| [ QuizWit ] in KIDS 글 쓴 이(By): pomp (위풍당당) 날 짜 (Date): 1998년 12월 29일 화요일 오후 05시 05분 07초 제 목(Title): Two-envelope paradox resolved! 이거 뭔 소린지 해석/해설/의견 부탁합니다. Two-envelope paradox resolved! Author: Joshua Juran Email: i-dont-like-spam@bloody.vikings Date: 1998/12/27 Forums: rec.puzzles ------------------------------------------------------------------------- In his book _Satan,_Cantor,_and_Infinity_, Raymond Smullyan presents "The Envelope Paradox" (p. 189). The paradox involves a simple game: There are two envelopes, one containing twice as much money as the other. You get to choose an envelope, open it, and then optionally trade it for the other. How often should you trade? Smullyan actually presents two paradoxes based on this game. The first involves probability: Suppose the envelope you pick contains $100. There is an equal probability that the other envelope contains $50 or $200, so by trading you will lose $50 or gain $100. Clearly, the odds favor trading. But, you would have applied the same logic had you instead picked the other envelope first! Picking the other envelope *before* you open one of them doesn't change the probability of winning -- why should trading *after* you've opened one make a difference? Discussion: There is a flaw in this logic. Let us suppose that n (the amount of money in the envelope you open) is $100. The other envelope contains either kn or n/k, where k is a coefficient -- previously, 2. But in this version of the game, it's 10,000. So, the other envelope either contains $1,000,000, or one cent. (If $100 is not a big deal to you, multiply n and k by the same factor until n is big enough.) Are you willing to risk throwing away your prize for a penny? Do you truly believe that there is a 50% chance you'll win a million? Let's take another look: I'm running this contest out of my own bank account, and you know I don't have a million dollars. So if your envelope contains $100, you know the other one has a penny. The chance that it contains $1,000,000 is zero. Does this contradict the description of the game? No. You have an equal probability of picking the bigger or smaller prize. But this does not mean that having picked one, the probability that the other will be bigger is 50%. If you picked the bigger one, then that probability is zero. If you picked the smaller, then it's 100%. You just don't which until you open the second envelope. At issue is that the 50% chance is a 'blind chance' -- probability based on no information. When you obtain partial information, that skews the probability. For example, the "Monty Hall" scenario: If you pick door 1 and Monty exposes door 2 as a goat, there is now a 2/3 chance that door 3 has the car, so you should trade. In my example, there are only two doors. One door has one more car than the other. You picked door 1, and Monty shows you the car behind it -- and offers to let you trade to door 2. Are there two cars behind door 2, or just another goat? The Paradox Reduced ------------------- Says Smullyan's Sorcerer: "I have given this problem to several experts on probability theory; some were as puzzled as I, and others gave me an explanation in terms of there being no such thing as a probability measure on the infinite set of positive integers. But I suspect that probability is not the heart of the matter, and I have thought of a new version of the paradox which doesn't involve probability at all." (Smullyan, p. 190) He presents the following propositions: Proposition 1: The amount that you will gain, if you do gain, is greater than the amount that you will lose, if you do lose. Proposition 2: The amounts are the same. "Obviously, the two propositions cannot both be true..." (190). We shall see. Smullyan's plan is to prove that at the same time they are contradictory, both in fact are true, and a diabolical plan it is, indeed! Smullyan's logic follows: Let n be the amount in the envelope you pick. The other envelope therefore contains 2n or n/2. So, either you will gain n, or lose n/2, proving Proposition 1. Now, let d be the difference between the amounts in the envelopes (which is also the smaller amount). Clearly, whether you gain or lose, the gain or loss is d, the difference between the amounts, proving Proposition 2. Discussion: Any paradox results from an unwarranted assumption which pollutes the problem space, resulting in a contradiction. The first problem is that the game description is incomplete. What is missing is the cost of admission. How can you lose at all when you didn't pay anything and you're guaranteed to get a positive amount? In his proofs, Smullyan bases gain and loss on the amount in the originally chosen envelope, so let the price of admission be that amount. So, here's how this works: You pick an envelope and open it, and you pay the amount in the envelope to