[bid_em_up]

Hannie, on Jul 18 2007, 01:19 PM, said:

Oh, and to the people who want to switch, I always found this an entertaining thought: if you are going to switch anyway, why open the first envelope? Choose which one you will pick and then quickly pick the other one!

And then swap back and forth indefinitely as it is now equally likely you have now picked the wrong one.

[Fluffy]

Hannie, on Jul 18 2007, 06:13 PM, said:

Here is a similar problem:

[i]There are two identical boxes, each box contains two identical envelopes. In the first box the envelopes contain $2,500 and $5,000, in the second box the envelopes contain $5,000 and $10,000.

I don't think this is similar problem.

If I run a simulation upon this it would be correct to switch:

2500$ -> switch you will win +2500$ or +7500$
5000$ -> switch you will win +5000$ or lose -2500$
10000$ -> Do not switch. (-5000$ or -7500$)


Wich relates to the initial problem, the higher you have, the less you want to switch.

so what am I missing?

[bid_em_up]

Hannie, on Jul 18 2007, 01:17 PM, said:

bid_em_up, on Jul 18 2007, 01:02 PM, said:

helene_t, on Jul 18 2007, 12:53 PM, said:

Han is right. Fluffy's reference to an upper limit is close but not quite. Suppose the probabilites of the first envelop containing x dollars are
$ prob
1 1/2
2 1/4
3 1/8
4 1/16
5 1/32
6 1/64
.. etc

This allows for no upper limit but Han's argument still holds.

The problem with these statements is that the envelopes in question contain a finite (and definite amount, $5000). All the other "probabilties" are meaningless.

It is not meaningless. You will only be able to understand the problem if you consider the complete distribution of probabilities into account. I hope my previous post (where all irrelevant amounts are discarded) clarified this.

Yes it is.

1 1/4(N+1) is still greater than N+1.

and it doesnt matter what amounts are in the envelopes, as long as the amount in the other envelope is always exactly 1/2 A or 2A.

I won't argue math with you, though. But logically, it does not make sense. At least not to me.

[Fluffy]

what Han and bid_them_up are talking about I don't understand

What I understand is that you can pick a random number equally probable to any other between X and Y

But you cannot do that between X and Infinite.

[luke warm]

well i'm not a mathematician but (someone said this above) my thinking is that, given the original problem justin gave, i can either lose 50% of the value i already have or gain 100%... is that correct or not? if so, i'll switch... as for han's question: why not just pick an envelope and immediately 'change your mind', why not indeed?

[bid_em_up]

Fluffy, on Jul 18 2007, 01:35 PM, said:

what Han and bid_them_up are talking about I don't understand

What I understand is that you can pick a random number equally probable to any other between X and Y

But you cannot do that between X and Infinite.

I 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?

It doesnt matter what number you pick is what I am saying.

Given that infinity is infinity, you can let infinity = A.

But....if A = infinity, you can always come up with 2 times infinity (because its infinity, it can always be twice as large as itself, or 1/2 as much as it originally was). You are using infinity as a finite number in this case (even though it really is not).

Logically it makes sense (to me) to let infinity be equal to N+1, in which case the problem still breaks down to:

A = N+1
B = 1/2(N+1) or B = 2(N+1)
2B = 2 1/2 (N+1)
B= 1 1/4 (N+1)
1 1/4(N+1) > N+1

Han is way above my head though in his examples.

[helene_t]

Fluffy, on Jul 18 2007, 08:30 PM, said:

Hannie, on Jul 18 2007, 06:13 PM, said:

Here is a similar problem:

[i]There are two identical boxes, each box contains two identical envelopes. In the first box the envelopes contain $2,500 and $5,000, in the second box the envelopes contain $5,000 and $10,000.

I don't think this is similar problem.

If I run a simulation upon this it would be correct to switch:

2500$ -> switch you will win +2500$ or +7500$
5000$ -> switch you will win +5000$ or lose -2500$
10000$ -> Do not switch. (-5000$ or -7500$)


Wich relates to the initial problem, the higher you have, the less you want to switch.

so what am I missing?

