4 ms·

I read through the common resolution but I just can't get why you need all this. The problem seems to be at the start: you select one of two amounts, then you s

by dools 13y ago

I read through the common resolution but I just can't get why you need all this. The problem seems to be at the start: you select one of two amounts, then you say the other amount is either double or half what you chose. That means you have 3 amounts in the equation: 0.5x, x and 2x. But in reality there are only 2 amounts: x and 2x. So you have to state the problem like: you choose an envelope. The other envelope either contains x or 2x. It can't contain 0.5x because that amount never existed in the first place.

Osmium 13y ago
Unless you picked 2x to start with, in which case the other envelope does contain half of what you chose... i.e. there are two envelopes, x and 2x, and you choose one. There is a 50% chance that the envelope you didn't choose is half the value of the one you picked (you pick 2x, and the other is x), or there's a 50% chance the envelope you didn't choose is twice the value of the one you picked (you pick x, and the other is 2x). I think it's misleading to introduce this "third" quantity of 0.5x... but perhaps I'm missing something subtle, in which case I'd be happy to be corrected!
- dools 13y ago
  I think it's misleading to introduce this "third" quantity of 0.5x Yes, it is misleading, that's my point :) Put it another way: you have two quantities in two envelopes, A and B. If you choose B, what is the probability that the other envelope contains a 3rd quantity, C? The probably of that is zero, because C was not present in the initial two envelopes. Still another way of putting it: I will place $10 in an envelope, and $20 in another envelope, then ask you to choose one. What are the chances that the other envelope contains $40? The chance of that is zero, because the only two amounts present at the start of the trick were $10 and $20.

4 ms·

I read through the common resolution but I just can't get why you need all this. The problem seems to be at the start: you select one of two amounts, then you s

by dools 13y ago

Osmium 13y ago
Unless you picked 2x to start with, in which case the other envelope does contain half of what you chose... i.e. there are two envelopes, x and 2x, and you choose one. There is a 50% chance that the envelope you didn't choose is half the value of the one you picked (you pick 2x, and the other is x), or there's a 50% chance the envelope you didn't choose is twice the value of the one you picked (you pick x, and the other is 2x). I think it's misleading to introduce this "third" quantity of 0.5x... but perhaps I'm missing something subtle, in which case I'd be happy to be corrected!
- dools 13y ago
  I think it's misleading to introduce this "third" quantity of 0.5x Yes, it is misleading, that's my point :) Put it another way: you have two quantities in two envelopes, A and B. If you choose B, what is the probability that the other envelope contains a 3rd quantity, C? The probably of that is zero, because C was not present in the initial two envelopes. Still another way of putting it: I will place $10 in an envelope, and $20 in another envelope, then ask you to choose one. What are the chances that the other envelope contains $40? The chance of that is zero, because the only two amounts present at the start of the trick were $10 and $20.