DorothyGale
DorothyGale
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August 12th, 2010 at 8:20:52 PM permalink
A friend reminded me of this take on the Monte Hall problem ... or is it ???



The Conundrum

A man is presented with two envelopes full of money. One of the envelopes
contains twice the amount as the other envelope. One the man has chosen
his envelope, opened and counted it, he is given the option of changing
it for the other envelope. The question is, is there any gain to the man
in changing the envelope?

It would appear that by switching the man would have a 50% chance of doubling
his money should the initial envelope be the lesser amount and a 50% chance
of halving it if the initial envelope is the higher amount. Thus, let x be
the amount contained in the initial envelope and y be the value of changing it:

y = 0.5*(x/2) + 0.5*(2x)

Let痴 say that the initial envelope contained $100 (so that x = $100). There should
be a 50% chance that the other envelope contains either 2 * $100 = $200 (2x)
or a 50% chance that the other envelope contains (1/2) * $100 = $50 (x/2).
In such a case, the value of the envelope is:

$125 = 0.5*($100/2) + 0.5*(2*$100)

This inequality can be shown by simplifying this equation:

y = (0.5)*(x/2) + (0.5)*(2x) = (5/4)x

This implies that the man would, on average, increase his wealth by 25% simply
by switching envelopes! How can this be?



--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
Wizard
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Wizard
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August 12th, 2010 at 9:08:51 PM permalink
Oh boy, I love this problem! I've spent many an hour discussing and writing about it.

I'll keep it brief for now, just to get the ball rolling. First, if the person offering the envelopes is going to be uncomfortable or unable to put $200 in one of them, then don't switch. However, if I may, let's assume the person offering the envelopes could easily afford to part with 2x the amount of your envelope. Do you still switch?

I've had people much more educated in math and science disagree with me on this, at least on the reason, but I say that with zero information about the wealth or behavior of the envelope stuffer, then your odds are the same switching or not switching.

If you argue that you should switch, should that argument hold BEFORE you open your envelope? Assume your envelope has $x. Wouldn't the expected value of the other envelope be .5*(x/2) + .5*2x=1.25x? If it were, and the host let you switch back and forth as much as you wanted without opening the envelopes, then indeed, the expected value argument would lead you to keep switching infinitely.

I contend that you can't formulate an expected value without knowing the possible outcomes and probabilities of the other envelopes. If your envelope has $100, then I contend it doesn't mean there is not a 50/50 chance the other one has $50 or $200. It is not a random variable. There is either a 100% chance is has $50 or a 100% chance it is has $200.

Another way to think about it is to call the difference in the envelopes $y. By switching you will either gain or lose $y with 50/50 chance. Thus switching is a breakeven bet.

That is enough for now. I'll await more comments before going further.
It's not whether you win or lose; it's whether or not you had a good bet.
NicksGamingStuff
NicksGamingStuff
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August 12th, 2010 at 9:15:45 PM permalink
These math "problems" are really interesting to me, its like the numbers say one thing but "logically" (or a standard human thought) they should be another. I find my self following the argument and understanding what is being said, but it does not quite fully process to that "a-ha" moment for me, I guess thats why my degree is not in math!
Doc
Doc
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August 12th, 2010 at 9:36:27 PM permalink
I think that without opening/counting at all the man should just switch the envelopes back and forth as many times as possible and eventually open one of them to find himself fabulously wealthy.




Edit:

Dagnabit! Once again, the Wizard posted while I was typing! I need a utility that will tell me who else is in the process of posting.
mkl654321
mkl654321
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August 12th, 2010 at 10:03:07 PM permalink
It is trivial that no gain can be realized by switching, as if that were true, then the recipient could switch again and again ad infinitum, increasing his expected value each time until each envelope held a trillion skillion jillion dollars. Which would be good, as the taxes on whatever amount he received when he finally got sick of switching and opened one or the other would instantly cure the federal deficit.

