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TomasB
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Both Jonathan and Daegs seem to be missing the point.

Jonathan, you say that the statement "let x be the dollars in the envelpe in my hand" could be the thing creating the paradox, yet that statement is undoubtedly true if you in fact open the envelope and find x dollars. So the question is _why_ does the paradox occur from such a simple and true statement if that in fact is the statement causing the paradox?

Daegs, everyone _knows_ that it should not matter if you switch or not (if you do not have a clue of the distribution). Therefore there must be some error in the reasoning that tells you to switch since the reasoning clearly shows that you _should_ switch. So the puzzle is to find the flaw in the reasoning that tells you to switch, _not_ find a reasoning that shows that it doesn't matter if you switch. Again - we _know_ that it is true that it shouldn't matter if we switch.

/Tomas
Josh the Superfluous
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Congrats Daegs! I think you are right.

In all fairness to Daegs. He does answer the given puzzle. The particular reasoning discussion was added by TomasB and Mr. Martin has yet to comment if his puzzle was to be modified with TomasB's discussion.

If we all agree with Daegs gut level reasoning as the solution to puzzle 1, and agree to logically dispute Tomas' flawed logic puzzle (post #2) and MagicJohns second post, I think we'll all get along.

(Jonathan's right too)
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"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2
Talarspeed
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You guys are too deep for me!

Switch it!
Don't switch it!

Either way, you are now richer by X amount of dollars!
You're better off than when you entered the room! Go to the local bar and celebrate your good fortune!
magicjohn2278
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Quote:
On 2006-03-24 23:18, Daegs wrote:
Let's say we always switch:
1/2 we pick $1, see it and switch-> WIN - $100
1/2 We pick $100, see it and switch-> LOSE - $1
Let's say we always stay:
1/2 we pick $1, see it and stay-> LOSE - $1
1/2 we pick $100, see it and switch->WIN - $100

So 2 out of 4 we lose, 2 out of 4 we win.



If any of this is doubted just go through it again above, where it is seen its always 1/2 chance of winning or losing.

Adding anymore math or logic to a solved problem just creates problems.


So in 2 out of 4 cases you would lose $1 and in 2 out of 4 you would win $100 - so you you seem to have proved that switching is a good idea? Risking losing $1 on a 50/50 chance of winning $100 seem to be the best thing to do! - But why?
Josh the Superfluous
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Did anyone add a calculation to factor in what you would do if the envelope you opened contained an odd number (e.g. $1000.01).

I think the pay-off odds should be kept seperate from the odds of paying-off (50/50)

(I just reread John's above post. OOOOW My head hurts.)


Posted: Mar 25, 2006 9:23am
-----------------------------------------
MagicJohn You are also risking 50 cents if x= 50cents. A bad bet on a different scale.
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
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landmark
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So I posed this question to some high school math teachers; the general consensus was this:

There's no paradox. Both envelopes are good. Given the information you have, switch as much as you want. Stop switching when you want. Go home with some money.

They were not troubled by this at all.

I was troubled that they were not troubled.

But maybe Jon is right-- maybe there is not enough information yet to induce a paradox.

Yet Jon, in your counterexample, you say a voice comes on and says "If it's tuesday. . ."
Well then, a rational person would try to figure out if it was Tuesday and decide on a switch based on that answer. If one had no idea what day it was, then a rational person would not switch as there is only a 1/7 chance that it is Tuesday etc.,

In the same way, the info that there is either double or half the money in the other envelope seems to point to switching--but perhaps this is the flawed premise--perhaps this information isn't information at all. Perhaps in some way this information is similar to a statement like "the other envelope is twice as wide or half as wide."

I don't really believe this, but perhaps this is a path to investigate.

Jack Shalom
Daegs
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Oh right, I only skimmed post after first...anyway
Quote:
Let's say the envelope you hold has X dollars in it. That means that the other envelope has 2X dollars or X/2 dollars in it with equal probability. Your expected winnings of switching would be 0.5*2X + 0.5*X/2 = 1.25X while if you keep your envelope your expected winnings is X. So you are better off switching.


