Kkwan has literally been repeating the same “your envelope is X and the other is either 2X or 1/2X with 50-50 probability” nonsense for almost a year now, despite having been shown time and time again that mathematics doesn’t work that way when dealing with unknown probability distributions. Isn’t it time for everyone to move on?

Is it not credible to envisage that for any finite positive value of X, there are two possible situations: either (X, 2X) or (X, 1/2X) and the probability of either situation is 1/2?

No! Mingy. The thread must go on. If Kkwan were not caught up in this nightmarish perpetuity, he might be using his superior intellect to take over the world. That must not happen.

What superior intellect?

When I posted this thread, I never expected it to go on “ad infinitum”. The TEP seemed to be an intriguing puzzle which should have some clearcut solution.

However, the TEP appears to have no definite solution because it is so crafted with an inherent ambiguity and symmetry which defies elucidation.

What is clear is that if one switched envelopes, it should be once or any odd number of times and the gamble to switch is worth taking for the potential gain/loss of 100%/50%.

Not to elicit panic, but has anyone noticed that Kkwan has not contributed to this thread in the last few days? ... What is he up to?

He will be all right. He only contributes so now and then, repeating the same stuff over and over again. The best possible case of course would be that he is busy switching envelopes to get rich. Maybe he will smash us all by buying the provider of this forum and close it down…
Or he is sitting at some corner of a street, unwashed, unshaved, left by his wife and family, switching two envelopes again and again, murmuring “it should work, it should work, my argument is airtight, I will be rich, I will be rich…”

Double or half. Switch to get rich! No casino or bank wants to play the TEP with me.

Well if those are the only 2 options, I hope that it is the former, though I don’t wish CFI to be closed down. Maybe if he gets exhorbitantly wealthy, he will keep the forum going, and particularly this thread, so as to continuously throw it in everyone’s face that he was right all along.

GdB and you have fertile imaginations.

With the TEP as it is, it is unjustified to assume only one possible situation when there are two possible situations.

Thus, double or half on switching is a gamble worth taking.

Although it is true that you have the smaller or larger amount with a probability of 50/50, it is false that the amount in your envelope is the smaller or larger amount with 50/50 probability.

It’s a subtle distinction and no surprise the puzzle teases the brain. We can apply it when we get to 8.

You don’t gain on average by switching because although you have the smaller amount half of the time, that amount is, on average, half of the amount you have when you have the larger amount.

If it is true that the probability that you have the smaller or the larger amount is 1/2, it follows that if the amount in the selected envelope (denoted as A), is either the smaller or the larger amount (because you don’t know whether you have the smaller or the larger amount) and that the probability of either is 1/2.

Thus, 2 is true although it is not so explicit.

Consequently 3, 4, 5 and 6 follow with no contradictions.

The reason why 2, 3, 4, 5 and 6 are valid is total ignorance which is inherent in the TEP.

You do gain on average by switching if half the time (A, 2A) are in the two envelopes and half the time (A, 1/2A) are in the two envelopes. This is expressed by the expected value in the other envelope as in 7:

And

8. This is greater than A, so I gain on average by swapping

So, it beneficial to switch once or any odd number of times.

If it is true that the probability that you have the smaller or the larger amount is 1/2, it follows that if the amount in the selected envelope (denoted as A), is either the smaller or the larger amount (because you don’t know whether you have the smaller or the larger amount) and that the probability of either is 1/2.

Thus, 2 is true although it is not so explicit.

Hi Kkwan, are you open to a solution if it’s shown to you?

If 2. were true the rest follows but 2. is false and your reasoning over 2, is fallacious.

As I said: you do not use the knowledge you have, namely that the amounts are fixed. Show me where you use this information in your calculation.

And you did not answer my question: do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

The amounts in the two envelopes are fixed, but you don’t know whether it is (X, 2X) or (X, 1/2X).

For any two amounts, it is simpler to describe them as either (X, 2X) or (X, 1/2X) instead of (X, Y).
So, the potential gain/loss is X /1/2X.

C’mon: only TEP pairs of amounts can be written as (X,2X), or (125A,250A). It is not easier to describe them as such, it is wrong. (2,6) is not a TEP pair, but by switching you gain or loose 4. (4,8) is a TEP pair, and by switching you gain or loose 4. So whatever the pair, so even whatever the TEP pair, your gain and loss are always the same.

