Sunday, 13 February 2011

The 2 Envelope Problem

You are on a gameshow – a pilot episode, so you have no idea about the level of prizes likely to be offered.  You answer a couple of questions correctly and are told to chose one of two envelopes. Inside the envelope you chose is $300 and the host tells you that one of the two envelopes has twice the amount of money in it than the other one.  He asks you if you wish to swap for the other envelope – i.e. instead of $300 you could win $150 ($150 less than you currently have) or you could win $600 ($300 more than you currently have).  It would seem to be a 50:50 likelihood of either outcome.  50% chance of losing $150 and 50% chance of gaining $300 – anyone familiar with gambling, or mathematics, will tell you that on average you will make a profit if you take this gamble therefore you should swap to the other envelope.

The problem with this reasoning is that it would apply whichever envelope you picked – the odds of your picking the envelope with the higher amount in it is 50%. swapping envelopes should still give you 50% chance of getting the higher amount.  Is there any gain in swapping?

So far as I know, there is no consensus over a solution to this paradox.  The "Cover's Strategy" is obviously rubbish if given a little thought - of course it works as a strategy, but it does not explain the paradox.
If an explanation does not exist then which predicate is false?  This has troubled me for over 10 years.


HavenNyx said...

Or.... Either way you have more money than the $0 you started off with, so does it really matter which envelope you choose? Either way, you win.

Anonymous said...

You've said the show is the pilot, and I am the contestant. You mention no other contestants. So it's just me (for now).

The game implies one of two outcomes. Either the game will increase the extremes, such as next set will be $1200 vs $75 until a zero point (failing the game) or there is a limited number of rounds/amount of money you can have.

If the goal is to make the best profit, I would take the $300 and go have a fun weekend of their dime. However, since this is a game show, the pilot no less, and I'm one vain SOB and would like to be on TV, I'd most likely make it my goal to last as long as possible. I have money, outside of the context of this game, and $600 is a drop in the bucket. I'd rather get caught up in the drama since I'm not worrying about paying rent on time, you know?

The safest bet is no bet. But safe is boring.

The gain for me would not be monetary, but excitement. And that would be a guaranteed gain.

ResCogitans said...

you're both thinking about the social context of the situation, which is fine if the question is "what would you do?". However, the point of the contrived situation is to frame a mathematical / statistical / game theory paradox where the question is "what should you do?"

if you play a game with a fair coin where if you flip a head you will win $1 and if you flip a tail you get nothing, then how much would you pay to play? 50c for you to break even if you play the game repeatably. The maths behind games like that is very simple and applicable to the 2 envelope paradox (google it - it's famous!) and yet there is no consensus about how to resolve it...

Anonymous said...

So what was your hope in illustrating the point? Or were you simply venting?

You can break it down out of social context to pure math, but that's not how I approach things. I imagine that's not how you approach life either. Risk is fun. To take the human element out of it is to take the reason for proposing the question in the first place out of it.

It's not a simple matter of, is this mathematically advantageous or not? If it was, there would be a consensus. Gambling (which is exactly what this is) is about having fun and taking risks. Otherwise, why bother? You wouldn't. It's because you, nor I, am a machine.

Anonymous said...

Are not machines*


My first point in the third body of text is, when you put it in those terms, where it can be either or, with pure chance as the factor, not good odds versus bad odds, then the practicality of applying it out of social context is mostly nil. You already know the chances, and social context becomes everything after this point. Not something to cull for the sake of objectivity.

ResCogitans said...

Basically i was venting - it really bugs me:
If an explanation does not exist then which predicate is false? This has troubled me for over 10 years.

Actually i do approach most of life like an equation: a cost/benefit analysis. That is something you have talked about before I believe so I'm a little surprised at your comment.

As to whether we are machines... well where do you draw the line between a biological construct and a machine?

Anonymous said...

Yes, I do calculate life as cost/benefit. But I don't eliminate the human element of "fun". It has a high value in that it eliminates boredom, something you should be well acquainted with as well. If you don't well, then I find that even more odd...