play the game. Of course, you can keep the envelope, incurring no gain or loss -- a wash. If you trade, you return that envelope but you still pay that much to play. Then you receive back either double or half that amount, so your net gain equals the price or your loss equals half of it. Proposition 1 stands. But once again, when you subtract the price of admission from the amount in the other envelope, it's plus or minus d, the difference. Proposition 2 stands. The solution follows: Suppose your envelope contains $100. You now have the option to play a $100 game in which you will win $100 or lose $50. The amounts are different. Now, if the other envelope contains $50, then you either lose $50 or gain $50. The amounts are the same. Here is the crucial distinction -- when you win $50, you are only plaing a $50 game! Proposition 1 compares winning and losing in a $100 game, but Proposition 2's win is a smaller game, so of course the winner's prize is smaller. The "amounts" in Proposition 2 are not the same as the amounts in Proposition 1, and the false statement is not either of the propositions, but Smullyan's assertion that they "obviously" contradict! Now that the paradox has been dismantled (with only the slightest trace of smugness), thanks to Smullyan's reduction of the problem to a simpler form, let us reexamine the original question of probability: Is it advantageous to trade envelopes, or not? If there is, in fact, equal probability that the other envelope has more or less money, I believe you should trade every time, just as with Monty Hall. But in order for the probability to be 50%, it must be just as likely (to say nothing of feasible) for the other amount to be 2n as n/2. Again, if we replace the coefficient 2 with 10,000, the game runs into real-world considerations. The starkest is that there is only a finite amount of money in the world, which imposes limits on what amounts you can put in the envelopes. A player may be aware of these limits, which will skew the probability. For example, if you happen to know that Monty Hall's car is not behind door 1, then you can pick door 1 first and win for sure. And if you know that the game administrator doesn't have a $999,900 to give away, you can save yourself $99.99. Perhaps a more abstract reward system (like points) could be used where there's no limit on resources, but I believe there's a fundamental problem with calculating the probability: You can't make a random choice from an infinite set. The probability of choosing any particular element is *literally* zero, so it's impossible to choose it at random. This argument applies to all elements -- no element at all from the set can be chosen randomly -- so the choice is impossible, and therefore any choice from an infinite set is not random, but biased in some way. Eat your heart out, Zeno. :-) Josh -- -- Joshua Juran Metamage Software Creations =) Tools for Wizards wanderer.at.metamage.com http://www.metamage.com/ * Creation at the highest state of the art * Re: Two-envelope paradox resolved! Author: bbartholomew Email: bbartholomew@voltdelta.com Date: 1998/12/28 Forums: rec.puzzles --------------------------------------------------------------------------- In article <i-dont-like-spam-2712980202310001@apple.metamage.pdq>, i-dont-like-spam@bloody.vikings (Joshua Juran) wrote: > In his book _Satan,_Cantor,_and_Infinity_, Raymond Smullyan presents "The > Envelope Paradox" (p. 189). The paradox involves a simple game: There > are two envelopes, one containing twice as much money as the other. You > get to choose an envelope, open it, and then optionally trade it for the > other. How often should you trade? > (snip snip) Your explanation was good, but you missed the grand finish. Yes, the paradox rests upon a faulty calculation of expected value, based upon the unstated assumption that the distribution of money in envelopes is flat and infinite. Of course, there is no such thing as a flat, infinite, probability distribution. So you are left with the problem of "calculating" expected value when you don't know anything about the distribution. With any possible distribution of values into the envelopes, the strategy of always switching will give the same expected return as that of never switching.But neither approach is optimal. The simple, but truly wonderful answer, is to GUESS! Pick a value. If the first envelope exceeds that value, keep it. If it does not, trade it.No matter what you pick, or what actual values are contained in the envelopes, this strategy will do no worse than the "always" or "never" approaches. If you should be so fortunate as to guess a value which lies between possible upper and lower amounts, you have improved your return. -- Bruce Bartholomew Riverside, CA -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ... & circumstance |