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 138938.  Thu Jan 25, 2007 3:32 pm I'm just going to send you here http://www.math.hmc.edu/funfacts/ffiles/20001.6-8.shtml Also some envelope 'facts' http://www.envelopesexpress.com/envelope_facts.asp

 138966.  Thu Jan 25, 2007 4:50 pm If you gave me two envelopes and one had twice as much money in as the other, I wouldn't switch. However, my reason would be that I would still have more money than I had in the first place because you'd given it to me, so I really wouldn't mind what was in the other envelope. I'm not greedy, you know.

 138968.  Thu Jan 25, 2007 4:55 pm Do you know, I think we could develop this idea into some kind of game show. Perhaps boxes instead of envelopes, and more of them than two ...

 138997.  Thu Jan 25, 2007 5:51 pm Just shows that maths puzzlers are a strange breed. From a pure practical pov then either way you look at it you're still gaining no matter which envelope you keep. As a mathematical exercise it illustrates futility more than paradox because of course you could keep switching envelopes and still be left with a 50/50 chance of getting either \$x or \$2x. However, you won't get either until you stop switching envelopes so from that same practical pov, just open one. In that game show alluded to it's a bit more complex and includes an offer on your box, presumably based on the probability that it contains a decent amount along with some personality reading and bluff thrown in. The basic premise is still the same though, that whatever you do you'll come away better off than when you went in. The only reason it drags out is because of greed, gambling or both. I can't decide whether I'd play along or opt for a reasonable deal if I were in that position. My reasoning is that I'd walk away at least a penny richer in any event so do I play purely for fun or to score higher. I suppose my mood at the time would swing it one way or the other.

138999.  Thu Jan 25, 2007 5:57 pm

 suze wrote: Do you know, I think we could develop this idea into some kind of game show. Perhaps boxes instead of envelopes, and more of them than two ...

Yes but you'd have to find some washed up, poinless, overpaid nearly was to present it.

 139000.  Thu Jan 25, 2007 5:59 pm This already featured in a game show (in a slightly different form). It is known as the "Monty Hall Dilemma" after an American Game show called "Let's Make A Deal", which is hosted by the aforementioned Mr. Hall. The problem became quite notorious in the realms of mathematics because a mathematician called Marilyn vos Savant published a discussion of the problem which said that people should accept the swap when offered. This caused an uproar from a lot of people who considered that they were better mathematiicans than Ms. vos Savant*. However, they were forced to eat their words. You may read about it here: http://www.cut-the-knot.org/hall.shtml or here http://mathforum.org/dr.math/faq/faq.monty.hall.html It is important to realise that this only works if the person offering the swap knows what is behind the doors. It wouldn't apply in "Deal or No Deal" because the banker who offers the swap does not know the contents of the boxes. Marilyn has a cool website which features a lot of logical paradoxes and fallacies. http://www.marilynvossavant.com/ *They should have considered that the woman is called "Savant" so is obviously going to be smarter than anyone else**. **Except for Susie Dent of course.

139071.  Fri Jan 26, 2007 3:59 am

Or here: post 15226

 139107.  Fri Jan 26, 2007 6:47 am I'm struggling to see the relevance of the Monty Hall problem here. In that case you are effectively given more information by the host who knows where the prize is and takes out one of the other options before asking you if you want to switch. The important fact here is that two out of three times you will pick wrongly to start with and so are guaranteed to win if you switch on those occasions. As you say, the Deal or No Deal case is not the same as the boxes are opened as you choose them and no-one knows the contents of the sealed boxes. The Envelope paradox is based on a flawed calculation and again no-one knows the contents.

139177.  Fri Jan 26, 2007 9:52 am

 djgordy wrote: It is known as the "Monty Hall Dilemma" after an American Game show called "Let's Make A Deal", which is hosted by the aforementioned Mr. Hall. The problem became quite notorious in the realms of mathematics because a mathematician called Marilyn vos Savant published a discussion of the problem which said that people should accept the swap when offered. This caused an uproar from a lot of people who considered that they were better mathematicians than Ms. vos Savant*. However, they were forced to eat their words.