When the AI for machines is advanced enough to to start disregarding rules for a laugh, a chuckle, or a smile, then I'll say we have reached uncanny valley in regards to sentience.

ResCogitans said...

as coincidence has it... for my next post i was planning to write something about laughter.

i do have fun, and can be very spontaneous so long as it is during a time i have set aside for spontaneous activity ;)

Anonymous said...

You made me laugh.


This isn't laugh designated time. That's after lunch.

Odatafan said...

The not-so-clever
solution is this: switching at $300 is indeed a good idea. Switching at $300/64 is obviously a good plan. But if you see a value so large that it surprises you, then it's a bad idea to switch. If the amount X in the envelope surprises you because it is very large... that's because you guess than the average envelope contains much less. The fact that you saw X is consistent with one of the following:

The show is richer than I thought, and the prizes are X and X/2.

The show is even richer yet, and the prizes are X and 2X.

Clearly, the first one is more likely.

ResCogitans said...

yes, there are strategies in real life to deal with it, but my problem is with the mathematical paradox of it. if the envelopes have an infinity of possibility then by symmetry, your expectation value when you swap should be zero, so p.2$ - (1-p).$ = 0. But this means that you seem to be twice as likely to have initially chosen the lower amount to the higher...
i definitely believe that symmetry means your expectation value means swapping sums to zero, but then that leads to...

read your post on it and i'm none the wiser - you've only succeeded in getting me all angsty over this paradox again! :p
your blog looks interesting - don't have time now but will check out more soon.

Odatafan said...

I got a good critique: consider that the game show is in an unknown currency, and then the numbers get translated into money-as-we-know-it only after we decide whether to switch or not. Against that, it seems hard for a player to guess a probability distribution over all possible values. If the player were able to guess a distribution, then that distribution would lead the player to think "I have the higher value" with sufficiently high probability the the expected value of switching is 0 or less. Here's a mathematical argument to that effect:

If you see a value so large that it surprises you, it can be a bad idea to switch. If the amount X in the envelope surprises you because it is very large... that's because you guess than the average envelope contains much less. The fact that you saw X is consistent with one of the following:

> The envelopes are better than I thought, and the prizes are X and X/2.

> The envelopes are much better than I thought, and the prizes are X and 2X.

If your "surprise curve" says that odds of those two conditions are, say 9:4 in favor of the first case, then you should not switch, as the costs of switching down exceed the benefits of switching up. Now, suppose you are surprised to see $3000 (or suppose the number "a googol of game currency" surprises you), but you figure "I would be more surprised to see twice as much, but not twice as surprised," then you are assuming that the distribution of
"what the value could be" is very flat. But, in fact, that distribution is so flat that it doesn't have a finite integral. That is: the distribution of "what a value could be" has to be
rather tight in order to be integrable. You believe in integrability if you believe "the probability that the prize is between 1 and a million coins is greater than 0.00001" or any
similar statement, with the parameters "1", "a million" and "0.00001" replaced by anything.

I think that one reason this paradox is compelling because whatever your distribution, you should definitely switch from a median value. This is balanced in the expected-value calculation by the great sums you would lose by switching from a high value. Another reason is that in the absence of any information, we may be completely unwilling to use a distribution at all.

ResCogitans said...

so imagine YOU are on a gameshow and win one of the envelopes. the host doesn't even tell you how much is in it, but still asks if you want to swap to the other one.
Do YOU think there is any point in swapping?

Odatafan said...

> ... the host doesn't even tell you how much is in it, but still asks if you want to swap to the other one. Do YOU think there is any point in swapping?

No, I think there is no point to swapping before seeing the value.

ResCogitans said...

he tells you there are 2000 gameshow points in it - points are redeemable at the end of the show. this is the pilot episode so you don't know whether 2000 is a lot or not.
at this point:
0. when given the initial choice of envelopes you have a 1/2 probability of choosing the higher amount.
1. as you don't know what points are worth, there is still no point in swapping.
2. if there is no point in swapping, then the expectation value of swapping is 2000.
3. if the expectation value is 2000 then 2000 = p.1000 + (1-p).4000
4. the probability of swapping and getting 1000 is 2/3.
5. this calculus holds whatever value was in the first chosen envelope.
6. when given the initial choice of envelopes you have a 2/3 probability of choosing the higher amount.

