[PairTheBoard]

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I think what you may be missing here is that my EV calculations are based on the fact that A and n are independent.

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ok. I think I just fell into another version of the Two Envelope Paradox flawed thinking. n is not a random variable. But if I choose n based on A I turn it into a random variable. In fact, I do so in a way that actually worsen my results, as you point out here:

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Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.


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Another way to look at it is in an example. Say y=10. Say I look in the envelope and see 10. Say I then choose the random decision based on n=200. Will my results be better than Never-Calling-Off? Well, no. If I repeat the experiment for this fixed y and look at all the times I see 10 in the envelope, I will be calling off the bet some of the time which is poor because the Never-Calling-Off strategy wins every time I see 10 in the envelope. This even though the n=200 Decision strategy works better than the Never-Calling-Off strategy for y=10. That's because when you repeat the experiment for that EV you get to see A=20 half the time.

Introducing the function f is a nifty way to express the decision and generalize.

It looks like

2f(y) - f(2y) >= 1 for all y

would imply

f(2^n *y) <= 2^n(f(y)-1) +1

with right hand side going negative for large n, unless
f(y)=1

I'm going to have to think about all this a little bit. I think it has shed a little more light on the Two Envelope Paradox.

PairTheBoard