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Two envelopes problem

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Re: Two envelopes problem

#21

Post by thematrixturtle »

kazinator wrote: post #16

It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will b…

You are missing the point. The problem of whether you should actually switch is obvious and trivial (you should not). The challenge here is explaining why the plausible 'proof' that you should switch is wrong.

Re: Two envelopes problem

#22

Post by Longwelwind »

kazinator wrote: post #16

It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will b…

>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.

Re: Two envelopes problem

#23

Post by Schroedingersat »

scrapheap wrote: post #9
jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

It becomes a lot clearer if you increase the number of goats.

I have a deck of a billion cards, one entitles you to the car, the others have a picture of a goat. You pick one at random but do not get to see it.

Of the remaining 999,999,999 cards, I discard 999,999,998 goats.

Do you want your initial card, or the remaining card in my deck?

Or to frame it another way, do you want the car iff you picked it first or iff you did not pick it first?

Re: Two envelopes problem

#24

Post by thekiptxt »

scrapheap wrote: post #9
jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

I didn’t question the math, but what made me intuitively believe the Monty Hall problem was this: Me: I am thinking of a number between 1 and a million. If you guess correctly, you get a prize.

you make your guess Me: Ok, I will reveal that the number is either $YOUR_GUESS or 452,871. You may keep your original guess or switch to 452,871.

Of course, you would switch, because the chances you guessed correctly originally were 1 in a million.

Re: Two envelopes problem

#25

Post by thekiptxt »

sjburt wrote: post #3

I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic paper and yet the conclusion is so obviously wrong. Decision problems are tricky and in non intuitive ways.

This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen.

Or code that “obviously” has a bug only after it caused an issue in production.

Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”

Re: Two envelopes problem

#26

Post by kazinator »

Longwelwind wrote: post #22
kazinator wrote: post #16

It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will b…

>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 anot…

I did solve the paradox: it emanates from falsely looking at the switch as doubling or halving the money.

If the envelopes contain, say, $50 and $100, it looks like a doubling/halving. But nothing materially changes if we add $1000 to each one to make $1050 and $1100. The probabilities don't depend on what is in the envelopes, and the win/loss is just a spread of the difference between their values.

Why would anyone, while solving some problem, worry about some unspecified more complicated situation which supposedly resembles that problem?

Give me the specific of that more complicated problem and I will knock it out of the ballpark using whatever reasoning fits that problem.

Re: Two envelopes problem

#27

Post by antisthenes »

kazinator wrote: post #16

It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will b…

No post body was provided.

Re: Two envelopes problem

#28

Post by CJefferson »

kazinator wrote: post #16

It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will b…

I'm not sure it's that simple.

let's imagine the first envelope contained between 1 andand 1000 dollars, the second double the first.

if you open your envelope and it contains a whole number of dollars between 1 and 1000, then you should swap, id take "250 or 750" over 500.

Re: Two envelopes problem

#29

Post by jordo »

scrapheap wrote: post #9
jdkee wrote: post #2

Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

I always think of the Monty Hall problem with 100 doors. And with the game show host revealing 98 'goat' doors thus leaving two doors left (the original guess, and another one). You'd switch then right? :)

Re: Two envelopes problem

#30

Post by tsimionescu »

dllthomas wrote: post #15
scrapheap wrote: post #9

Earlier quoted context omitted.

Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stic…

When Monty reveals a goat, your odds increase from 1/3 to 1/2 if Monty is picking randomly and just happened not to reveal a goat (and if he revealed a car we start from the top, or you automatically lose, or you automatically win, or...), but to 2/3 if Monty will reliably reveal a goat. I think it's hard to say that you have no extra information in that latter case. It's true that we can add other options (pick both…

The GP knows all this, and agrees it's better to switch (look at the second paragraph). They just dislike the common formulation (that is given in terms of the extra information from the boot being opened), and find another formulation more compelling (your original choice is one door, while the switch is actually to the best prize from the other two doors, not simply to another one door).
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