Smart math people, help?

[Echognome]

P ( our envelope contains the smaller amount | our envelope contains $M )

 

is not necessarily the same as the a priori probability (which is 1/2 of course). We're trying to work out the expectation from switching given that our envelope contains $M, and in order to do this we need to know the probability that our envelope contains the smaller amount given that it contains $M. That is, the a priori probabilities are useless, what we need to know are the conditional probabilities.

Why is the a priori probability 1/2?

 

Why are the a priori probabilities useless?

 

The last time I checked P(A|B ) = P(A,B )/P(B ). Without the a priori probability of B, it seems rather hard to calculate.

 

Here the P(B ) = P(our envelope contains $M). We don't have that defined in our problem in any way. Note the conditional probability does not exist if that probability is 0. I've already stated above why it can't be uniform.

[david_c]

Here the P(B ) = P(our envelope contains $M).  We don't have that defined in our problem in any way.

Indeed - and so we can't calculate the expectation in the original problem. The fallacy in many people's approaches is to assume that using 1/2 must work, when in fact the probabilities are completely undefined. (Though one thing we do know, as you say, is that the probabilities can't all be 1/2.)

 

But you can resolve this problem, by specifying a distribution which we use to choose the amounts in the envelopes, and still be in the situtation where switching is best no matter what amount you found in the envelope.

 

Why is the a priori probability 1/2?

I was making the assumption that, after the amounts in the two envelopes had been decided, you would pick which one you were going to open at random.

[MikeRJ]

Some one may already have said this in adifferent way, but here goes..

 

Suppose I have not opened an envelope, what do I know?

The lowest value envelope contains x dollars.

The higher value contains 2X dollars.

My average expectation selecting an envelope randomly is 1.5X dollars.

If I changeand get X dollars I will lose 0.5X dollars against my average expectation.

If I change and get 2X dollars I will gain 0.5X against my average expectation.

On average changing will gain as much as it loses – as one would expect.

 

So why does this change if I know the sum of money in the envelope I have selected? Now if I do not change I have an ACTUAL expectation, not an AVERAGE expectation - I have more information.

 

If I open an envelope containing $5000 and I change I will end up with either 2500 or 10000 (average with $6250), therefore I should change.

 

The reason this works is that my envelope can never contain 1.5X dollars – only X or 2X.

 

Of course in the situation where you know what you have in the envelope, you will never have the chance to change again.

 

Mike

[P_Marlowe]

B is indeed wrong, or rather, the set-up of the problem isn't possible.

 

Before you can answer the problem, you would have to describe how likely each combination is. The problem assumes that every combination is equally likely, i.e. $2,500+ $5,000 is just as likely to $5,000+$10,000, and likewise for every other combination. But this isn't possible.

 

To see why this isn't possible let's assume that we do have such a distribution of probabilities. The chance that the the lowest amount lies between 1 and 2 must be some positive number, let's say "x". But then the chance that it lies between 2 and 4 must also be "x". And the chance that it lies between 4 and 8 must also be x. But then the chance that the lowest amount lies between 1 and 2^n is x + x + x +.. = n*x. And for large n this is larger than 1! This isn't possible.

 

I'm playing bridge atm so I'll try to do a better job later.

Hi, I'm talking in a pure math sense, I don't care about the applied form. It is completely hypothetical in my mind in a world where infinity exists. There is no upper limit. This is also the solution Taleb offers but it seems flawed to me but obviously you know more than me. I am now reading the wiki article on this.

Hi,

 

No upper limit is not the same as "infinity" exists.

 

In a world where infinity exists, you cant compare infinity

with infinity / 2, more precise infinity is in a sense the same

as infinity / 2.

 

A example (has nothing to do with the original problem,

but may help explain the above statement):

 

Imagine two hotels with infinite rooms belonging to the same

hotel chain. Every hotel is completely booked up, i.e. no free

room.

Now in the 2nd hotel they needes to repair the electric, so every

guest had to move out.

The portier of the first hotel said: "No problem they can move into

our hotel. The guys in our hotel move to the room with the number

they get, if they multiply their original room number with two

(2n - n being the original room number).

The guys from the other hotel move to the room with the

number they get, if they multiply their original room number

with two and subtract 1 (2n-1)."

Everybody followed order, and everybody found his own room.

 

And the original hotel was still completly booked out.

 

With kind regards

Marlowe

[helene_t]

If I open an envelope containing $5000 and I change  I will end up with either 2500 or 10000 (average with $6250), therefore I should change.

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

 

I still think my example with the weights of the males and females must be easier to reconcile with intuition.

 

After you observed that the first person weighted 100 kg, the probability that he's the male is more than 50%.

 

Before you made the observation, the probability is of course 50%. This means that your average gain in percentages, if you switch, is positive. The fallacy is to consider this average percentage gain to be something attractive. It's not. You're not interested in percentages, you're interested in dollars. The average probability of 50% is comprised on some high male-probabilities in case it turned out to be a heavy person and some low ones is it turns out to be a light person. If it's a 40-kg person that you correctly acess to be the female you can win a lot of percents by switching, but those are percents of a small number so the amount of dollars you gain is not more than the smaller percentage of a higher amount in case the first person turned out to be a 100-kg male. So although your expected percentage gain is +25%, your expected dollar gain is $0.

