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OT: Non-Wrestling: Best math brain teaser ever, for regular people

Dogwelder

Well-Known Member
Aug 1, 2013
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Please do not just google the solution. The fun is in the thinking and trying! :)

Problem

Basic setup: You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead.

The switching argument: Now suppose you reason as follows:

I denote by A the amount in my selected envelope.

The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

The other envelope may contain either 2A or A/2.

If A is the smaller amount, then the other envelope contains 2A.

If A is the larger amount, then the other envelope contains A/2.

Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

So the expected value of the money in the other envelope is:

(1/2)(2A) + (1/2)(A/2) = (5/4)A

This is greater than A so, on average, I gain by swapping.

After the switch, I can denote that content by B and reason in exactly the same manner as above.

I will conclude that the most rational thing to do is to swap back again.

To be rational, I will thus end up swapping envelopes indefinitely.

As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. Thus, in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction.
 
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Instead of trying unwind your expression of the problem I’d do it in terms of value of switching.

The amount in one envelope is A, and in the other is 2A. You have one envelope in-hand to start. There’s a 1/2 chance that you start with 2A, and the value of switching is -A. There’s a 1/2 chance that you start with A and the value of switching is +A. Therefore the expected value of switching is 1/2(A - A)=0.

The sleight of hand in your expression revolves around the asymmetry introduced by using the 2x / 1/2x expression.
 
Instead of trying unwind your expression of the problem I’d do it in terms of value of switching.

The amount in one envelope is A, and in the other is 2A. You have one envelope in-hand to start. There’s a 1/2 chance that you start with 2A, and the value of switching is -A. There’s a 1/2 chance that you start with A and the value of switching is +A. Therefore the expected value of switching is 1/2(A - A)=0.

The sleight of hand in your expression revolves around the asymmetry introduced by using the 2x / 1/2x expression.
Yeah. You’re not wrong. But as stated in the problem, you need to teach an engineering student which of the numbered steps is wrong, and exactly why. Otherwise, how would he avoid making the same mistake next time, when he works for Airbus and when he might cause airplanes to fall out of the sky?
 
Yeah. You’re not wrong. But as stated in the problem, you need to teach an engineering student which of the numbered steps is wrong, and exactly why. Otherwise, how would he avoid making the same mistake next time, when he works for Airbus and when he might cause airplanes to fall out of the sky?
Don't let him accept envelopes of cash.
 
The written explanation seems sound on the surface, but you can't use 2A and 1/2A within the same formula as they refer to two different scenarios. You have to either use 2A and A, or A and 1/2A.

Correct or not that's the best I got - TGIF

 
Pretty sure this played out already.
Princess Bride Poison choice
Speaking of poison:
Danny Kaye: The pellet with the poison is in the vessel with the pestle, the chalice from the palace has the brew that is true, right?
Mildred Natwick:Right, but there's been a change. They ... broke the chalice from the palace.
Danny Kaye:They ... broke the chalice from the palace?
Mildred Natwick:... and replaced it with a flagon.
Danny Kaye:A flagon?
Mildred Natwick:... with a figure of a dragon.
Danny Kaye:A flagon with a dragon.
Mildred Natwick:RIGHT.
Danny Kaye:But, did you put the pellet with the poison in the vessel with the pestle?
Mildred Natwick:Noooo, the pellet with the poison is in the flagon with the dragon, the vessel with the pestle has the brew that is true.
Danny Kaye:The pellet with the poison's in the flagon with the dragon, the vessel with the pestle has the brew that is true.
Mildred Natwick:Just remember that.
 
Yeah. You’re not wrong. But as stated in the problem, you need to teach an engineering student which of the numbered steps is wrong, and exactly why. Otherwise, how would he avoid making the same mistake next time, when he works for Airbus and when he might cause airplanes to fall out of the sky?
None of the steps are numbered, either. Another paradox?
 
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Please do not just google the solution. The fun is in the thinking and trying! :)

Problem

Basic setup: You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead.

The switching argument: Now suppose you reason as follows:

I denote by A the amount in my selected envelope.

The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

The other envelope may contain either 2A or A/2.

If A is the smaller amount, then the other envelope contains 2A.

If A is the larger amount, then the other envelope contains A/2.

Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

So the expected value of the money in the other envelope is:

(1/2)(2A) + (1/2)(A/2) = (5/4)A

This is greater than A so, on average, I gain by swapping.

After the switch, I can denote that content by B and reason in exactly the same manner as above.

I will conclude that the most rational thing to do is to swap back again.

To be rational, I will thus end up swapping envelopes indefinitely.

As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. Thus, in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction.
all wrong, one has $10.00 and the other has $20.00... start over?
 
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