The x in that case wasn't referring to the smaller value but to the value of the other envelope. I'll rephrase it to make it clearer: Let x be the amo...
Because Sleeping Beauty is certain to be questioned during the experiment. Compare with my alternative scenario here where the probability of being qu...
The argument is: Let x be the amount in one envelope and 2x be the amount in the other envelope. Let y be the amount in the chosen envelope and z be t...
Not really, as we can only consider this from the perspective of the participant, who only knows that one envelope contains twice as much as the other...
I'm trying to understand your use of conditioning, so can we start at the very beginning? Let x be the amount in one envelope and 2x be the amount in ...
I think I showed in the OP why this isn't the case. The EV calculation commits a fallacy, using the same variable to represent more than one value. Th...
If you're in subcase 1 and you have £10 and you switch then you lose £5, if you're in subcase 2 and you have £10 and you switch then you gain £10. Eac...
I don't know how you get 0. If you're in subcase 1 then you lose £5 by switching. If you're in subcase 2 then you gain £10 by switching. There is a 50...
I flip a coin but don't look at the result. The probability that it landed heads is 1\over2. You both seem to be mixing up the participant's subjectiv...
Yes, and I think it is perfectly correct to say, in case A, that the probability that the other envelope contains £20 is 50%. As I mentioned before, k...
They are aware. They are told before the experiment starts that one envelope contains twice as much as the other. They open their envelope to find £10...
I don't understand what you're saying or how this is any different to the envelopes. I put a coloured ball in one envelope and a coloured ball in anot...
If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you...
I don't see how your cases solve the problem. If there is a 50% chance that I am in (5, 10) and a 50% chance that I am in (10, 20), and if I have £10,...
The framing is the paradox. I pick one of two envelopes at random. One is twice the value of the other. Given that the probability that I picked the m...
I'm not sure what you mean. Perhaps you could answer the questions I posed earlier? If I flip a coin and don't look at the results then what is the pr...
I've mentioned this before, but from the Wikipedia article: It doesn't matter if you can frame the situation in such a way that there is no rational r...
I think learning the value of your envelope is an uninformative posterior and so gives you no information with which to reassess the prior probability...
Then the paradox arises. The probability that the other side is half the value is 1/2 and the probability that the other side is twice the value is 1/...
Before I flip a fair coin, what is the probability that it will land on heads? I say 1/2. After flipping the coin, but before looking at the result, w...
I honestly don't understand your interpretation of probability. This seems very straightforward. Maybe a different example. I have a red ball hidden i...
That’s not what happens in this example. I am shown two envelopes, one containing £10 and one containing £20, and I freely choose one at random. I don...
I agree that there is an error with the calculation of the expected value. That's what I explain in the OP. My argument with you is over the assigned ...
An interesting variation taken from here: To be clearer with what each volunteer is considering, it is: P(1 and Heads or 2 and Tails or 3 and Heads or...
No I didn't. I showed that the argument which purports to show that switching is rational commits a mathematical fallacy, and that there is no rationa...
There is no opened envelope: All I know is that one envelope contains twice as much as the other and that I picked one at random. I don't know what's ...
I tell you that one envelope contains £20 and the other envelope contains £10. A true random number generator was used to determine which of envelopes...
It just assumes that: P(A = the smaller envelope) = P(B = the smaller envelope) = 1/2 Which is correct. If either A is smaller than B or B is smaller ...
Does it not stand to reason that the probability that the coin landed heads in this example is less than the probability that the coin landed heads in...
I don't agree with this at all. I have two envelopes. I have put £5 in one envelope and £10 in the other envelope. You use a true random number genera...
So by your logic we can't even talk about the probability of a coin toss landing heads being 1\over2? I don't think that's at all reasonable, or even ...
Why not? I know that one envelope contains twice as much as the other. I pick one at random. What is the probability that I picked the smaller envelop...
The same supposed paradox occurs even if we know the possible values. Assume that one envelope contains £10 and the other envelope contains £20. Let y...
It's explained in the OP: Let x be the amount in one envelope and 2x be the amount in the other envelope. Let y be the amount in my envelope and z be ...
I stand by the claim that the probability that the other envelope contains twice as much as my envelope is equal to the probability that the other env...
Which is 1. I know that I will be interviewed if the coin lands heads. Consider a simpler version of the experiment. If heads then I will be interview...
Yes, that's the left hand side of the theorem that we're trying to solve: P(Heads|Questioned). We use the known values on the right hand side to deter...
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