Envelope Problem

[sagefr0g]

think about it. a fair game would be either u double ur hundred or lose it. that would be the normal odds for a 50 50 wager.

look at the original post, quoted below:

"There are two envelopes, one has twice the money as the other. You open the first envelope and it has $100 in it. Do you switch? "

there is no wager. the game (or transaction) isn't fair. the player (chooser of the envelope) can't lose anything, but can only win. there is no cost for the player to partake in this bet or transaction (except time). it's a game or transaction skewed in favor of the player, always! and no, the player isn't playing with the house money when making the decision, and even if it was, it wouldn't matter because the player is still gonna make money, always, lol. it's just an imaginary win, win situation, imaginary because no one in their right mind would offer such a transaction unless it was part of a game that in the long run did in fact initially cost the player more money than either envelope would ever contain.
therefore it doesn't matter if one stays or switches, excepting for the case where the game could be played in perpetuity, wherein then staying instead of switching would take less time, which would mean more plays, which would over the long run earn about twice as much money for the player.

 

[ZenKinG]

All of you have a lot to learn. Read my post about this jesus. This isn't about EV, it's about CE.

100 > 25 in EV.

 

[London Colin]

All of you have a lot to learn. Read my post about this jesus. This isn't about EV, it's about CE.

100 > 25 in EV.

Oh for crying out loud! It's a thought experiment, with a long history, designed to show how it is possible to compute an answer for the EV that is patently absurd, and thus pose the question 'what went wrong in this calculation?'.

Whether or not our notional player goes through with the swap, having (wrongly) decided it has a higher EV is utterly irrelevant.

If it makes things easier for you, substitute cents for dollars in every post you read in this thread, or imagine yourself to be a billionaire faced with this puzzle.

 

[Joe Mama]

2 Envelopes.
Envelope #1 has A $
Envelope # has 2xA $
You are given an envelope at random, you have a 50% chance of having 2xA $, and a 50% chance of having A $.
You switch envelopes, you now have a 50% chance of switching to A $, and a 50% chance of switching to 2xA $.

Would you be willing to pay 5% A $ to make this switch?

 

[Meistro]

This is a very simple calculation. You have a 50% probability of receiving $200 (value $100) and a 50% probability of receiving $50 (value $25). The combined value of switching is $100 + $25. Clearly, $125 is more valuable than $100.

The true odds for this would be $200 or $0. That would be a break even game.

Let's say that a casino had a roulette wheel, with no zeros just the 36 numbers, where if you bet red and it landed on red you received $100 payout as well as your $100 bet back of course, but if you bet red and it landed on black, you would only have to give up half of your $100 bet for a loss of $50.

You are going to try to tell me that this would be a break even game? But it is a perfect analogy for the switching game.

 

[Meistro]

This is a paradox because the long term EV of switching is $0, but the short term of switching only once is + 5/4th of your bet. It is easy enough to demonstrate if you start with $100 ten times, and then switch, five times getting +100 and five times getting -50. You end up with +$250 profit, just by getting the expected result, an even number of wins and losses.

Now if you start with $100 and then switch 10 times doubling or halving the resulting sum, and get the expected result, you will break even. But if you do the same thing twice, and do not get an exact amount of wins and losses each time, but end up out of two trials with an even number of wins and losses, you will have made money. The value in switching comes from the possibility of asymmetric results and the fact that you gain more $ when you win than you lose $ when you lose.

 

[ZenKinG]

all of you are so caught up in EV, you are completely ignoring CE.

THIS IS A CE PROBLEM NOT AN EV PROBLEM.

JESUS

 

[London Colin]

look at the original post, quoted below:

"There are two envelopes, one has twice the money as the other. You open the first envelope and it has $100 in it. Do you switch? "

there is no wager. the game (or transaction) isn't fair. the player (chooser of the envelope) can't lose anything, but can only win. there is no cost for the player to partake in this bet or transaction (except time).

By fair, he means zero EV. You can consider the amount in the first envelope you open to be the wager. The game is 'switch or no switch?'.

In a fair game, there is no point in playing because there is, on average, nothing to be gained or lost.

And, to re-iterate, the game of 'switch or no switch?', played with 'indistinguishable' envelopes, is self-evidently a fair game. There is, on average, nothing to be gained from switching. 'Nothing to be gained from switching' is pretty much the dictionary definition of what the word 'indistinguishable' means.

It's often cast as a 2-player game. Each is given one of the envelopes and has the opportunity to ask his playing partner to swap. Following Meistro's reasoning, they would each conclude that they gain a positive EV by making the trade. But this is (or should be) obvious nonsense. It's a zero sum game. The amount that one player wins is always exactly balanced by the amount that the other loses.

One envelope contains X, the other 2X. The EV of a randomly chosen envelope is (X+2X) / 2 = 1.5 X.
When they make the trade, one of them goes from 2X to X, making a loss of X, the other goes from X to 2X, making a gain of X. With a 50:50 chance of gaining or losing X, the EV of the trade is zero.

