Probability question
Joe Avery
Suppose we have a coin with probability of heads equal to p (not
necessarily fair). We perform an infinite number of experiments
simultaneously with one toss for each experiment. Question 1: can we
claim that the event heads will show at least once in this infinite
number of experiments of one trail each? Question 2: what about if
there is an infinite number of trials for each of the infinite
experiments? Can we claim that heads shows up infinite times?
Thanks and I apologize in advance if this is a stupid question.
Mike
Questions re infinity can't be stupid.
If p>0, then an infinite number of trials in a single experiment will
produce an infinite number of heads (and tails, if p is also <1).
Ray Koopman
For any given p, the probability of at least one head in n trials is
1 - (1-p)^n, which can be made as close to 1 as you want by taking n
big enough (assuming, of course, that p > 0).
If 0 < p < 1 and n is infinite then both the number of heads and the
number of tails will be infinite.
Joe Avery
> On Apr 26, 3:58 pm, Joe Avery <joe_avery_2...@yahoo.com> wrote:
>
> > This isn't homework (I'm too old for that)
>
> > Suppose we have a coin with probability of heads equal to p (not
> > necessarily fair). We perform an infinite number of experiments
> > simultaneously with one toss for each experiment. Question 1: can we
> > claim that the event heads will show at least once in this infinite
> > number of experiments of one trail each? Question 2: what about if
> > there is an infinite number of trials for each of the infinite
> > experiments? Can we claim that heads shows up infinite times?
>
> > Thanks and I apologize in advance if this is a stupid question.
>
> For any given p, the probability of at least one head in n trials is
> 1 - (1-p)^n, which can be made as close to 1 as you want by taking n
> big enough (assuming, of course, that p > 0).
Thank you all guys. I appreciate. However, I cannot see how the above
formula is obtained. My book says that the probability of an event
with probability p<1 to show 1 times (Bernouli trials) is np(1-p)^
(n-1). Am I missing something?
John
The formula you give is for exactly 1 head and only 1 head. You asked
for at least 1 head.
In order to calculate at least one head you calculate the probabilty of
all tails (1-p)^n. Then the probability not all tails is 1-(1-p)^n.
Not all tails = at least one head.
Joe Avery
> Joe Avery wrote:
> > On Apr 27, 4:33 am, Ray Koopman <koop...@sfu.ca> wrote:
> >> On Apr 26, 3:58 pm, Joe Avery <joe_avery_2...@yahoo.com> wrote:
>
> >>> This isn't homework (I'm too old for that)
> >>> Suppose we have a coin with probability of heads equal to p (not
> >>> necessarily fair). We perform an infinite number of experiments
> >>> simultaneously with one toss for each experiment. Question 1: can we
> >>> claim that the event heads will show at least once in this infinite
> >>> number of experiments of one trail each? Question 2: what about if
> >>> there is an infinite number of trials for each of the infinite
> >>> experiments? Can we claim that heads shows up infinite times?
> >>> Thanks and I apologize in advance if this is a stupid question.
> >> For any given p, the probability of at least one head in n trials is
> >> 1 - (1-p)^n, which can be made as close to 1 as you want by taking n
> >> big enough (assuming, of course, that p > 0).
>
> > Thank you all guys. I appreciate. However, I cannot see how the above
> > formula is obtained. My book says that the probability of an event
> > with probability p<1 to show 1 times (Bernouli trials) is np(1-p)^
> > (n-1). Am I missing something?
>
> The formula you give is for exactly 1 head and only 1 head. You asked
> for at least 1 head.
Thanks John. You are correct. Precision means a lot in math I should
have known.
>
> In order to calculate at least one head you calculate the probabilty of
> all tails (1-p)^n. Then the probability not all tails is 1-(1-p)^n.
> Not all tails = at least one head.- Hide quoted text -
Understood. What about exactly one head when n goes to infinity? Is
that defined?
I don't know if this makes sense but can we also determine the
probability that heads shows up infinite times from these formulas, as
n goes to infinity?
Joe
>
> - Show quoted text -
John
I can't see a problem with defining it as the limit as n->inf of the
formula you give. e.g. 0.
> I don't know if this makes sense but can we also determine the
> probability that heads shows up infinite times from these formulas, as
> n goes to infinity?
>
That is a little more tricky. My intuition (and Kolmogorov's 0-1 law)
tells me that the answer is infinite heads with probability 1.
Finite heads would mean that eventually you would get all tails. e.g an
infinite sequence with no heads which our earlier results suggested had
probability 0.
But this looks like a double limit and I really don't know what I'm
talking about so hopefully someone else will explain it properly.
danh...@yahoo.com
The strong law of large numbers asserts that as n->oo, (number of
heads in n trials)/n -> p with probability one. For p>0, the numerator
must -> oo (with probability one).