Since reading this I've been doing a bit of digging about the Monty Hall problem and discovered some things I was previously unaware of, so I thought I'd share them with the group.

Firstly, Gordy is incorrect to refer to Marilyn vos Savant as a "mathematician". She is described by wikipedia as a "magazine columnist, author, lecturer, and playwright" and, whilst apparently holding a very high IQ*, does not have any formal mathematical qualifications.

Secondly, whilst many people did indeed claim that Ms vos Savant's mathematical treatment of the problem in her column in Parade magazine (September 9th 1990) was in error, some of these can be excused since the problem, as posed, was not precise enough. In particular, it did not specify whether the host was always going to open one of the doors or whether he would always offer the contestant the chance to swap (in the Let's Make a Deal show hosted by Monty Hall, one of the doors was opened but the contestant wasn't allowed to swap. It was merely a device for building tension, and could have lead to some confusion about the problem as stated). It didn't even specify whether or not the host was bound to reveal a goat rather than a car. There are certain ways to phrase the question which would ensure that changing your mind would guarantee that you would always lose.

Thirdly, whilst the problem became famous when featured in Ms vos Savant's column, it was originally published, and correctly solved, by a biostatistician called Steve Selvin. He published a letter in American Statistician, Feb 1975, V. 29, No. 1, p. 67 under the title "A Problem in Probability" in which he not only fully described the problem, but also correctly concluded that the contestant was better off changing their mind. Since this conclusion is so counter-intuitive at first sight, Professor Selvin, like Ms vos Savant, received many letters of complaint claiming he was in error, causing him to publish a second letter two issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134) to answer his critics and point out that his mathematics was, indeed, correct.

I remember being posed this problem while at university and, like most people, couldn't wrap my head around such a counter-intuitive result**. Eventually we ran a computer simulation (on a BBC B micro!) of the problem and that showed me that changing your mind was indeed the right option. It was only then that I began to realise the true nature of the solution.

* discussions about what IQ actually proves left for another time

** the result was not explained to me in detail, merely presented as a fact that changing your mind was the best option

139204.  Fri Jan 26, 2007 11:32 am

 Quote: If you always change your first call, the result will always be that if you guessed right first time (which you will one time in three) then you'll get it wrong the second time, whereas if you guessed wrong the first time (which you you will two times in three) then you'll get it right the second time. So you'll get it right 2/3 of the time. IE: if the right answer is A, and you guess A, the host will reveal one of the remaining two and you will choose the other one: wrong. If the right answer is A and you guess B the host will reveal C and you will change your guess to A (the only one left): right If the right answer is A and you guess C the host will reveal B and you will change your guess to A (the only one left again): right.
which is what I call logic, rather than maths - but it seems to me to be a satisfactory description of what's happening.

 139207.  Fri Jan 26, 2007 11:38 am Some people would say that mathematics is a branch of logic. Obviously, I must apologise to Ms. vos Savant for having called her a mathematician. No slight was intended.

 139312.  Fri Jan 26, 2007 5:00 pm To whom?

 139349.  Fri Jan 26, 2007 7:10 pm I would say that, once you've figured it out, the logic is rather straightforward - so it's hard to see why she is supposed to have endured all this flak.

139532.  Sat Jan 27, 2007 5:36 pm

 themoog wrote: I'm struggling to see the relevance of the Monty Hall problem here.
I'm not surprised. These are two completely different problems. The flaw in the two-envelopes problem is that you need some prior distribution in order to make an informed decision as to which envelope to pick. If you don't realise this, the problem could seem paradoxical. The Monty Hall problem is a straightforward exercise in probabilities. It could seem paradoxical because the respective probabilities that the prize is behind the 2 closed doors are 1/3 and 2/3, whereas it might seem that the situation is symmetrical and they should be 1/2 each.