And kkwan, you did not answer my points again:

1. Show me where you use the fact that after the envelopes are closed, the amounts are fixed. You only say the amounts are fixed, but you do not use it in your calculation. That means you use an incomplete description of the TEP situation.
2. Do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

Your avoiding to react on these point shows that you see where the problem lies, but just do not admit it. Otherwise you would have a meaningful reaction on them.

Double or half. Switch to get rich! No casino or bank wants to play the TEP with me.

I will. No problem.

But you only get your gain after 100 TEP plays where you won 1/4 more than the sum of your first picks. Otherwise you pay me, let’s say, 3/2 of the sum of your first picks.
And if you do not dare to do that: why do you know you will fail?

The reason why 2, 3, 4, 5 and 6 are valid is total ignorance which is inherent in the TEP.

Total ignorance?

Following I know:
1. The amounts are done in the envelopes, the envelopes are closed, so the amounts are fixed
2. One amount is twice the other.
3. I pick one of these envelopes, and no other.

You do not use 1 and 3. You make yourself more ignorant then you really are.

2. Do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

Your avoiding to react on these point shows that you see where the problem lies, but just do not admit it. Otherwise you would have a meaningful reaction on them.

Kkwan does have a meaningful reaction, he says we should consider two of the possible pairs.

Your solution amounts to “no we shouldn’t”

But of course what Kkwan is looking for is why we shouldn’t.

Kkwan does have a meaningful reaction, he says we should consider two of the possible pairs.

No, he hasn’t. In the first place there are infinite possible pairs. But of all the possible pairs you have only one, not two. And that is something you know, and should include in your calculation. kkwan doesn’t. To say you have a certain amount, and the other amount is half or twice that amount is not a complete description of TEP. It leaves out that you started with one pair, and that when you picked one amount there is only one other possible amount, and not two.

So you think you can deal with the problem by just leaving out some information you do have? It definitely belongs to the solution, as done here. How would you put in the fact that you started with just one and fixed TEP pair?

You see that the outcomes differ, depending on taking this into account or not.

Hi Kkwan, are you open to a solution if it’s shown to you?

If 2. were true the rest follows but 2. is false and your reasoning over 2, is fallacious.

I think I’ve pinpointed the problem.

Of course. However, any proposed solution must explicitly show me why step 2 is false.

Let’s consider what is the gist of step 2:

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

What step 2 states is that, A denoted as the amount in the selected envelope in step 1, could be either the smaller or the larger amount and the probability of either situation is 1/2.

The total probability of what A is, is 1.

This approach is reasonable with total ignorance of the amounts in the two envelopes.

What you say is:

it is false that the amount in your envelope is the smaller or larger amount with 50/50 probability.

It is false if and only if you know the relative value of A which you don’t (with total ignorance) in the context of the TEP.

C’mon: only TEP pairs of amounts can be written as (X,2X), or (125A,250A). It is not easier to describe them as such, it is wrong. (2,6) is not a TEP pair, but by switching you gain or loose 4. (4,8) is a TEP pair, and by switching you gain or loose 4. So whatever the pair, so even whatever the TEP pair, your gain and loss are always the same.

Either (X, 2X) or (X, 1/2X) are valid pairs as per the TEP. They are distinct pairs for any finite value of X.

1. Show me where you use the fact that after the envelopes are closed, the amounts are fixed. You only say the amounts are fixed, but you do not use it in your calculation. That means you use an incomplete description of the TEP situation.
2. Do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

Your avoiding to react on these point shows that you see where the problem lies, but just do not admit it. Otherwise you would have a meaningful reaction on them.

The amounts are fixed but you don’t know whether they are either (X, 2X) or (X, 1/2X). It is superfluous and misleading to introduce Y when X is sufficient in the context of the TEP. Occam’s razor.

OTOH, with X and Y, unless you know the relative values of X and Y which you don’t, X-Y or Y-X makes no sense.

However, If Y=2X, then (X, Y) is equivalent to (X, 2X) and if Y=1/2X, then (X, Y) is equivalent to (X, 1/2X).