[helene_t]

Trumpace, on Jul 19 2007, 10:08 AM, said:

I think you are mistaken.

If probability of an event is zero, it does not mean it is impossible.

Sorry but I know what I'm talking about. "Impossible" and "has zero probability" are synonymous.

If something "impossible" happens, it obviously wasn't impossible after all. The prior probability might have been 0.00000000000000001 or less, but not zero.

Of course probabilities often come from simplified mathematical models which rule out a lot of possibilities for convenience. So if some computer program tells you that a certain event has zero probability according to established theory, the event may still happen in practice.

[Gerben42]

Sorry but I've had this discussion like 5 times already, don't be mad if I won't join in

[ochinko]

Oh boy, I'm getting dizzy just thinking of the possibilities.

If I trade my $5000 for the other envelope I either lose $2500 or win another $5000. With every two swaps my (average) net gain is $2500.

So if I am allowed unlimited number of swaps, I'll be a millionaire within a day by just swapping the two envelopes back and forth. Will this strategy be hampered by the fact that after the first swap I already know what is in the other envelope?

[P_Marlowe]

helene_t, on Jul 19 2007, 03:11 AM, said:

Trumpace, on Jul 19 2007, 10:08 AM, said:

I think you are mistaken.

If probability of an event is zero, it does not mean it is impossible.

Sorry but I know what I'm talking about. "Impossible" and "has zero probability" are synonymous.

No, it is as simply as that.

"Impossible" and "has zero probability" is only synonymous,
if the space of events which can happen is finite, i.e. only
in cases where there exist a finite number of outcomes. (*)

Thats it, if you dont accept this, you will have problems understanding
stochastic.
I dont have a clue about stochastic, as far as it was possible,
I avoided stochastic like I am trying to avoid hell, it was hard,
believe me, because I did study math.

With kind regards
Marlowe

(*) Pick a arbitary natural number and I guess which number oyu
pick. I have zero possibility to guess, which natural number you
picked.
And please dont argue with the limitness of space and time to write
the number somewhere, so that it would later be possible
to compare. It is just a experiment of the mind (but there are real
problems connected with this).

PS: From a mathematical point of view events with zero probability
are necessary to get the mathematic going.
You may remember Mass theory, where measurement of certain
set is zero, but the set is not empty, this is related, just another look
at the whole thing.

[P_Marlowe]

Gerben42, on Jul 19 2007, 03:12 AM, said:

Sorry but I've had this discussion like 5 times already, don't be mad if I won't join in

whimp (hopefully spelled right).

But I can understand it, and will stop
the discussion as well, which shows how
whimpish I am as well.

With kind regards
Marlowe

[Trumpace]

helene_t, on Jul 19 2007, 03:11 AM, said:

Trumpace, on Jul 19 2007, 10:08 AM, said:

I think you are mistaken.

If probability of an event is zero, it does not mean it is impossible.

Sorry but I know what I'm talking about. "Impossible" and "has zero probability" are synonymous.

If something "impossible" happens, it obviously wasn't impossible after all. The prior probability might have been 0.00000000000000001 or less, but not zero.

Sorry, but from a pure math perspective they are not synonymous.

Ideally we would like them to be synonymous, but that is not the case.

(It is similar to: probability of 1 does not imply a certain event.)


For instance consider the measurable subsets of [0,1] as the events of the sample space, with probability of the event corresponding to the measure. We can show that all the axioms of the probability spaces are satisfied. Consider the sets of measure zero. (eg countable subsets). These have zero probability, but are not "impossible".

[Trumpace]

P_Marlowe, on Jul 19 2007, 03:26 AM, said:

Gerben42, on Jul 19 2007, 03:12 AM, said:

Sorry but I've had this discussion like 5 times already, don't be mad if I won't join in

whimp (hopefully spelled right).

But I can understand it, and will stop
the discussion as well, which shows how
whimpish I am as well.

