you left out the key part of the sentence, " when you already have $100" what matters is the percentage gain or percentage loss. not the dollar amount. a 100% gain is exactly the same thing as 50% loss. Do you understand that?
I did. And I understood it.
Here is the problem as stated by wiki in the link:
“ You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? ”
A paradox is explained:
It seems obvious that there is no point in switching envelopes as the situation is symmetric. However, because you stand to gain twice as much money if you switch while risking only a loss of half of what you currently have, it is possible to argue that it is more beneficial to switch.[1] The problem is to show what is wrong with this argument.
It is the symmetry between the two envelopes that create the paradox because you don't know how much is in either envelope. The paradox is explained:
10 steps shows logic like Miestro used to make switching the correct move but since you have no information about the contents of either envelop the logic would tell you to switch back.
This is greater than A, so I gain on average 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.
So the symmetry of not knowing anything about either envelope causes you to continue to apply the same logic to whichever envelop is the envelope you are not holding and switch to that envelope. But if you know the contents of one envelope the symmetry is broken and after the first switching decision you never decide it is better to switch. You know the envelope with the higher expected value which is $125. You see the key is the wiki problem had no information on either envelope which created a symmetry that never allowed you to mathematically believe your envelope had a higher EV than the other. The same logic that you used to justify the first switch would justify switching back. But knowing the contents of one envelop breaks this symmetry that equates both envelopes and no paradox exists. The unknown envelop either has $200 or $50, for an EV of $125. If both envelopes contain unknown amounts, the other envelope has the same EV as whatever envelope you are holding. Or to use the switching logic that says to switch to the unknown amount when you know how much is in one envelope, the envelope you aren't holding always has an EV of 1.25 times whatever is in your envelope when you don't know how much is in either envelope. So you can never feel the envelope you are holding has a higher EV no matter which envelope you are holding. Do you see how knowing the contents of one envelope breaks the symmetry in the unknown amount envelope problem and makes the envelope that doesn't have the $100 the clear choice since it has an EV of $125 and once you are holding that you know the other envelope has $100 in it.
Last edited by Three; 02-18-2018 at 06:41 PM.
But if you really can gain by switching, then you would expect that after four wins in a row followed by four losses in a row, which is a typical long term result, that you would have a higher amount. But you wouldn't.
100, 200, 400, 800, 1600, 800, 400, 200, 100
And of course since you are multiplying and dividing the order doesn't matter. You could win 3 times, then lose 2 times, then win 1 time, then lose two times, and you would still end up back in the same place.
What made it click for me was imagine that you won one and then lost one repeatedly. You would just go to $200 than back to $100. No gain.
Last edited by Gramazeka; 02-18-2018 at 07:31 PM.
"Don't Cast Your Pearls Before Swine" (Jesus)
Sigh. Are you just yanking my chain? If you didn't switch 4 times you would have $400. If you switched 4 times and got an even distribution of $50 and $200 outcomes you would have two $50 returns and two $200 returns for a return of $500 for the 4 switches. $500 is more than $400, drum roll please, 5/4ths of $400. Wasn't 5/4ths of $100 the EV of switching.
But not the correct answer.
https://www.maa.org/external_archive...n_0708_04.html
Actually in the movie, it was a different, but similar problem. There are THREE choices: one door has a new car, and the other two have a goat. You pick door #1. The host opens door #3, and there sits a goat. Now you have a choice; do you stay with door #1, or do you choose #2? Here, it is correct to switch.
Its actually a simpler problem. In the beginning, each door has 1/3 chance of having the car. You picked #1, which means 2&3 combined have 2/3 chance. Well, once it's revealed that #3 is empty, door #2 now has 2/3 chance, and your origional choice still has 1/3 chance. So you switch.
Didn't love the movie, but I do love this scene:
https://m.youtube.com/watch?v=Zr_xWfThjJ0
Maybe, but many, many others already have. I even gave links.
I love math problems like this. When I was about 10, I got hooked on the hotel riddle. Three guys go in together to rent a room. The room is $30. They each pay $10. Later, the desk clerk comes to their room and says they overpaid; the room only cost $25 and hands them 5 $1 bills back. Now each person has $9, with two left over. But 9x3 is 27, and 27+2 is 29. Where did the other dollar go?
Before anyone tries to explain this to me (I'm looking at you, Three) I finally figured it out, and I do understand it. But I literally lost sleep over this one.
Oh the perils of fourth grade...
the math proofs are wrong because there is never any ev of $125.. that is this mistake.. $125 was never a possible outcome so you cant have those two boxes modeled that way. you have $100 and you can either lose 50% or gain 100% which is a net ev of exactly 0 not + $25 as explained:
it is not better to gain $100 than lose $50 when you already have $100.
50% loss is exactly equal to 100% gain because if you lose that bet, now you need 100% gain to get back to the original $100 and 400% instead of 100% to get to the $200 from the original $100 we had. The ev to switch is $0,not +$25 and the advantage in edge to switch is also exactly zero. From perspective of a card counter we dont tend to like risking money without an edge so we should never switch. but we all would anyway because its more fun and there zero positive or negative expectancy.
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