139535.  Sat Jan 27, 2007 5:58 pm

I didn't say that the Envelopes Problem was the same as the Monty Hall Dilemma. I brought up the Monty Hall Dilemma because Suze said:

"Do you know, I think we could develop this idea into some kind of game show.

Perhaps boxes instead of envelopes, and more of them than two ..."

Having said that though, there are some similarities with the Monty Hall Dilemma:

 Quote: As with the Monty Hall Problem, if you really want to analyze the situation, you have to start by looking at the way the scenario was set up. Let L denote the lower dollar value of the two checks. The other check thus has value 2L. Let P(L) be the prior probability distribution for the choice the host makes for the lower value in the envelopes. (This will affect the entire game. Of course, we don't know anything about this distribution. But we can see how it affects the outcome of the game. Read on. When you make your choice (C) during the game, you choose either the envelope containing the lower value (C=lower) or the one that contains the higher (C=higher). As the amounts are hidden from you, you choose entirely at random, with equal probabilities for the two options, so P(C=lower) = P(C=higher) = 0.5 During the game, the value (V) of the content of the chosen envelope is revealed to be a certain value M. Given this information, what is the posterior probability that the chosen envelope contains the higher or lower value? That is, what is P(C|V=M), the probability that you chose the envelope containing the lower/higher value, given you now know what V is? This is the probability you need in order to compute any expected gain. The correct expected gain calculation is: (2M)P(C=lower|V=M)+(M/2)P(C=higher|V=M) - M The paradox above arose because you assumed that P(C=lower|V=M) = P(C=higher|V=M) = 0.5 Let's see why this cannot be the case. (In what follows, remember that L, V, C are variables and M is a numerical constant.) By Bayes' Theorem: P(C|V=M) = P(V=M|C)P(C)/P(V=M) . . . (1) Taking the first of the two cases, where you choose the lower value (V=L), we have P(V=M|C=lower) = P(L=M) The second of the two cases, where the chosen envelope contains double the lower value, is P(V=M|C=higher) = P(L=M/2) Substituting each of these two identities in to (1) gives P(C=lower|V=M) = P(C=lower)P(L=M)/P(V=M) . . . (2) and P(C=higher|V=M) = P(C=higher)P(L=M/2)/P(V=M) . . . (3) From (2), P(C=lower|V=M) is the same as P(C=lower) only if P(L=M) is the same as P(V=M). If it is not then in the calculations of expected gain you have used the incorrect probability. Specifically, you have used the prior probability, P(C=lower)=0.5, of choosing the lower value, rather than the posterior probability P(C=lower|V). The same argument works for P(C=higher|V), starting from (3). Under what circumstances could we have P(L=M) = P(V=M)? Since P(V=M) must normalize the distribution, we have P(V=M) = P(C=lower)P(L=M) + P(C=higher)P(L=M/2) that is, P(V=M) = 0.5 P(L=M) + 0.5 P(L=M/2) . . . (4) From (4), to have P(L=M) = P(V=M) we would need P(L=M/2) = P(L=M) for all values of M. This is an infinite uniform "distribution". But no such distribution exists (it cannot be normalised). Hence it is impossible to have any prior distribution for which P(L=M) = P(L=M/2) is satisfied for all M. As a result there is no posterior distribution for which, given any V, we could have P(C=lower|V) = 0.5. The result you will get in the game depends on the prior distribution for the amounts in the envelope. For example if the prior distribution P(L) were uniform between 0 and \$30,000, and you found \$40,000 in the envelope, then you would expect to lose if you swap, whereas if the prior were uniform between \$30,000 and \$100,000, you would expect to gain. To summarize: the paradox arises because you use the prior probabilities to calculate the expected gain rather than the posterior probabilities. As we have seen, it is not possible to choose a prior distribution which results in a posterior distribution for which the original argument holds; there simply are no circumstances in which it would be valid to always use probabilities of 0.5.

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