With kind regards
Marlowe

wimp

[helene_t]

Trumpace, on Jul 19 2007, 10:32 AM, said:

Sorry, but from a pure math perspective they are not synonymous.

For instance consider the measurable subsets of [0,1] as the events of the sample space, with probability of the event corresponding to the measure. We can show that all the axioms of the probability spaces are satisfied. Consider the sets of measure zero. (eg countable subsets). These have zero probability, but are not "impossible".

After having done some googling I realize that the word "impossible", in relation to continous probabilty measures, actually refer to the empty set rather than to sets with zero measure. I.e. the word "impossible" is used for the event itself rather than for the observation of the event. So I was wrong. Sorry.

Anyway, I hope you still agree that null events cannot be obseved, e.g. if you ask me to think of a number between zero and one then any number I might think of, has positive probability. (The original problem was Han's boxes, each of which contained two envelopes. The point is that if there's an infinity of boxes, some must have higher probabilities than others. Of course that's a rather trivial problem because the set of boxes is countable so they could all have possitive probability, it isn't necesary for this argument to consider super-countable sets).

[cherdano]

Jlall, on Jul 18 2007, 11:02 AM, said:

Hannie, on Jul 18 2007, 12:00 PM, said:

This is a well known paradox. Let me think about how to explain it best, I'll edit this post.

I knew you would know !H Maybe it woulda been well known to me if I hadn't dropped out of school

Unfortunately fun and kind of useful stuff like this (some intuition about probabilities is certainly worth having in many situations) is hardly ever taught in school or even college.

[david_c]

frouu, on Jul 19 2007, 08:03 AM, said:

david_c, on Jul 18 2007, 08:53 PM, said:

[For the smart math people: just take a very slowly decaying distribution. For example, let the probability that envelopes contain $(2^n) and $(2.2^n) be k.(1-epsilon)^(|n|).]

In this case, when analysing whether it's right to switch for a particular amount M, the calculations are so close to what they would be if the probabilities were 1/2 that it makes no difference. The conclusion is

For every amount M you see in the envelope, your expectation if you switch is greater than M.

(In fact, greater than 1.2M, say. We can get any multiple less than 1.25.)

Is this a paradox? It shouldn't be - it's true. Do you believe me?

I don't believe you because you're restricting the outcomes of the experiment to {k*2^n} sequence and you don't a priori know what "k" is.

Well, two points:

(i) You've misread the example. k is a normalising factor for the probabilities, to make sure they sum to 1.

(ii) Yes it's a discrete distribution. That's the easiest example to write down. However, you could find a continuous distribution with the same properties if you like.

[bid_em_up]

helene_t, on Jul 19 2007, 01:35 AM, said:

Not necesarily since the probability that the other envelope contains $10000 may be less than 50%. Or more. That's the whole point.

Ok, I think I understand what you are attempting to get at now.

The original problem says:

Quote

2 people were having the following discussion:

A: Say you have 2 envelopes filled with money. The value of one envelope is half the value of the other. These values can approach infinite $.

But it does not say that there is an equal chance of having twice as much money as there is as having 1/2 as much money.

Its like if I had two barrels of money (don't I wish), each with 100 envelopes in them. In barrel one, all the envelopes contained $5000. In barrel two, 99 of them could have $2500, and one could have $10000, so the percent chance of drawing $2500 is 99%. In this case you should not switch. If it was 99-1 in favor of $10000, then you should.

We are not actually told that the odds of $2500 and $10000 being in the other envelope is 50-50, even though when I originally read it, I assumed it was implied. But in reality, we are only told that one envelope will contain exactly 1/2 the amount of money in the other. Since we don't know what the odds are of $2500 or $10000 being in the envelopes in the other barrel, the answer to the question is, "it is impossible to decide." or "not enough information to make an informed decision".

Am I getting closer to understanding it?