0 <> 6

which statement is false?

Odatafan said...

> which statement is false?

** 6 does not follow from 12345.

That's a very compelling argument. You asked where the argument breaks down. I'd say that, in particular, 6 does not follow from 5.

From p(y|x) for various x you want to infer p(y). For let y be the case "the other envelope is better," and let x be the condition "the value in the envelope is X." If we had an integrable prior distribution of envelope values, we could reason using the "law of total probability" that
p(y) = sum_x p(y|x) dX. When the distribution of X is not integrable, do you want to reason according to a law such as the following?

Law?: if p(y|x) is constant as x ranges over all possible conditions, then the total probability p(y) takes that same value.

When X is not integrable, we can gather conditions x into pairs and find that the probability p(y|pairs of x) is twice as good; then p(y) would take on a larger value by that same law. That law is not valid.

I agree that 1+2+3+4+5+6 contradict 0. I think that 1..5 are clever reasoning -- teasing out the consequences of "indifference." I don't think 1..5 are logically necessary at all. You have made a compelling argument about
the implications of indifference: what to do if you don't know what gameshow points are worth to you, and you only know that more points are worth more to you.
I think that once you make the assumption that utility is linear in gameshow points and that p(y|x) is constant , then I think the chain of reasoning 1..5 is correct, leading inexorably to the conclusion 5. I wish that anyone reading this argument would become slightly more open to rejecting the assumption of indifference. In order to *strengthen* that effect, one could go on reasoning after 5, reaching problematic conclusions about small amounts of money and for large amounts of money. I will try to do that now.

** Nasty stuff follows from 12345 at small values.

Is there a minimal value? 1...5 constrains the distribution of envelopes so that for all x a value of x/2 is always twice as likely as 2x. So, there is no minimal amount of money in this game, and the probability of tiny amounts outweighs the probability of large amounts. Perhaps part of the paradox is being pushed into the gap between 0 and arbitrarily small values. As a result, gameshow points cannot translate into real money. We could suppose that there is a constant c, such that gameshow points times c = currency in your hometown. But currency in your hometown doesn't exist in arbitrarily small amounts. So there has to be a minimal gameshow value. But by arguments 1..5 there is none.

** Nasty stuff follows from 12345 at large values.

The implied distribution seems to be X^(-0.5), since for that distribution the chance of switching down is twice the chance of switching up. This distribution is not integrable. Not only does it put tremendous weight on small values, also puts weight at very large values (or at infinity).

How is the game show supposed to generate its random envelope values? It must have, in fact, an integrable distribution. More likely, it has a bucket full of envelopes -- a discrete distribution. The customer may well not know it, but some such distribution must exist in order to create the game show.

I had a few hundred extra characters to say, so I put it all at

Best Regards,

ResCogitans said...

I'll grant you that point 5 is definitely suspect, but surely point 4 is the same as point 6?

Of course there could be a probability distribution for how high an envelope value can be, but the probability factors of 1/3 and 2/3 are hardwired in and so such a distribution would have to be very specific - whereas in reality it never will be.

With moral paradoxes I'll happily conclude that the premise of morality is wrong. And it is probability paradoxes like this that lead me towards concluding that 'probability' itself is a shaky concept at times.

I defy you to come up with a definition of the word 'probability' such that i couldn't come up with a scenario of the form "the probability of event X is p" which isn't covered by your definition!

Odatafan said...

Hi, sorry I didn't cite this blog when carrying the argument over to the other page. It cites it now. I answer at length there; to summarize:

Are points 4 and 6 the same? I think 4 is a conditional probability and 6 is a "total" probability. They can definitely be different. In the context of an integrable distribution, they are related in a simple way: p(y) is the weighted average of p(y|x) as x ranges over conditions representing possible values.

What is a notion of probability? I think that it begins with an integrable distribution. Wikipedia's article "probability theory" begins that way.

Best Regards,