I don't need to refute the math, though I'm certain it is refutable; the effort would be pointless. It's like the ancient Greek "paradox" where a runner ran half the distance to the finish line, then half of the remainder, and so forth, so that he never actually got there--he just kept halving the remaining distance. I remember reading about this when I was a very little kid--my reaction was "that's ridiculous", as obviously at some point the runner would complete ALL of the remaining distance, not just half of it. However, I could not refute the math until I learned about summing an infinite series, much later, when I was a not-so-little kid.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
mkl654321
mkl654321
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August 12th, 2010 at 10:14:13 PM permalink
Quote: Wizard


Another way to think about it is to call the difference in the envelopes $y. By switching you will either gain or lose $y with 50/50 chance. Thus switching is a breakeven bet.



Because I'm an English major and not a math major, I'll try to deal with this problem with words rather than numbers...

It seems to me that whether you
pick and open,
pick, switch, and open,
pick, think about switching and don't, then open the original one,
pick, switch, then switch back, standing there in indecision so long that the sun heats up one envelope and sets fire to it, thus forcing you to choose the remaining envelope,
or the guy offering you the envelopes kicks you in the balls and snarls, "Pick one already or I'll kill you" and you hastily grab one,
the ultimate choosing of the envelope is random, no matter how much fiddling and waffling goes on before one of the frickin' envelopes is finally opened. Therefore, the expected value of switching is +0.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Doc
Doc
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August 12th, 2010 at 10:57:01 PM permalink
Quote: mkl654321

It's like the ancient Greek "paradox" where a runner ran half the distance to the finish line, then half of the remainder, and so forth, so that he never actually got there--....

I remember hearing a variation of this used to describe the difference between a mathematician and an engineer. They are each presented with the same problem: You are 10 feet away from a wall. Against the wall and facing you is a beautiful, naked, willing, young lady. Every one minute, you are allowed to advance one-half the distance toward her. Will you ever get there in a finite amount of time?

The mathematician's response, supposedly, is that with the ever-decreasing steps, one can only approach and never actually reach the objective in a finite amount of time.

The engineer's response, supposedly, is, "Yes, I will get there, for all practical purposes."




Yes, that's sexist.
weaselman
weaselman
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August 13th, 2010 at 5:37:14 AM permalink
The chances of doubling and halving the amount are not 50/50, because you cannot pick *any* amount with equal probability. It would have to be either an equal probability on a finite interval (or a finite number of fixed amounts) or a non-constant distribution like a bell curve, so that some values are much more likely to occur than the others.
In any case, if you know the distribution, you should be be able to tell the odds of you gaining or loosing by switching, after looking at the original amount, and most of the time, they won't be 50/50. And if the distribution is unknown, there is nothing to be gained by knowing the first amount.
"When two people always agree one of them is unnecessary"
DJTeddyBear
DJTeddyBear
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August 13th, 2010 at 5:38:03 AM permalink
Quote: mkl654321

...standing there in indecision so long that the sun heats up one envelope and sets fire to it...

...the guy offering you the envelopes kicks you in the balls and snarls, "Pick one already...

LOL!

I could not have said it better myself.

---

I agree that it's a 50/50 situation.

NOT 50% that the other envelope will be double or higher or whatever, etc., but 50% that you picked the higher valued envelope to begin with.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Doc
Doc
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August 13th, 2010 at 6:12:23 AM permalink
Quote: DJTeddyBear

NOT 50% that the other envelope will be double or higher or whatever, etc., but 50% that you picked the higher valued envelope to begin with.

Two part confused-by-the-conundrum posting---

Part #1: I do not yet see why "50% that you picked the higher valued envelope to begin with" does not lead to "50% that the other envelope will be the lower valued one." I think it is more likely that the probabilities are being used improperly to calculate an expected value.

Part #2: The only twist I have come up with is that the amount of money that can be in an envelope is not a continuous function; money comes in discrete units. If you examine one envelope and find that the money in it could not be evenly divided into half (eg., $247.59), then you must have gotten the smaller amount and should swap. I'm not sure how to follow through on this notion in the event that your envelope contains a nice even amount.

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