[barmar]

P_Marlowe, on Jul 19 2007, 03:24 AM, said:

"Impossible" and "has zero probability" is only synonymous,
if the space of events which can happen is finite, i.e. only
in cases where there exist a finite number of outcomes. (*)

Thats it, if you dont accept this, you will have problems understanding
stochastic.
I dont have a clue about stochastic, as far as it was possible,
I avoided stochastic like I am trying to avoid hell, it was hard,
believe me, because I did study math.

With kind regards
Marlowe

(*) Pick a arbitary natural number and I guess which number oyu
pick. I have zero possibility to guess, which natural number you
picked.

When dealing with infinities I think you also have to deal with limits. I.e. rather than saying that the probability of guessing the number is zero, you say that the probability approaches zero as the maximum allowed guess approaches infinity. "Limit(x) = 0" is not the same as "x = 0". In fact, what it means is that you can make x as close to 0 as you wish (by altering some other parameter), but it never actually reaches it. In this case, there are no "impossible" things, just "very improbable".

[helene_t]

bid_em_up, on Jul 19 2007, 05:36 PM, said:

helene_t, on Jul 19 2007, 01:35 AM, said:

Not necesarily since the probability that the other envelope contains $10000 may be less than 50%. Or more. That's the whole point.

But it does not say that there is an equal chance of having twice as much money as there is as having 1/2 as much money.

Its like if I had two barrels of money (don't I wish), each with 100 envelopes in them. In barrel one, all the envelopes contained $5000. In barrel two, 99 of them could have $2500, and one could have $10000, so the percent chance of drawing $2500 is 99%. In this case you should not switch. If it was 99-1 in favor of $10000, then you should.

We are not actually told that the odds of $2500 and $10000 being in the other envelope is 50-50, even though when I originally read it, I assumed it was implied.

In general, if you don't know what the odds are you must assume them to be 50-50. If you know that one cage contains a lion and the other contains a tiger and you open one of them, the chances as stated are 50-50. It could be that you have more information (there is a green cage and a red cage and you open the red one and the red one is more likely to contain the tiger), but that would have to be stated.

So with the envelopes, the a priori chance of choosing the one with more money is 50%. But after you open it and count the money in it, knowledge of the amount of money in that particular envelop might give you more information. For example, if you know that the maximum possible amount is $6000, and you count $5000, you know that you picked the one with the large amount.

And now comes the solution to the paradox: it is possible that some particular amounts don't give you more information. For example, it could be that an amount of $1756 leaves the probabilities unchanged 50-50. But it can be proved that in general, knowing the amount in one envelope will alter the chance that that envelope is the one with the large amount. It is impossible to design the experiment such that no amount will alter the odds. In particular, the odds must be altered such that although the percentages you gain by switching will on average (computed before you count the money) be 25%, the average gain in dollars must be zero. Again, this zero gain is an average that is computed before you open the envelope. If $1756 leaves the odds 50-50 and the first envelope happens to contain exactly $1756, then you should switch.

I know this sounds contra-intuitive. It was my hope that the example with the weights of the males and females was easier to understand,

[barmar]

The problem with the male/female weights example is that it includes an implicit condition that wasn't stated: the common sense knowledge that women rarely weigh 100 kg, but this is a common weight for men. This alters the probabilities significantly.

But the original problem doesn't have a condition like this, unless you assume it's happening in the real world where there's a finite amount of money. But the paradox still exists in the pure mathematical realm, where there's no lower or upper limit to the numbers that can be in each envelope.

[helene_t]

barmar, on Jul 19 2007, 06:42 PM, said:

The problem with the male/female weights example is that it includes an implicit condition that wasn't stated: the common sense knowledge that women rarely weigh 100 kg, but this is a common weight for men. This alters the probabilities significantly.

But the original problem doesn't have a condition like this, unless you assume it's happening in the real world where there's a finite amount of money. But the paradox still exists in the pure mathematical realm, where there's no lower or upper limit to the numbers that can be in each envelope.

