If my £10 envelope is Envelope X then switching to Envelope 2X gains me £10 and if my £10 envelope is Envelope 2X then switching to Envelope X loses me £5. — Michael
You are playing a game for money. There are two envelopes on a table.
You know that one contains $U and the other $V, [but you do not
know which envelope is which or what the numbers U or V are]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?
You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other $2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?
You actually have 2 more complicated cases. The first case is when you have X and the amount in the other envelope is X/2, the second case is when you have X and the amount in the other envelope is 2X.
Consider the first case. The envelopes now contain X and X/2. If you have X and switch you gain -X/2, if you have X/2 and switch you gain X/2, which are equally probable, so they cancel in the expectation giving 0.
The second case. The envelopes now contain X and 2X. If you have X and switch you gain X, if you have 2X and switch you lose X, these are equally probable, so they cancel out in the expectation giving 0. — fdrake
You can interpret this article however you wish. — Jeremiah
Suppose we know only that the two amounts in the envelope are different (and no longer that the larger amount is double the smaller one). No information is provided of how the different amounts were chosen. We are allowed to open one randomly selected envelope. Does there exist a strategy that will lead us to accept the larger amount with a probability greater than ?
The answer is yes. This may come as a surprise to an unprepared reader, but the strategy is very simple.
Let and denote the different amounts in the first and second envelopes, respectively, and let be a fixed positive number. We consider the following strategy:
Choose , if and only if ; choose otherwise.
...
If the smaller amount equals , say, and the larger , then a randomized choice yields an expected value of . Strategy (6), however, yields an expected amount of .
Having an envelope means being in case A or being in case B. Opening the envelope does nothing to inform you of whether you are in case A or case B. Let's assume case A and case B are equally likely. That is, it's equally like that A (your envelope contains X and the other contains 2X) or B (your envelope contains X and the other contains X/2).
If you are in case A, having an envelope means that the envelope you hold is either X or 2X.
If you are in case B, having an envelope means that the envelope you hold is either X or X/2.
Assume you're in case A. If you have the lower valued envelope X and switch, you gain X. If you have the higher valued envelope 2X and switch you lose X. It is equally likely that you have X or 2X, so the expected gain conditional on being in case A is:
0.5(X-X)=0
Assume you're in case B. If you have the lower valued envelope X/2 and switch, you gain X/2. If you have the higher valued envelope X and switch, you lose X/2. So the expected gain conditional on being in case A is:
0.5( 0.5X - 0.5X) = 0 — fdrake
Where did I switch between stuff? — fdrake
You always observe X in your envelope. What was the probability of obtaining U=X? Not the conditional probability of getting X given that you opened the envelope containing it. — fdrake
Imagine that we have two envelopes, one filled with X and one with 2X.
If I have an envelope containing X, and switch to an envelope containing 2X, I gain X from the switch.
If I have an envelope containing 2X, and switch to an envelope containing X, I lose X from the switch.
Does that make sense to you? — fdrake
Would you agree that {5,10,20} are the possible values for things in the envelopes? — fdrake
If 5 and 20 are permissible values for the envelopes, does it stand to reason that I could receive an envelope which is 4 times another envelope? — fdrake
Then what's the probability of receiving an envelope containing 10? — fdrake
In your set up. There are three possible values for envelopes, {5,10,20}. Imagine that all envelopes are still closed. And you are given an envelope. What's the probability of having 10? — fdrake
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