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The Magic Cafe Forum Index » » Puzzle me this... » » Two envelopes (0 Likes) Printer Friendly Version

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Andrei
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Magicjohn, to say that the chances are 50/50, is to miss the point of what we've been discussing.

You are twice as likely to pick values in the middle range of given envelopes, than you are to pick values in the extremes. Thus, if you switch a value which is not in the middle range, chances are you will get something in the middle range, rather than something even more extreme.

That is the whole idea of the paradox. You THINK it's 50/50, but in reality, it isn't.

Andrei
magicjohn2278
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I know that it is coming up to April 1st, but surely no-one believes this!

Just because you are more likely to pick a low value for X doesn't mean that that when you switch you are less likely to go down!

If Z = 300

Values 1 to 100 can appear twice (in either envelope) switch and go up or down
Values 100 to 200 only appear once (in either envelope) switch and go down

Consider each envelope
Values for Envelope A
100 to 200
1 to 100
1 to 100

Values for Envelope B
1 to 100
1 to 100
100 to 200

Match up the values with the corresponding envelope
Value Envelope A……Value Envelope B
100 to 200…………….1 to 100 ---- A is high B is low
1 to 100………………..1 to 100---- A and B could be high or low
1 to 100-------------------100 to 200 A is low B is high

So it is more likely that when we pick a value for X it will fall in the range 1 to 100, but nevertheless when we switch, the probability of switching for more or less is 0.5 each.
TomasB
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"1 to 100-------------------100 to 200 A is low B is high "

How do you figure that if one envelope has 5 (it's from 1 to 100) the other envelope can have between 100 and 200? Seems like that can never happen yet you have it as a case there.

/Tomas
Andrei
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John, please re-read Tomas' post above with the middle third bit and the probability dispersion of the envelopes.

Andrei
magicjohn2278
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What I am saying is that when the value for B is in the middle third then the value for A has to fall into the low third.

ie if there are 100 instances where B is in the Middle third, there are 100 instances where the value of A falls in the low third. And in every case, if you pick A and switch, you will switch for more.

(OK, I've slipped up here I should be referring to the cases where B is in the top third. then A must fall in the bottom third (of the finite total) - but the maths will still work out the same!) - If you want me to clarify then I will.
magicjohn2278
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Clarification (and corrections!)

Lets label the “thirds” of Z …. Z1, Z2 and Z3 (Z1 is bottom third, Z3 the high third.)

The argument is that if you pick x from a value in Z3, switch and you can only go down. True.

If you pick X from Z1 or Z2 then you can go either up or down. True.

As there are more values that fall into Z1 or Z2 than in Z3 you are more likely to pick one of these. True.

You are twice as likely to pick a value for X that can go up or down, than one that will go down. Not quite true!


There are two envelopes

Values from Z3 can appear once in either envelope. Picking these values and switching will result in a loss.
There are an equal number of corresponding values from Z1 that can appear once in either envelope. Picking these values and switching always results in a gain. (the corresponding value from Z3)
There are an equal number of values from Z1 and Z2 that can appear in either envelope picking these values will result in a gain 50% of the time and a loss 50% on the time.

There is a 1/3 chance that we will pick a Z3 value, switch and we always lose.
There is a 2/3 chance that we will pick a Z1 or Z2 value, switch and we can win or lose.
But when we switch our Z1 or Z2 value, we will only lose ¼ of the time.

1/3 times we pick Z3, switch and lose
1/3 times we pick a Z1 that results in a switch and gain
1/3 * ½ times we pick Z1 or Z2 switch and gain
1/3 * ½ times we pick Z1 or Z2 switch and lose
Expectation of switching and winning
(1/3 * 1)-(1/3*1)+(1/3*1/2)-(1/3*1/2)=0
TomasB
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Quote:
On 2006-03-30 06:20, magicjohn2278 wrote:
Clarification (and corrections!)

Lets label the “thirds” of Z …. Z1, Z2 and Z3 (Z1 is bottom third, Z3 the high third.)

The argument is that if you pick x from a value in Z3, switch and you can only go down. True.

Sure that's true, but didn't you miss some cases? Can you, for example, select the biggest value in Z2 and still double it without getting out of range? Your reasoning is all over the place so I can't get heads nor tails about it.

The easiest way to think about it is not where you can go from each value, but how you can _reach_ each value. Focus on a value and reason in how many ways that value can be placed in an envelope. Either you have seleced double that value to be placed in an envelope and this is half of that selected value, or you decided directly to place this value in an envelope (and placed half of it in the other envelope). If the value you are focusing on is too big, you can only select to place it in an envelope in one way, by selecting it immediately. There is no double value that can be selected that would generate this value in an envelope.

/Tomas
magicjohn2278
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There is only one value in Z2 than you can double without going out of range, that is Z/3 - perhaps the range for Z2 should start at (Z/3)+0.000001..

I don't understand you reasoning in this matter either!

... however, I think we are all getting a little bored with this and it's probably time to move on to something else! Smile
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