Wait a minute, I thought the person in question is swapping the envelope they have (with A amount in it) for another one…..not adding to something they already have.
Isn’t that the situation?
It is.
But in GdB’s equation A does not refer to the amount in the envelope, necessarily, it instead refers to the smallest amount.
That is kkwan’s criticism.
That’s the puzzle, why make A the smallest amount, rather than the amount in your envelope?
If A is the amount in your envelope then you do have a 50/50 chance of double or half and so should switch.
By asking If I were to pick A would that be the best selection. The answer is no because there is double or half A in the other envelope with a 50/50 probability, so B is best.
It works just as well from the start.
The same argument will apply vice versa, so A is best.
Right.
We again have an absurd situation, in this case A is better and B is better.
I think if we answer what’s wrong with that, we’ll be able to apply the same reasoning once we’ve selected an envelope.
You are offered double or half depending on the toss of a coin.
So if you get double, the total of the two amounts, what you have + what you will have is 3A
If you get half the total of the two amounts, what you have and what you will have is 1.5A .
Isn’t this magic?
No it isn’t, and you know that. You changed the problem (as I did in one of my previous postings). In kkwan’s formulation, the contents of the envelopes is fixed, the amounts are already in the envelopes. One contains A and the other 2A. In your reformulation it is not fixed, and so can be changed. Your formulation and kkwan’s are not equivalent.
You changed the problem (as I did in one of my previous postings).
Yes.
In kkwan’s formulation, the contents of the envelopes is fixed, the amounts are already in the envelopes. One contains A and the other 2A. In your reformulation it is not fixed, and so can be changed. Your formulation and kkwan’s are not equivalent.
I think you might be on to something GdB but it isn’t clear.
Let’s change my example a bit.
I have £10
I’m offered double or half depending on whether a coin, hidden in a box, that has already been tossed, is a head or a tail.
It occurred to me that philosophically, both envelopes must be given equal weight regardless of the amount they contain. It IS a Schrodinger problem !
Both envelopes will be given equal weights if they have the same amounts in which case the question and the advantage of switching or not switching does not arise at all, i.e. there is no problem.
It is not a Schrodinger problem, though it is apparently intractable.
Assume one envelope has A dollars in it, the other 2A.
But if A is an unknown quantity, how can you assume one envelope has A dollars, the other 2A?
By doing so, you have considered A as a known quantity.
Ok, you’ve got to be kidding with that.
The entire REASON that labels like ‘A’ or ‘x’ or ‘y’ are used to begin with is to represent unknowns! Otherwise, they’d use a specific number.
And how can we assume that one envelope has A dollars and the other 2A? (in other words, one has twice the amount as the other) because YOU stated that very thing! GdB even bolded the part where you did.
*mind boggled*
kkwan, you seriously can’t be as thick as the above comment of yours makes you appear…....
In algebra, unknowns are generally denoted by by A’s, x’s or y’s, but having denoted A as the unknown amount in the selected envelope one cannot assume that other envelope has 2A because we do not know whether A is the larger or smaller amount.
It is equally possible that the other envelope has A/2, if A is the larger amount.
Both possibilities satisfy the known information that “the amount in one envelope is twice that in the other”.
Hence, by assuming “one envelope has A dollars in it, the other 2A” is to consider only one possibility but not the other. This is tantamount to having knowledge that A is the smaller amount.
The problem with the entire puzzle is that you cannot lose. You either win double or half of what’s in A, but you win in either case, which gives you liberty to switch.
An honest choice would be A has a certain amount, and B has either double or nothing. Now the problem becomes honest. You will have a certain amount in hand, but you can double it or lose it and end up with zero. Will you choose to gamble (go for broke) or keep what is in A?
Now what are the odds?
But in GdB’s equation A does not refer to the amount in the envelope, necessarily, it instead refers to the smallest amount.
That is kkwan’s criticism.
That’s the puzzle, why make A the smallest amount, rather than the amount in your envelope?
If A is the amount in your envelope then you do have a 50/50 chance of double or half and so should switch.
GdB assumes A is the smaller amount which is not justified because we have no information whether A is the smaller or the larger amount.
OTOH, there is no such assumption in the argument for switching, where A is considered as the amount in the selected envelope with two situations of equal probabilities:
1. If A is the smaller amount, the other envelope contains 2A
2. If A is the larger amount, the other envelope contains A/2
Both situations satisfy the only known information that “one envelope contains twice as much as the other”.
