I'd switch conditionally as explained here to reap the .25 gain. — Michael
Yeah, the effectiveness of the conditional switching strategy is a red herring that is irrelevant to the puzzle, so I didn't want to touch it. The thing to address now is that this strategy is not even actually what's recommended by Michael's position. If he is right, you ought to switch no matter what. — Snakes Alive
But it's also clear to me there's a conceptual muddle in the argument for switching in the non-iterative case, so I can't get past that. — Srap Tasmaner
Stop repeating this, we get what the argument is, you've said it 5,000 times. — Snakes Alive
Jeremiah
1k
We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.
My claim is then that the only possible outcomes for B is X or 2X.
Proof:
For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.
You are handed A and you see Y inside.
There are two cases here:
Case One
A=Y=X
By definition of A and B if A=Y=X then B=2X therefore B=2X
Case Two
A=Y=2X
By definition of A and B if A=Y=2X then B=X therefore B=X.
Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X]
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Notice how that is for all of Y. A very important concept. Also notice how it says "where X is some positive real number." It absolutely does not matter how X was selected. — Jeremiah
You are handed A and you see Y inside.
There are two cases here:
Case One
A=Y=X
By definition of A and B if A=Y=X then B=2X therefore B=2X
Case Two
A=Y=2X
By definition of A and B if A=Y=2X then B=X therefore B=X.
Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X] — Jeremiah
We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.
My claim is then that the only possible outcomes for B is X or 2X.
Proof:
For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.
You are handed A and you see Y inside.
There are two cases here:
Case One
A=Y=X
By definition of A and B if A=Y=X then B=2X therefore B=2X
Case Two
A=Y=2X
By definition of A and B if A=Y=2X then B=X therefore B=X.
Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X] — Jeremiah
The sample space is [[10,20],[5,10]]. Notice how there are two 10s. Now show me the math which allows you to eliminate both of them. — Jeremiah
you're defining the envelopes according to "contains 2X [10,20]", "contains X [5,10]". — Dawnstorm
So we have two cases here and have no clue which one we are in. Earlier I defined these cases as amount X case R, and and amount 2X case S. We have a lot of variables flying around so let's try to be consistent here.
Now that we have listed all possible outcomes we can define our sample space as [R,S], well call this sample space 2.
Now remember by our definitions an event is a subset of our sample space.
In event R X=10 and since in event R B must be 2X then B = 20.
So the sample space of event R is [10,20].
In event S 10=2X and since in event S B must be X then B= 5.
So the sample space of event S is [5,10]. (order does not matter)
So our sample space, which we named as sample space 2, is [R,S] where R is the set [10,20] and S is the set [5,10] or we can express it as [[10,20],[5,10]] — Jeremiah
Sorry no, that was not my intent. In the event of R you have A=10 and B=20. In the event of S you have A=10 and B=5. These are mutually exclusive events, which means in the case of R the amount 5 does not exist at all, and in the event of S the amount 20 does not exist at all. So one of those sample spaces is feeding you false information. The only way to avoid this is to treat X as the unknown variable it is. — Jeremiah
I suppose I could also have thrown in this:
— Srap Tasmaner
But does the reasoning make sense? — Dawnstorm
You have two different values of X and so this is misleading.
Where you have P(U=2X∣Y=10)(2X), the actual value of X is 10, given that you've defined the smaller envelope as having £10.
Where you have P(U=X∣Y=10)(X), the actual value of X is 5, given that you've defined the larger envelope as having £10.
So it works out as:
— Michael
But let's say I'll put in £10 if it's heads and £100 if it's tails. — Michael
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