i'm no mathemetician, but you lost me at $12.50.....

i read the formula, but how would that value even be considered (by the contestant) in the context of the choices/payoff???

 

The value of $12.50 per se probably wouldn't be considered. It's just a handy way to summarize the fact that you stand to gain more by switching.

 

actually, i would have thought that the "handy" number would be $10.

$20 possible vs. $10 in hand (discarding the negative event of possibly ending up with only $5).

but again, i'm no mathemetician.....you guys are too smart for me....i'll stick to being an accountant

 

[QUOTE]Originally posted by magician

 

magician-
In re-reading some of your posts, it seems you are suggesting that the risk of losing half your winnings is out-weighed by the possibility of doubling your money. Without even getting into that, the fact is the envelpe you pick the first time dictates the envelope you pick the second time, doesn't it?

To discuss the merits of letting greed overcome common sense, consider opening the envelope and finding $2 million. You stand to lose $1 million, and possibly gain $2 million. It might make sense to switch envelopes, but keep in mind there are only two. You might very well have selected the max amount on the first try. Are you really willing to throw away a million dollars on the off chance the people running the contest are willing to give away $4 million? The higher the value in the first envelope, the less it makes sense to swap, if that is your strategy.

 

[QUOTE]Originally posted by no really
1.

 

[QUOTE]Originally posted by no really
magician-
In re-reading some of your posts, it seems you are suggesting that the risk of losing half your winnings is out-weighed by the possibility of doubling your money.

 

[QUOTE]
Originally posted by magician
I never said that your comment about having a 50% chance of getting more money assumed anything other than the fact that there are two envelopes.

 

First, there is not one (possible) payoff; there are two: N and 2N using your notation.

The flaw in your analysis is that you're on the outside knowing whether the player has chosen the more-valuable envelope or the less-valuable envelope, a certainty the player does not enjoy. This is why you can refer to the contents of one envelope as N and to the contents of the other as 2N. Clearly in that case the expected value is 3N/2.

The player is operating under an additional uncertainty: he does not know whether his envelope is the N envelope or the 2N envelope. He has to account for both possibilities. Because the envelopes are identical, we assume (as he does) that he originally had a 50% chance of picking the N envelope and a 50% chance of picking the 2N envelope--this is the Bayes' Theorem result. If his envelope, containing X, is equally likely to contain N or 2N, then the other envelope contains either 2X or X/2 with equal probability. My expected value calculation (5X/4) is correct given the player's uncertainty.

There is absolutely nothing wrong with your analysis from the point of view of an omnicient outsider. From the players point of view it is incorrect. That's a direct result of his uncertainty about what the other envelope contains vis-a-vis his own.

Please look at the problem from inside the box for a moment; you've gotten the analysis from outside the box correct.

 

[QUOTE]Originally posted by magician
First, there is not one (possible) payoff; there are two: N and 2N using your notation.

The flaw in your analysis is that you're on the outside knowing whether the player has chosen the more-valuable envelope or the less-valuable envelope, a certainty the player does not enjoy.