[PairTheBoard]

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PTB -
For #2. "If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?"

If "the amount" is fixed the answer is yes. However, if you have a 50% chance of winning but "the amount" is $1 when you win and "the amount" is $2 when you lose then your EV is 0.

How's that? Does that agree with common sense?


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Aaron -
This is a very good statement of the frequentist position. You're not allowed to compute the expected value and have a variable in the result. So you can't say having 0.5 chance of getting $X/2 and 0.5 chance of getting 2*$X has an expected value of 1.25*$X.


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You asked for a common sense answer to your question. I think my answer made very good common sense. I also think it makes very good common sense when doing a calculation involving a term $X that $X should stand for the same thing throughout the calculation, whether you call it a "variable" or anything else.

With envelopes containing N and 2N, "Switching" amounts to betting half your envelope amount at 2-1 odds that it is the smaller amount. You say this is a true paradox and both arguments are equally strong. Do you thing the strength of argument for the "Switchers" would help them win money taking Part 2 in the following Proposition Bet?

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PTB -
Look at this from a Gambler's point of view. You are told about the two envelopes and asked to choose one. You are offered these Proposition Bets.

1. Given 3-2 odds, would you be willing to bet $10 that your envelope contains the smaller amount?

2. Given 3-2 odds, would you be willing to bet the amount in your Envelope that it is the smaller amount?


1 is a good bet for you while 2 is not. Do you see why?


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PairTheBoard