 

Taking averages of percentages easily leads to nonsense. Suppose we both have $100. Now I give you $10 and ten minutes afterwards you give me $10 back. I first lost 10% of my $100 and then gained 11.1% of my $90. You first won 10% of your $100 and then lost 9.1% of your $110. My net gain was 1.1% and yours 0.9%. Halelula, an easy solution to the treasury deficit!

[P_Marlowe]

<snip>

think Hannie is saying that because you cannot assign a number to infinity, you cannot define 1/2 of infinity or two times infinity, the problem is not possible to solve mathematically. Is this correct, Han?

No, 2*Inf=Inf and Inf/2=Inf. That's not a problem. The problem is that if there's an infinite number of boxes, some must have higher probabilities than others. Since otherwise each particular box would have probability 1/Inf=0 and therefore it would be impossible to open a box, events with zero probability never happen.

 

I know this sounds very contra-intuitive, it took me a long time fully to understand it myself.

Events with zero probability can happen, but you should

not wait for them.

 

An simple example: It is possible that you may hit the precise

center of a circular target with an arrow, but the probability

is zero.

 

With kind regards

Marlowe

[helene_t]

Events with zero probability can happen, but you should

not wait for them.

 

An simple example: It is possible that you may hit the precise

center of a circular target with an arrow, but the probability

is zero.

No! Emphatically no! Suppose your digital ArrowHitMeter reports the distance from the center to your hit as 0.0000. Then all you know is that the distance was rounded of to zero by four digits after the period, i.e. it is between 0 and 0.00005. That range has positive width and presumably a positive probability. If the probability was zero it would not have happened.

 

This may sound like a silly measurement-technology problem but it's not. It's not even a physical problem, related to Heisenberg's uncertainty principle or some such. It's a fundamental principle in probability theory: If you have a random variable on a continous scale (such as distance from the center) the observations of that random variable are always ranges with positive width. You cannot pick a real number. You can pick an integer, or you can pick a range of real numbers.

 

Of course you can ask me to pick a "real" number and I'll be happy to think of sqrt(2) or pi or some such. But that's an illusion. I can only pick from the countable subset of the real numbers that can be expressed by mathematical formalism (or whatever language I think in). So effectively I'm picking an integer and then thinking of some transform of that integer which happens to be a non-integer number.

 

One could argue that even though the excact value of a real-valued random variable cannot be observed, the excact value could still be "out there" is some sense. That is a deep philosofical issue which we have had in a couple of other threads. For the purpose of this thread I think it suffices to say that probability theory is concerned with events that can be observed, at least in principle.

[frouu]

 

[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.

 

say in one experiment you see 200 in the first envelope, you're assuming your outcomes are {25,50,100,200,400,800,..} so your k was 25. But you didn't know what this was. You'd be surprised in another experiment if you observed 300, which is not in your outcome set, and you didn't assign a probability for it.

[Gerben42]

The main problem with the whole dilemma is that people use a different metric for the worth of money. For someone like me, winning $1M or winning $2M is roughly the same, however winning $500k is probably much of the same also. Here I would switch as the worst that can happen is really great already!

 

Similarly, if my first envelope contains $10, I will also switch. $20 or $5 is also not as much or as few that I think much differently about it.

 

But inbetween for most people there is an amount where he will NOT switch because the amount is such that losing half it is worse than winning twice the amount.

[helene_t]

The main problem with the whole dilemma is that people use a different metric for the worth of money.

No, that's not the issue. The assumption is that you maximize expected gain. Whether that assumption is realistic or not is irelevant. The problem under discussion here is the mathematical one.

[Trumpace]

Events with zero probability can happen, but you should

not wait for them.

 

An simple example: It is possible that you may hit the precise

center of a circular target with an arrow, but the probability

is zero.

No! Emphatically no! Suppose your digital ArrowHitMeter reports the distance from the center to your hit as 0.0000. Then all you know is that the distance was rounded of to zero by four digits after the period, i.e. it is between 0 and 0.00005. That range has positive width and presumably a positive probability. If the probability was zero it would not have happened.

 

This may sound like a silly measurement-technology problem but it's not. It's not even a physical problem, related to Heisenberg's uncertainty principle or some such. It's a fundamental principle in probability theory: If you have a random variable on a continous scale (such as distance from the center) the observations of that random variable are always ranges with positive width. You cannot pick a real number. You can pick an integer, or you can pick a range of real numbers.

 

Of course you can ask me to pick a "real" number and I'll be happy to think of sqrt(2) or pi or some such. But that's an illusion. I can only pick from the countable subset of the real numbers that can be expressed by mathematical formalism (or whatever language I think in). So effectively I'm picking an integer and then thinking of some transform of that integer which happens to be a non-integer number.

 

You are mistaken.

 

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

[helene_t]

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 :lol:

[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? :lol:

[P_Marlowe]

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]

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

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]

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]

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 :lol:

[helene_t]

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]

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 :lol:

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]

 

[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]

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:

 

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]

"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]

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.