Han's two-box problem is a special case of the original problem. He's postulating the particular probabilities:
(2500;5000): 50%
(5000;10000): 50%

Given those particular probabilites, you can compute the solution. In this case it happens to be: switch if the first envelope contains $5000 or 2500, otherwise don't switch.

You can try with some other probabilites. Suppose, for example, there are are 2000 boxes, 999 of witch contain (2500;5000) and one contains (5000;10000), the remaining 1000 being irelevant now that you know the first envelop contains $5000. In that case you should not switch if the first envelop contains $5000.

Quote

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.

[BebopKid]

I think when looking at when to switch, this question must be asked:

Is reducing my take by 50% acceptable?

When I watch game shows, I see people frustrated at losing who could have left with a win.

The only reason I would want to look at the odds of increasing my take by 100% is if I can accept the result of 50% of my take.

When we're greedy, we tend to look at the odds of increasing our take, first.

I think that's the purpose behind the question.

[bid_em_up]

helene_t, on Jul 18 2007, 01:49 PM, said:

Quote

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'm talking envelopes. There are only two of them.

It doesnt matter if the envelope A has $1, $5, $50, $1,000,00 or $N+1 in it.

The other envelope will contain 1/2 or two times that amount. I don't get the part about some must have have higher probabilities than others.

But thats ok, I don't really want to either.

[goobers]

bid_em_up, on Jul 18 2007, 01:02 PM, said:

helene_t, on Jul 18 2007, 12:53 PM, said:

Han is right. Fluffy's reference to an upper limit is close but not quite. Suppose the probabilites of the first envelop containing x dollars are
$ prob
1 1/2
2 1/4
3 1/8
4 1/16
5 1/32
6 1/64
.. etc

This allows for no upper limit but Han's argument still holds.

The problem with these statements is that the envelopes in question contain a finite (and given amount of $5000 in the first envelope, call it Envelope A). All the other "probabilties" are meaningless.

The only thing that matters in this problem is:

Does Envelope B hold 1/2 as much as Envelope A or does it hold twice as much as Envelope A?

B must equal either 1/2 A or 2A, and A is equal to $5000 in the problem as Justin has proposed it. So there is either $2500 or $10000 in envelope B, no other numbers matter.

Person B is saying his Expected Value is greater if he switches envelopes. I happen to agree. His expected value is the sum total of what he receives by selecting B.

(B = 1/2A) + (B = 2A) =

(2B = 2 1/2) =

(B = 1 1/4A)

His expected value (on average) is 1 1/4 A.

Lets say he does this 4 times, and meets the 50/50 odds.

He draws 2500, 2500, 10000, 10000 = 25000/4 = 6250 which is greater than the 5000 he started with.

If he does it 1000 times, 10000 times or a million times, his average expected value still works out to be 6250, which is greater than his current amount of $5000.

QED.



(Yes, I am sure there are flaws in my math arguments, but the logic is the same).

There's something wrong with this.

You say

(B = 1/2A) + (B = 2A)

but that isn't correct. Your value for A is inconsistent. It should either be 1/2A and A, or A and 2A, but not 1/2A and 2A.

I think.

This problem is giving me a headache.

[Echognome]

I know they formulated a harder version of the problem, but at least to me, it's clear we cannot assume that the envelopes are equally likely as in the set up they haven't defined a clear prior distribution of envelopes. They only state that there are an infinite number of them. Thus they cannot be equally likely or we would not have a probability function.

[helene_t]

Maybe it's easier to understand if we replace the boxes and envelops with something for which we actually have a feeling for what the probabilities are.

Suppose you're told that in heterosexual couples the male is 50% heavier than the female (of course this is not always the case but let's assume it to be the case just for the sake of argument). Now you pick a random person and you can chose either to receive the person's weight (dollars per kilogram) or the person's spouse's weight. By the envelope argument you may reason that you're better of chosing the spouse, which can't be true of course.

