The Two-Envelopes game Here are some "solutions You picked an envelope at random, so you were just as likely to pick the one with more dollars as the one with fewer. Therefore if you see $8 then the other envelope is equally likely to contain either $4 or $ 16. On average, the other envelope contains(4+16)/2= 10 dollars, which is more than the $8 you see. So you should clearly switch to the unopened envelope However, a similar argument applies no matter what amount you see initially So you should always switch, regardless of the amount in the first envelope Thus, the best strategy is to pick one envelope and then, without even bothering to look inside, take the amount of money in the other. But that's absurd You were just as likely to pick the envelope with more dollars as with fewer So, on average, the amount of money in the envelope you picked is the same as the amount in the unopened envelope. So staying or switching makes no difference But what if you saw $1? In that case, you could be certain that the other enve- lope contained $2. So your strategy would make a difference Look at the problem from my perspective. Clearly, I should not put $1 in either envelope; that would give you a big advantage. If you opened the envelope with $l, then you would know to switch. But then if you see $2 in an envelope, you know that the other envelope must contain $4 since we just ruled out $1. So you also have a big advantage in this situation. My only choice is to never put $2 in an envelope either. And I cant use $3; if you saw that, you'd know the other envelope held $6. And if you see $4, you know the other envelope must contain $8. Apparently, I can t run the game at all! Were revisit this problem later in the term when we study probability2 Proofs The Two­Envelopes Game Here are some “solutions”: • You picked an envelope at random, so you were just as likely to pick the one with more dollars as the one with fewer. Therefore, if you see $8, then the other envelope is equally likely to contain either $4 or $16. On average, the other envelope contains (4 + 16)/2 = 10 dollars, which is more than the $8 you see. So you should clearly switch to the unopened envelope. However, a similar argument applies no matter what amount you see initially. So you should always switch, regardless of the amount in the first envelope. Thus, the best strategy is to pick one envelope and then, without even bothering to look inside, take the amount of money in the other. But that’s absurd! • You were just as likely to pick the envelope with more dollars as with fewer. So, on average, the amount of money in the envelope you picked is the same as the amount in the unopened envelope. So staying or switching makes no difference. But what if you saw $1? In that case, you could be certain that the other enve￾lope contained $2. So your strategy would make a difference! • Look at the problem from my perspective. Clearly, I should not put $1 in either envelope; that would give you a big advantage. If you opened the envelope with $1, then you would know to switch. But then if you see $2 in an envelope, you know that the other envelope must contain $4 since we just ruled out $1. So you also have a big advantage in this situation. My only choice is to never put $2 in an envelope either. And I can’t use $3; if you saw that, you’d know the other envelope held $6. And if you see $4, you know the other envelope must contain $8. Apparently, I can’t run the game at all! We’re revisit this problem later in the term when we study probability