Note, yet again, that all the values that could be in the cases are known from the start. There is no speculation about possible outcomes — Srap Tasmaner
What’s the expected value of his box? — Michael
average total value — Srap Tasmaner
The resolution of the apparent paradox is that the probabilities are not 50:50 for most values of Y.there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5. — Michael
"if Y = X then Z is 2X or X/2" only adds up to 3X in one instance, the rest results in false conclusions as it contradicts the premise that the total should always be 3X. — Benkei
In the above you have an inherent contradiction in your conditionals as X and 2X are both 10. — Benkei
1. If R = heads then ...
2. If R = tails then ...
1 and 2 have contradictory antecedents. But I'm not saying that both the antecedent of 1 and the antecedent of 2 are true. One is true and one is false, with a 50% chance of each being true. And the same with my example with the envelopes:
1. If X = 10 then ...
2. If 2X = 10 then ... — Michael
1. If X = 10 then ...
2. If X = 5 then ... — Michael
If X = 10 then the other is 20
If X = 5 then the other is 10
In both cases you have now assumed you're opening the smaller envelope and the other HAS to be the bigger envelope. — Benkei
Why do you need to include Y, Michael? — Jeremiah
It does not provide updated information for the original uncertainty of X or 2X
Now to revisit the idea of knowing the amount in the envelope: I think using amounts like £5/10/20 is misleading because £5 intuitively feels like a throwaway amount that anyone would be happy to lose. Instead, what if your chosen envelope contained a cheque for £10 million? Would you throw away £5m chasing an additional £10m that may not even exist?
And here it gets interesting for me because... given a £10 envelope, I really would switch because a £5 loss is nothing. Given a £10m envelope, I'd stay.
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