Again, this is all supposed to be blindingly obvious! The intent behind this puzzle is not to cast doubt on all of the above. The intent is that we take the line of reasoning that Meistro is so wedded to and, -

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

 

[gronbog]

This is a paradox because the long term EV of switching is $0, but the short term of switching only once is + 5/4th of your bet. It is easy enough to demonstrate if you start with $100 ten times, and then switch, five times getting +100 and five times getting -50. You end up with +$250 profit, just by getting the expected result, an even number of wins and losses.

This is absurd. There is no such distinction as short and long term EV.

If you play once, whether you switch or not, you have a 50/50 chance of ending up with the larger envelope.
If you play n times, whether you switch every time, or never, or randomly, you will expect to end up with the larger envelope n/2 times.

These two statements are mathematically and physically consistent.

To say that you will make a profit if you run the single switch scenario n times is clearly absurd. It is the same as running the n trial scenario once, which even you conclude will break even.

Please take Colin's advice and read up on, and understand the dynamics of this well known and already solved problem.

 

[Meistro]

Let's say that a casino had a roulette wheel, with no zeros just the 36 numbers, where if you bet red and it landed on red you received $100 payout as well as your $100 bet back of course, but if you bet red and it landed on black, you would only have to give up half of your $100 bet for a loss of $50.

You are going to try to tell me that this would be a break even game?

 

[London Colin]

Meistro, please help us out and list all of the academic papers that you have so far read on the two envelope problem.

Point out any that agree with what you have been saying.

Point out what you believe to be the specific errors being made in those that agree with what we are telling you. Don't just deride their conclusions; home in on exactly where you feel they are making a misstep in the lines of reasoning that lead them to these conclusions.

 

[Meistro]

Why are you incapable of answering a simple question?

 

[London Colin]

Why are you incapable of answering a simple question?

I took it to be a rhetorical one.

OK. No. That's obviously not what we are telling you.

We are telling you that by presenting this as an equivalent situation you are further demonstrating that you haven't understood the original situation. (Either that, or you just enjoy pursuing a ridiculous argument, to see how far you can take it.)

We are telling you that the internet is awash with academic information, which you can, if you are genuinely interested, use to gain the understanding you currently lack. (You don't even need to use google, I've given you the links!)

We are, by now, pleading with you to consult these resources, rather than continue to simply argue back and forth as if we are trading opinions on an open question.

 

[Meistro]

I do not see how you can admit that my hypothetical roulette game has positive EV and yet simultaneously maintain that switching from a $100 envelope, which is basically the exact same situation, does not. Your position is contradictory. You have the same potential results, the same probability, what exactly is different?

 

[sagefr0g]

hmm, at the risk of stirring the pot, lemme apologize for going ahead and saying, that imho the academics may have missed something (at least as far as my limited, half witted reading goes). that something, being potential. potential in the sense that the envelope transaction could be played more than once, even though it’s highly unlikely that it would ever happen once, lol.

that being the case, there is a ‘time’ advantage for choosing to stay, one would build earnings at a faster rate by staying rather than switching, since switching takes longer. rather than switch. imho one should not switch due to the possibility of being able to make the transaction more than once.

 

[gronbog]

Your roulette game is not the same as the envelope situation. Please read and respond to the points in the existing literature. You might also consider responding to the points made by me Colin and other posters rather than simply repeating your own solution over and over again hoping it becomes correct. Your logic for the envelope problem is attractive and intuitive, but incorrect. That's the what makes the problem intriguing.

 

[Meistro]

Your roulette game is not the same as the envelope situation.¨


oh, how are they different?

In both cases you have a 50 50 chance. In both cases one side pays $200 the other pays $50. They are the exact same. If the roulette game is positive ev then so is switch from a $100 envelope.

 

[Joe Mama]

I do not see how you can admit that my hypothetical roulette game has positive EV and yet simultaneously maintain that switching from a $100 envelope, which is basically the exact same situation, does not. Your position is contradictory. You have the same potential results, the same probability, what exactly is different?

When I get something like this and toss it around in my head, I may wake in the middle of the night or early the next morning and see it clearly and say AHA! Pleasant dreams!

 

[London Colin]

Your logic for the envelope problem is attractive and intuitive, but incorrect. That's the what makes the problem intriguing.

Indeed. And, in case it got lost among the back-and-forth flurry, I'll repeat that I find it particularly intriguing that this incorrect logic can be made correct by a fairly subtle change in the rules of the game -

But the interesting thing I take away from that paper is that you can make switching favourable if you recast the problem slightly -

Two envelopes are labelled A and B. An amount is placed in envelope A, and a coin is tossed to determine whether to place half or twice that amount in envelope B.

Now you have an asymmetry. If you are offered envelope A, you should ask to switch to envelope B.

The envelopes are no longer indistinguishable. You still don't know which is the larger; you just have information about the methodology that was used to fill each of them (and which is which).

Intuitively, it seems like nothing significant has really changed, and yet this is sufficient to make the 5/4 EV calculation correct.

 

[Meistro]

The envelopes were never indistinguishable, we already opened one and it contains $100. Did you even read the OP?