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Two envelopes problem

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Re: Two envelopes problem

#101

Post by wccrawford »

shannifin wrote: post #98
ghusbands wrote: post #85

Earlier quoted context omitted.

Calculating an expected value is meaningful and regular and does affect decisions. If you were told that you could pay £1 to flip a coin and heads would pay out 50p and tails would pay out £2, then playing it you would expect to make £1.25 (£0.25 profit) per play, on average. Many people interpret the envelope problem as being the same situation.

> Many people interpret the envelope problem as being the same situation. It just seems obvious to me that it's not the same situation. That is, I think the "paradox" is ultimately about the temptation to treat these as similar problems in the first place.

Right, the difference here is that the calculation that determines the expected payout is the same as the calculation that determines how much it costs to switch. With the coin, the cost to play is set and is lower than the expected outcome.

Re: Two envelopes problem

#102

Post by xigoi »

eru wrote: post #40
dllthomas wrote: post #17

Earlier quoted context omitted.

I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A . But with any actual distribution (... I think?) it does , at which point... no paradox. It's true that it's not clear (at least to me , but perhaps more generally) what distributi…

Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)

Who said there is a uniform distribution on the natural numbers?

Re: Two envelopes problem

#103

Post by jph »

andrewshawcare wrote: post #97

Once you pick an envelope, you no longer stand to only gain money. You can lose money and that has to be reflected in the potential value of each envelope. After the first selection, you must express the envelope value as the potential of what each envelope holds (the probabilities from the initial selection) which makes selecting again a wash. Let A = 50 Envelope 1 is 100 Envelope 2 is 25 First selection 1/2(100) +…

Excellent answer. Worth adding to the wiki page.

Re: Two envelopes problem

#104

Post by xigoi »

robocat wrote: post #78

> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch b…

Please don't use preformatted text for quoting, it's painful to read.

Re: Two envelopes problem

#105

Post by xigoi »

CaptainNegative wrote: post #36
Nihmie wrote: post #18

Earlier quoted context omitted.

Is there actually some way to guess with better than 50% chance? I'm pretty sure this could be reduced to the secretary problem where there's a pool of 2 candidates, which yields an optimal hire with probability 50%. Maybe you play word games and say, "I guess the other number is not higher," or "I guess the other number is not lower." Since the number you observed was produced once, there is some non-zero probabilit…

The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number. Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and…

But couldn't you make the same argument for guessing that it's higher iff the number is greater than 42? Therefore, if the number is 9, both strategies tell you to do different things, yet in each case, your probability of winning is >½. There must be a hole in the logic somewhere…

Re: Two envelopes problem

#107

Post by Archelaos »

dbt00 wrote: post #91
na85 wrote: post #33

Earlier quoted context omitted.

Well if debugging is the process of removing bugs from code, then logically the act of writing the code must be called "bugging" and the probability of having a bug in any new piece of code asymptotically approaches 1 with increasing SLOC ;^)

Every program has bugs, every program has inefficiencies. Therefore every program can be eventually reduced down to a single instruction, which will be incorrect.

> Every program has bugs

Where is the bug in this programm?

  END

Re: Two envelopes problem

#108

Post by lisper »

andrewshawcare wrote: post #97

Once you pick an envelope, you no longer stand to only gain money. You can lose money and that has to be reflected in the potential value of each envelope. After the first selection, you must express the envelope value as the potential of what each envelope holds (the probabilities from the initial selection) which makes selecting again a wash. Let A = 50 Envelope 1 is 100 Envelope 2 is 25 First selection 1/2(100) +…

> Let A = 50 Envelope 1 is 100 Envelope 2 is 25

No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2.

A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.

Re: Two envelopes problem

#109

Post by OJFord »

robocat wrote: post #78

> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch b…

Of course, if ever invited to play such a game, what you really want to do is ditch the contents of your wallet on your way. Can't lose!

Re: Two envelopes problem

#110

Post by ossopite »

chii wrote: post #62
ossopite wrote: post #48

Earlier quoted context omitted.

that's an excellent treatment. I'm curious: this video picks a particular example of a distribution of the amounts and shows that the expected profit from switching is not defined. Can one somehow prove that for all possible distributions, either the expected profit from switching is 0 or it is undefined?

the point of the video is that given a distribution, the total expected profit cannot be defined, because the infinite sum of the probabilities of each case don't have an order, and also adds up to positive infinity and negative infinity. it doesn't really matter what the distribution of the amounts are, as long as there's an infinite number of possibilities in the distribution (ie., it's not a finite amount of possi…

> as long as there's an infinite number of possibilities in the distribution

how would one prove that claim? it's not obvious to me. it seems like the specific conditions under which the series of terms in the total expectation has a well defined sum is important, but I didn't fully grasp those conditions

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