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).
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.
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.
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?