The solution is that it depends on the weight of the first person. If the first person weights 100 kilogram you reason that it's probably the male so you'd better stay put. If the first person weights 40 kg you reason that it's probably the female so you'd better switch.

[bid_em_up]

goobers, on Jul 18 2007, 02:00 PM, said:

Your value for A is inconsistent. It should either be 1/2A and A, or A and 2A, but not 1/2A and 2A.

I think.

This problem is giving me a headache.

In the problem as given, you are told one envelope is half the value of the other. You are then allowed to open one of the envelopes that contained $5000. Call this envelope A. But you do not know if it was originally A or originally B.

A is $5000. You are told this.

The amount of unopened envelope B must be either $2500 (1/2 of A) or the amount in the other envelope must be $10000 (making A = 1/2 of B, or 2A).

The problem with your idea is you are assuming you always opened envelope A, when in fact you do not know which one it originally was.

[Fluffy]

helene_t, on Jul 18 2007, 06:49 PM, said:

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.

I tried to set up a simalation for that (picking any number of random numbers and making a full equally probable sett of infinite numbers), and I knew it would be impossible, I didn't try it hard ebcaue I knew the answer before I started actually. It seemed very intuitive to me.

Anyway, my point is that when trying to set up a simulation (I am computer engineer, not mathematician), I will have to analice every case.

I've solved some problems this way, when I try to 'digitalice' the real world, I come down with the answer. A good example are the 'ropes and pieces' puzzles, when 2 things seem to be attached to each other, but actually if you move them the right way you can separate them without breaking anything.

If you try to simulate them you will have to determine wich movements are possible and wich not. But by the time you find wich movements are possible you don't need any simulation to solve the problem.

[sceptic]

easy this one, you are going to walk away with either 2500 or 10000, you will not walk away with 5000 unless that is an amount of money that would make a difference to your life

I would think this is a pshycological issue not a maths one

[han]

bid_em_up, on Jul 18 2007, 01:47 PM, said:

I 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. I don't even know what this infinity you talk about is.

[han]

Echognome, on Jul 18 2007, 02:03 PM, said:

I know they formulated a harder version of the problem, but at least to me, it's clear we cannot assume that the envelopes are equally likely as in the set up they haven't defined a clear prior distribution of envelopes. They only state that there are an infinite number of them. Thus they cannot be equally likely or we would not have a probability function.

I like this explanation.

[Blofeld]

Han is of course correct.

The fallacy is this. Although you have picked between the two envelopes randomly, it is in general false that the other envelope will have more money in 50% of the time.

To see that this cannot be true, imagine you open envelope A and find 1 cent in it. Envelope B can't have half a cent in, so must have 2 cents - and the probability it has more money is 100% rather than 50%.

This is just an illustration of the fact that our intuition in this area is often wrong, and relies on the accidental fact that money isn't infinitely subdivisible. If you allow for arbitrarily small amounts of money, the argument doesn't hold. But the one that Han gave prevents any distribution of possible amounts of money to envelopes that always has a 50% chance of having more money. In fact, what the "paradox" shows is precisely that such a distribution cannot exist!

[han]

Fluffy, on Jul 18 2007, 01:30 PM, said:

Hannie, on Jul 18 2007, 06:13 PM, said:

Here is a similar problem:

There are two identical boxes, each box contains two identical envelopes. In the first box the envelopes contain $2,500 and $5,000, in the second box the envelopes contain $5,000 and $10,000.

I don't think this is similar problem.

If I run a simulation upon this it would be correct to switch:

2500$ -> switch you will win +2500$ or +7500$
5000$ -> switch you will win +5000$ or lose -2500$
10000$ -> Do not switch. (-5000$ or -7500$)


Wich relates to the initial problem, the higher you have, the less you want to switch.

so what am I missing?

It is similar in that you have $5,000 and the other envelope contains either $2,500 or $10,000 and it is equally likely. Here you [i]would want to switch. The original problem is not possible. It starts with an impossible assumption.