Joe Viscomi
great question and good discussion; i have followed along here and i
have no problem with the theoretic probability comments – however, i
do wonder if we’re not being somewhat careless in the language we’re
using --
the original poster asks, “Can we claim that heads shows up infinite
times?” and the answer to that question is NO – i.e., we have no
guarantee that heads will ever show – think ‘Gambler’s Ruin’ -- yes,
there are probabilities that heads will show, and yes ,the probability
that an infinite number of tosses will produce A PROBABILITY for
infinite heads and this calculation rapidly APPROACHES 1 but it will
never be 1 and is just a probability, not a certainty -- thanks.
cheers,
joe
Mike
"Joe Viscomi" wrote in message
I beg to differ. The number of heads can't be finite as then the
probability
would then go to 0 as the number of trials goes to infinity, contradicting
the hypothesis that the probability of tossing a head on a trial is 0 < p <
1.
(This is why questions re infinity can't be stupid. There are so many
interesting and counter intuitive aspects! The basic problem is that
few people truly understand infinity. It's almost is beyond comprehension.)
Joe Avery
"The strong law of large numbers asserts that as n->oo, (number of
heads in n trials)/n -> p with probability one. For p>0, the
numerator
must -> oo (with probability one). "
Joe Viscomi wrote:
"the probability that an infinite number of tosses will produce A
PROBABILITY for
infinite heads and this calculation rapidly APPROACHES 1 but it will
never be 1 and is just a probability, not a certainty"
Both interesting responses. Dan's response introduces the concept of
the certain event in a relative frequency interpretation of
probability involving infinite trials. Joe Viscomi denies that we can
talk about certain events when we talk about probabilities, even in
the case of infinite trials.
I must admit I have trouble with introducing the notion of the certain
event in interpreting the relative frequency definition of probability
that Dan presented. It sounds more of a hypothesis about the existence
of the limit rather than a proof that the limit exists because the
number of heads must be infinite. In other words, it cannot correspond
to an actual experiment that generates the limit.
So, can we turn a hypothesis about the existence of a limit to a proof
of a certain event (infinite heads) to guarantee the existence of the
limit?
Am I lost or what?
Mark
> Dan wrote:
>
> "The strong law of large numbers asserts that as n->oo, (number of
> heads in n trials)/n -> p with probability one. For p>0, the
> numerator
> must -> oo (with probability one). "
>
> Joe Viscomi wrote:
>
> "the probability that an infinite number of tosses will produce A
> PROBABILITY for
> infinite heads and this calculation rapidly APPROACHES 1 but it will
> never be 1 and is just a probability, not a certainty"
>
> Both interesting responses. Dan's response introduces the concept of
> the certain event in a relative frequency interpretation of
> probability involving infinite trials. Joe Viscomi denies that we can
> talk about certain events when we talk about probabilities, even in
> the case of infinite trials.
>
There appears to have been a bit of confusion about the meaning of "with
probability 1". This is a rigorously defined concept. An equivalent term
is "almost surely".
http://en.wikipedia.org/wiki/Almost_surely.
> I must admit I have trouble with introducing the notion of the certain
> event in interpreting the relative frequency definition of probability
> that Dan presented. It sounds more of a hypothesis about the existence
> of the limit rather than a proof that the limit exists because the
> number of heads must be infinite. In other words, it cannot correspond
> to an actual experiment that generates the limit.
>
I don't understand how you have an experiment with infinite coin tosses.
If it is a thought experiment the definition of "with probability 1"
handles the apparent paradox regarding certain.
> So, can we turn a hypothesis about the existence of a limit to a proof
> of a certain event (infinite heads) to guarantee the existence of the
> limit?
>
Yes for all events where the limit exists (i.e. almost surely) we know
we must have infinite heads. So we know the event infinite heads is true
almost surely, or with probability 1.
> Am I lost or what?
>
>
>
Measure theory helps ;o)
Joe Viscomi
that the coin may not be a fair coin --
the difficulty as i see it is are we talking about probability or coin
tosses - are we discussing abstract mathematics or "common"
mathematics (akin, e.g., to folk psychology in philosophy) --
in my opinion, there is a distinct difference between a function that
" ... approaches x" and a function that "equals x" --
so for an infinite number of coin tosses there WILL BE an infinite
number of probabilities (see Cantor's diagonal proof) - i'm not saying
a probability will not exist, i am saying that the probability DOES
NOT guarantee the occurrence of the event in question --
BTW: not to go too far afield but for a fun discussion, see Persi
Diaconis for discussion of whether or not a coin toss is a random
event or a determinant event - for the record, i side w/Mr. Diaconis,
that a coin toss is a determinant event -- thanks.
cheers,
joe
Mike
> so for an infinite number of coin tosses there WILL BE an infinite
> number of probabilities (see Cantor's diagonal proof) - i'm not saying
> a probability will not exist, i am saying that the probability DOES
> NOT guarantee the occurrence of the event in question --
You're right. For a finite number of trials there is no guarantee
that one head or any particular number of heads or tails will appear.
