Heihojin,
The value of Pseudo's Y solution is easy to explain. Assume that the stuffer has a probability distribution for stuffing envelopes. For example, suppose his distribution is
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1) 25% Low envelope = $5 Hi envelope = $20
2) 50% Low envelope = $1 Hi envelope = $6
3) 25% Low envelope = $10 Hi envelope = $11.
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If the guesser knew this probability distribution, then the optimal strategy is easy to figure out. If he knew the distribution, the guesser would keep the first envelope if it contained 6, 11, or 20 dollars and switch otherwise. But let's suppose that the guesser is Pseudo and he does not know the stuffer's distribution. He now picks a satisfactory $ amount Y. If the first envelope contains Y or more, he keeps it and if it contains less, he switches.
Suppose Pseudo picks Y=$1000. Then he will always switch and that is no better than random. His expectation would be (12.50 *.25 + 3.50 * .50 + .25 * 10.50) = $7.50.
Suppose he picks Y=$7. Then he will always switch if the first envelope had $1, $5, or $6. In case 1) he always wins. In the other two cases, the Y strategy has the same expectation as random switching. His expectation would be (20 *.25 + 3.50 * .50 + .25 * 10.50) = $9.38. Psuedo will do better than random whenever there is a chance that low < Y < hi. In this example, there is a 25% chance that low < Y < hi.
As far as I can tell there is no "optimal strategy" for the guesser if he does not know the probability distribution of the stuffer. But, there are strategies that always do better than random. In my previous post, I proposed switching with probability exp(-d) where d is the dollar amount in the first envelope you open. The expectation for that strategy with the example distribution is
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.25*.5* exp(-5 )* 20 + .25*.5* (1- exp(-5 ))* 5 +
.25*.5* exp(-20)* 5 + .25*.5* (1- exp(-20))* 20 +
.50*.5* exp(-1 )* 6 + .50*.5* (1- exp(-1 ))* 1 +
.50*.5* exp(-6 )* 1 + .50*.5* (1- exp(-6 ))* 6 +
.25*.5* exp(-10)* 11 + .25*.5* (1- exp(-10 ))*10 +
.25*.5* exp(-11)* 10 + .25*.5* (1- exp(-11))* 11
= $7.97
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I claim that the exp(-d) strategy always does better than the three strategies: always switch, always keep the first envelope, and random.
When you have a infinite discrete stuffer probability distribution, the math gets a bit harder. Suppose the stuffer will put L(i) in the lower envelope and H(i) in the higher envelope with probability P(i) where i = 1,2,3, .... Then the expectation is
expectation = Sum[ P(i) * E(i) , {i, 1, Infinity}]
where
E(i) = (L(i) + H(i))/2 for the random strategy, always switch, or always keep the first,
E(i) = H(i) if L(i)<Y<= H(i) and (L(i) + H(i))/2 otherwise for the Y strategy, and
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E(i) = 1/2 exp(-L(i))*H(i) + 1/2 (1-exp(-L(i)))* L(i) +
1/2 exp(-H(i))*L(i) + 1/2 (1-exp(-H(i))) *H(i)
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for the exp strategy.
The Y strategy is simple and always does as well or better than random. The exp strategy always does better than random. I expect there are strategies that always do worse than random also.