The error comes with "0.5*2X + 0.5*X/2 = 1.25X"... that is wrong.


You shouldn't use .5X, use X and 2X for your math since either way you are getting double that other. By comparing X/2 to 2X you are in effect doubling the bounds of other envelope.


So 1/2 of the time -> ".5*X + .5*2X = 1.5X"
Other 1/2 of time -> ".5*2X + .5*X = 1.5X"

its straight down the halfway mark.
TomasB
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A slightly different version of the game, just to confuse things. Smile

I tell you that I have placed a, to you, unknown amount in an envelope that I give to you. I then flipped a coin to decide if I should place half of that amount or double that amount in another envelope. I offer you to switch. Would you?

Here's the last time this was up for discussion here, by the way:

http://www.themagiccafe.com/forums/viewt......orum=101

/Tomas
magicjohn2278
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Daegs and Jonathan, what it all boils down to is;

I give you a sum of money.

If you wish,you may toss a coin - If it comes down heads, I double your money.
If it comes down tails, you give me just half of what you have.

Do you toss the coin? Is it worth taking the bet?
landmark
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Magicjohn,

That's an interesting simplification, but I don't
think there's any paradox there--flip the coin.



Jack SHalom
Daegs
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Don't flip... personally I'd rather have $100 for sure than a very real possiblity of $50.(even if I could win $200).

The key here is that you don't know whats in the envelope.
landmark
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Actually, Tomas' link to the prior discussion is worth reading. Arguments all the way around seem more cogent there Smile

Jack Shalom
magicjohn2278
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Quote:
On 2006-03-25 19:52, Daegs wrote:
Don't flip... personally I'd rather have $100 for sure than a very real possiblity of $50. (even if I could win $200).

The key here is that you don't know whats in the envelope.

But the point is that you CAN look in the envelope if you wish... even if you don't, you know that the other contains either double or half of what you are holding.

So is it to your advantage to switch?

... probably yes?


Posted: Mar 25, 2006 8:23pm
-----------------------------------------
Jack, the situation is exactly the same with the envelopes...

If you switch, you either lose only half what you have, or double it!

... but this can not be...?
Platt
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Quote:
On 2006-03-25 16:13, magicjohn2278 wrote:
Daegs and Jonathan, what it all boils down to is;

I give you a sum of money.

If you wish,you may toss a coin - If it comes down heads, I double your money.
If it comes down tails, you give me just half of what you have.

Do you toss the coin? Is it worth taking the bet?

Yes, it's clearly worth taking the bet. On the flip of coin, I'd take risking 1/2 to doubling my money any day. The paradox seems to have deepened. So is the above coin flip really analagous to switching the envelope? It can't be.


Posted: Mar 25, 2006 11:50pm
-----------------------------------------
Quote:
On 2006-03-25 19:52, Daegs wrote:
Don't flip... personally I'd rather have $100 for sure than a very real possiblity of $50. (even if I could win $200).

The key here is that you don't know whats in the envelope.

You would? Why? Unless you needed that $100 for a medical emergency that's illogical thinking. Particularly if you were given this option numerous times. You're risking $50 to win $100. Over time you'd make a killing opting to bet on the coin flip.
Sugar Rush is here! Freakishly visual magic. http://www.plattmagic.com
Jonathan Townsend
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Double or half?
Coin landed heads?
Amount in envelope... nope, that is impertinent.

Still... which data are pertinent?

Till BOTH are opened or there is sufficient data to compute expectations there is nothing to calculate. Let x be the number of washing machines on the street... or what you will still not pertinent data.

If you require proof, take a few singles into envelope pairs and ask the homeless on the street to pick an envelope. Watch what happens. Listen to their common sense answers. They wind up with a buck or two. And you might get free of an inefficient perspective.

