[PairTheBoard]

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But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either.

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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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I thought my idea sounded too good to be true. But I guess I'm not understanding how this is working. You said,

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For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.



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I understand you get screwed with small n. But in my idea you'd be choosing n large. I understand you can't insure it's the optimum n. But if you choose n larger than 1.64A^2 and A=y then you've chosen n greater than 1.64y^2 and you improve over Never-Calling-Off. And if A=2y, then you've still chosen n greater than 1.64y^2 and you still improve over Never-Calling-Off. In either case you've improved over Never-Calling-Off. What am I missing here?

PairTheBoard