable characters keeps persistently throwing a head. He comments that this could either be due to some kind of narrative fate or simply a demonstaration of the laws of probability, that each coin flip is as likely to come up heads as the last.
pan
Oh, and JayJay, the reason that the probability can be 0.75 rather than 0.5 is because in this case the probability is not memoryless. You need to remember the result of trial 1 in order to compare it to trial 2, so that you might know whether you got a head or not.
Having got a head, the probability of another is 1/2. But before you start they are not two unrelated events. Or to put in another way: having got a head, the probability of getting at least one head is now 1.
pan