BTW the "if it's Tuesday" thing from an earlier post was an example of impertinent data. To make a Monty Hall problem out of this, add an envelope and have fun. Smile
...to all the coins I've dropped here
Daegs
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A bird in the hand is worth two in the bush.... I'd rather have a sure $100 than a chance at $200 (with a chance of losing $50).... but that's just me(yes if you can do this over and over I'd be fine with going for it... but once the risk is not worth it, I cant even go out to eat with $50....)

Anyway, that is all different than the envelope problem.

The problem is the faulty assumption that there are 3 values in play, 1/2X, X and 2X.

There is only X and 2X!!! (or *only* X and X/2)

There is not $50, $100 and $200 in play, there is only either $50 and $100 or only $100 and $200.

It is bad logic, imho, to say that the other envelope contains either 2X or X/2... it doesn't, it contains only one of those.

Let's say you have two envelopes, one with X and one with Y(where Y = 2X).

You pick an envelope, there is a .5 chance that you picked X and a .5 chance that you picked Y.

If you picked X then the other envelope contains Y, if you picked Y then the other envelope contains X... at no time does X/2 come into it.

The other envelope contains ONLY the value it originally started with, it is not changing...

EDIT: I Give up, this truly is a paradox... you are right I concede, there is no right answer....

Look at it this way:

1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose.
2: The amounts in the Envelopes are Y and 2Y, by switching you either lose Y or gain Y, so the amount you gain is the same as the amount you lose.

But the envelope you open up can be either case as its just a constant.

Again I give up, this truly is a paradox.
magicjohn2278
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Daegs.. don't give up so easily! I was thinking about this this morning and am beginning to think there may be no paradox here, just bad maths!

I am given an envelope, and given to chance to switch for one containing half or double the one I hold with a 50/50 probability.

I think my options are:

Keep what I have and gain nothing.
Switch what I have and lose half.
Switch what I have and double my money.

which most people would see as a reasonable bet and go for the switch.

But am I confusing odds and probaility here? After all, the bet can only be made once, if we switch twice, there is a 100% proaility that nothing has changed.

We all know that switching will only give one of two results - you either win or lose some cash.

So are the possiilities above a fair representation of the options? - Possibly not.

Perhaps the options are these (actual figures for illustration only.)

Do nothing (Holding the $20)...No change
Do nothing (Holding the $40)...No change
Switch (holding the $20).......Gain $20
Switch (holding the $40).......Lose $20

50/50 chance of winning or losing $20 now - So where did my favourable gamble go???


?
Platt
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Quote:
On 2006-03-26 06:22, magicjohn2278 wrote:

Perhaps the options are these (actual figures for illustration only.)

Do nothing (Holding the $20)...No change
Do nothing (Holding the $40)...No change
Switch (holding the $20).......Gain $20
Switch (holding the $40).......Lose $20

50/50 chance of winning or losing $20 now - So where did my favourable gamble go???


?


The above information seems overly simplified. It seems to assume that if you're staring down the $20, the other envelope must contain $40. Or if you're staring down the $40 the other envelope must contain $20. What about the chance of the other envelope containing $80? Or if you're staring down the $20 the other envelope containing $10?

Now I'm further confused.
Sugar Rush is here! Freakishly visual magic. http://www.plattmagic.com
Jonathan Townsend
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First step in solving a word problem is to clearly state your working premise.

In this case, Let x be the value in dollars of the currently selected envelope.
Then the value in dollars of the money in the other envelope is EITHER x/2 or 2x and you have insufficient data to determine which. The rest of the analysis appears to be wheels spinning though without contact to the ground, and no useful conclusions are reached.

The part I find amusing is that folks forget to say "thanks" for the gift of what is inside the selected envelope.
...to all the coins I've dropped here
magicjohn2278
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Alternatively:

You hold either X or 2X..

Do nothing (Holding X)...No change
Do nothing (Holding 2X)...No change
Switch (holding X).......Gain X
Switch (holding 2X).......Lose X
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