However, for an infinite number of trials, assuming 0 < p < 1
where p = Prob{H} in a single trial, both the number of heads and
the number of tails must absolutely, positively, be infinite as well.
Don't need limits to prove it either: simple symmetry will do.
John
> i agree w/Mike, this is a great question, especially the stipulation
> that the coin may not be a fair coin --
>
> the difficulty as i see it is are we talking about probability or coin
> tosses - are we discussing abstract mathematics or "common"
> mathematics (akin, e.g., to folk psychology in philosophy) --
>
Applied mathematics I would have thought. It is common in practice to
use the continuous case as an approximation for a discrete but large
number of events as it is often makes the problem more tractable
mathematically.
> in my opinion, there is a distinct difference between a function that
> " ... approaches x" and a function that "equals x" --
>
Yes there is a difference between a sequence that approaches a limit and
the limit itself. But when we talk of infiinte coin tosses we are
talking about the limit not the sequence.
> so for an infinite number of coin tosses there WILL BE an infinite
> number of probabilities (see Cantor's diagonal proof) -
You mean there are uncountably infinte different possible outcomes of
infine coin tosses.
> i'm not saying
> a probability will not exist, i am saying that the probability DOES
> NOT guarantee the occurrence of the event in question --
>
Here you would need to define what guarantee means. This definition
would of course depend on how you wanted to use the concept. In
probabilty theory the choice has been made to replace the every day
concept of certain with almost surely but AIUI for practical purposes
the two are the same.
>
> BTW: not to go too far afield but for a fun discussion, see Persi
> Diaconis for discussion of whether or not a coin toss is a random
> event or a determinant event - for the record, i side w/Mr. Diaconis,
> that a coin toss is a determinant event -- thanks.
>
Do you mean deterministic. It is often useful to model deterministic
events as random when it is not practical (or we cannot) calculate
results deterministically. The long and short is for practical purposes,
deterministic or random, we don't care. If it looks like a duck and
quacks like a duck...
> cheers,
> joe
Joe Avery
>
> You're right. For a finite number of trials there is no guarantee
> that one head or any particular number of heads or tails will appear.
I guess we all agree to this
>
> However, for an infinite number of trials, assuming 0 < p < 1
> where p = Prob{H} in a single trial, both the number of heads and
> the number of tails must absolutely, positively, be infinite as well.
> Don't need limits to prove it either: simple symmetry will do.
Question: Infinite (number of heads and tails) with propability 1 in
the sense of 'almost surely" meaning that there is a tiny propability
that this will not be the case OR infinite (number of heads and tails)
in the sense that this is the certain event, i.e. infinite tosses with
probability p in a single trial give rise to a deterministic event
{infinite heads and tails} as the number of trials approaches
infinity?
In other words, do we conclude that infinity when conbined with a
random process entails a sort of determinism of this kind?
This is great. Great discussion, thank you all.
Mike
SNIP
> Question: Infinite (number of heads and tails) with propability 1 in
> the sense of 'almost surely" meaning that there is a tiny propability
> that this will not be the case OR infinite (number of heads and tails)
> in the sense that this is the certain event, i.e. infinite tosses with
> probability p in a single trial give rise to a deterministic event
> {infinite heads and tails} as the number of trials approaches
> infinity?
> In other words, do we conclude that infinity when combined with a
> random process entails a sort of determinism of this kind?
Yes, as there is absolutely no chance that an infinite number of trials
(with 0 < p < 1) would produce a finite number of heads or tails.
Infinity is better understood as a concept unto itself, rather than a limit
of some finite quantity or process. The gap between an ever expanding, or
increasing finite quantity (or an ever continuing finite process) and
infinity is as large as infinity itself. One does not approach infinity
from the finite. It is not a continuous process. There is no road
connecting the finite to the infinite. Even from the largest imaginable
finite quantity, one still has to make "a giant leap" (an understatement of
infinite proportions) to reach infinity.
So how DOES an ever expanding binomial distribution ultimately collapse into
the simple "determinism", i.e., the certainty of an infinite number of heads
(and tails) as the number of trials goes to infinity?
Beats me. The giant leap took us over that little detail.
A fascinating discussion of the nature of infinity is buried in this
article.
http://www.newyorker.com/archive/1992/03/02/1992_03_02_036_TNY_CARDS_000362534?currentPage=1
The detail skipped over by that leap also explains the famous "Two Envelop
Paradox" (and others) from game theory. Any google search will find many
articles attempting to explain the paradox but they all come down to
attempting to apply finite, real world thinking to the world of infinity, or
equivalently, carelessness in defining the domain of the game.
Joe Viscomi
Mike, we agree mathematically - the rub for me is langauge - too much
philosophy (on my part) ;-)
John; yes, ur correct, i did mean deterministic ...
great discussion; thanks.
cheers,
joe