That's exactly the reason why I chose male and female weights: we know that women rarely weight 100 kg.

The same applies to the envelope problem. You must know something similar, like "envelops rarely contain $1000000000000000" or whatever. Even in the mathematical realm. It is impossible to conduct and experiment, even a thought experiment, with an infinite number of possible outcomes unless some outcomes are deemed more likely than others.

[david_c]

helene_t, on Jul 19 2007, 05:34 PM, said:

And now comes the solution to the paradox: it is possible that some particular amounts don't give you more information. For example, it could be that an amount of $1756 leaves the probabilities unchanged 50-50. But it can be proved that in general, knowing the amount in one envelope will alter the chance that that envelope is the one with the large amount. It is impossible to design the experiment such that no amount will alter the odds. In particular, the odds must be altered such that although the percentages you gain by switching will on average (computed before you count the money) be 25%, the average gain in dollars must be zero.

This is all true, but it only gives you useful information in the case where the expected amount of money in the envelopes is finite. Since we are told that the amounts in the envelopes can "approach infinity", it is perfectly possible that the expectation is infinite.

In this case, it may well be the case that once you've opened the envelope, no matter what amount of money you find, the expectation from switching is higher than the amount you are currently looking at.

[jtfanclub]

david_c, on Jul 19 2007, 11:54 AM, said:

This is all true, but it only gives you useful information in the case where the expected amount of money in the envelopes is finite. Since we are told that the amounts in the envelopes can "approach infinity", it is perfectly possible that the expectation is infinite.

So, you open the envelope, and it has infinity dollars in it. You're told that if you switch envelopes, you have a 50% chance of it having half of infinity dollars in it, and a 50% of having twice infinity dollars. Do you switch envelopes?

If the expectation is infinite, then the entire question is meaningless.

[helene_t]

david_c, on Jul 19 2007, 06:54 PM, said:

Since we are told that the amounts in the envelopes can "approach infinity", it is perfectly possible that the expectation is infinite.

In this case, it may well be the case that once you've opened the envelope, no matter what amount of money you find, the expectation from switching is higher than the amount you are currently looking at.

It's true that the expected amount could be infinite. Think of a particular probability density function with infinite mean, for example p(2^n) = 2^(-n), n=1...Inf. and let the envelope with the small amount contain 2^n dollar and the other twice as much. Now if the envelop you open contains $16 you know that the prior probability of the small-amount envelop containing $16 was 1/16 while the prior probability that the small-amount envelope contained $8 was 1/8, so the posterior probability that you picked the large-amount envelope is 2/3 and the expected gain by switching is zero. So it doesn't matter if you switch or not.

Unless you open an envelop with $2. Now you know that the other contains $4 so you must switch.

What's your expected gain from switching? You stand to lose Inf and you also stand to gain Inf, and since Inf-Inf is undefined there is no answer.

On the other hand, for any particular amount in envelope A you stand to gain $0 except if envelope A contains $2, in which case you stand to gain $2.

This sounds a little strange. Did I do something wrong?

[david_c]

helene_t, on Jul 19 2007, 06:22 PM, said:

This sounds a little strange. Did I do something wrong?

No, you didn't. And like I keep saying, it is perfectly possible for it to be right to switch no matter what amount you find in the envelope. Does that sound strange too?

[david_c]

jtfanclub, on Jul 19 2007, 06:14 PM, said:

david_c, on Jul 19 2007, 11:54 AM, said:

This is all true, but it only gives you useful information in the case where the expected amount of money in the envelopes is finite. Since we are told that the amounts in the envelopes can "approach infinity", it is perfectly possible that the expectation is infinite.

So, you open the envelope, and it has infinity dollars in it.

No it doesn't.

The amounts in the envelopes are always finite, but they can be chosen in such a way that the expectation before you open an envelope is infinite.

If you don't understand how this is possible you need to learn some basic probability theory.