January 13, 2003
One Hundred Interesting Mathematical Calculations and Puzzles, Number 10
One Hundred Interesting Mathematical
Calculations Calculations and Puzzles, Number 10: The Grass Is Greener Paradox
Someone standing behind a table shows you two envelopes. She tells you that one has twice as much cash in it as the other. She tells you to pick one and open it. You do. You discover that it has, say, $50 in it.
She then asks you if you want to switch. You reason as follows. Either the two envelopes had $25 and $50 in them, or $50 and $100. There is a fifty percent chance that I picked the lower-cash envelope, and a fifty percent chance that I picked the higher-cash envelope. So there is a fifty percent chance the other envelope contains $100, and a fifty percent chance the other envelope contains $25. On average, since ($100+$25)/2 = $62.50, I stand to gain $12.50 if I switch.
You say that you want to switch. The woman offering the deal points out to you that if you had chosen the other envelope, the same chain of reasoning would have led you to prefer the envelope you are holding, so that you should keep the envelope you originally chose.
Which envelope should you choose, and why?
From John Allen Paulos's Innumeracy.
IMHO, the easiest way to see what is going on is to change the game slightly in a couple of ways. T
First, the woman announces not only that one of the envelopes contains twice as much money as the other, but that there is a maximum amount of money in the game--that the most that any of the envelopes contains is $800--and that there is one chance in 10 that the higher-cash envelope contains each of the following amounts: $800, $400, $200, $100, $50, $25, $12.50, $6.25, $3.125, $1.5625. Then it is absolutely clear what you should do: if you pick an envelope and it contains $800, stand pat: you have the higher-cash envelope. If the envelope contains $1.5625, you know to switch: you have the lowest-cash envelope. In between, the reasoning that led you to want to switch is correct--you know that since you didn't pick the highest-cash envelope, it is worth switching.
Second, the woman announces not only that one of the envelopes contains twice as much money as the other, but that the amount of money in the higher-cash envelope was arrived at through the following procedure: she started with a baseline amount, and flipped a coin. If the coin was "heads," she stopped: that's how much money is in the higher-cash envelope. If the coin was "tales," she doubled the amount of money in the envelope and repeated the procedure.
If the baseline amount was (say) $25, that gives the following distribution of probabilities for the higher-cash envelope:
And so forth.
Now if you pick an envelope and it has $50 in it, it is clear that you face a fair bet. It is twice as likely that you have the higher-cash envelope in a $50-$25 division as that you have the lower-cash envelope in a $50-$100 division. If you switch, you have to expect that 2/3 of the time you'll get $25 and 1/3 of the time you'll get $100--so that there is no expected gain from switching because the expected amount of money in the other envelope is $50, the same as the amount of money in the envelope.
Note, furthermore, that this particular probability distribution of money among envelopes--twice as much with half the chance--can only be achieved if the woman setting up the test has an infinite amount of wealth. Think about what the expected value of the money in the larger envelope is: $25 x 50% + $50 x 25% + $100 x 12.5% + $200 x 6.25% + ... = $12.50 + $12.50 + $12.50 + $12.50 + $12.50 + ... = infinity.
The source of the catch in the problem is the claim that you know nothing about the situation other than that one envelope has twice as much money in it as the other. That can only be true for all possible amounts of money in the envelopes if the tester has access to infinitely many dollars. If you know that the tester has finite resources for the test, the following decision procedure suggests itself: Is the amount of money in your envelope large relative to what you think the resources of the tester are? If so, then stand pat. You've probably been lucky. Is the amount of money in your envelope small relative to what you think the resources of the tester are? If so, then switch. You've probably been unlucky. (More precisely, perhaps, ask yourself the question: Is the chance that the original division was $X-$2X more than half as large as the chance that the original division was $X-$X/2?)
Posted by DeLong at January 13, 2003 05:24 PM
Assuming that the grass is greener is essentially ignoring the information you have on the finiteness on the value you could expect. The only way to entirely exclude information on the average amount is to expect it to be any number between zero and infinity.
Finding a finite amount of money in the first envelope, you then conclude you were extremely unlucky to badly underperforme the average. An average which you think is infinite with a probability very close to 100%. Therefore, silly enough, you switch envelopes.
What you should do after opening the first envelope is to admit that you then have information on the average expected value. You know it is finite. Finding 50$ your best guess is that the average is 50$, so you don't change. You conclude that the envelopes hold 25$ and 50$ with 66% chance or 50$ and 100$ with 33% chance.
There is an upper limit, in order to convey information on that, the amount you find have to have a better chance than 50/50 to actually be at this limit. The pessimist is the realist!
~". She tells you to pick one and open it. You do. You discover that it has, say, $50 in it.
She then asks you if you want to switch. "~
Is it part of the game design that the opportunity to trade envelopes must always be offered? Or is there any possibility that the player might (rightly or wrongly) infer that the game-master knows which envelope holds the greater prize and makes a CONDITIONAL offer?
The outcome table -- assuming conditional offers and a game master attempting to minimize payouts --
50% the player picks the lesser amount: game over.
50% the player picks the greater amount -->
2nd round, and ANY percentage of agreement
with the offer to switch reduces the payout.
If the player infers that the game master's offer might be conditional, then his incentive to stand pat on his first choice increases with his increasing suspicion.
If itís fair game in setting up Newcombís Problem to posit a more or less infallible alien predictor, I donít see why
itís so unreasonable here to stipulate in the set up of this problem that the Ďsomeoneí standing behind the table is an alien (or a god) with limitless resources. It certainly makes the problem more interesting, since the kind of solutions suggested so far, which are certainly good responses to the problem as phrased, now don't seem to help.
While everyone is compiling puzzles involving probability and philosophy, one of the nicest puzzles in this area is Bertrandís Paradox.
From a given circle, choose a random chord. What's the probability that the chord is longer than a radius?
Itís worthwhile trying to figure out just what this is before reading through that paper, and then comparing your results to the responses they discuss.
I read Brian's comment and involved with the random 'chord problem' with great interest. I still don't see how this relates to the random envelope problem.
What I think lies in the envelope-paradox is that it tricks you into using different methods in linking the average expected amount of money in the envelopes to what you expect from the first and the second envelope.
Maybe you start by feeling that 50$ is small in a situation like this, in the famous TV-show problem with two goats and a cadillac, or in lotteries in general 50$ is not much. Therefore you label this envelope the 'observation' envelope, and think up a way to 'infer' that the expected amount in any envelope actually is higher, in line with your original feeling. The second envelope on the other hand is the 'production-envelope', in wich you expect the expected amount, and you swap.
But if you had found let's say 50*10^14432 $ - would you really care to swap then?
The random chord problem wasn't meant to help here, it's just another fun problem in this area.
Getting back to the envelopes, I don't think thinking about absurdly high values like 50*10^14432 helps matters much. Indeed, even in practice, once the amount in my envelope gets that high I'd probably throw out my prior probability distribution about what resources the other agent had, and so I might be tempted by the argument to switch :)
The reason that thinking about the player with unlimited resources is helpful is that it forces us to confront a possibility (a very very unlikely one, but still a possibility) where some propositions we might have thought inconsistent are all true.
(1.1) If I find $25 in my envelope, I'll be in a position where it is in my interests to switch.
(1.2) If I find $50 in my envelope, I'll be in a position where it is in my interests to switch.
(1.n.) If I find $25 * 2^(n-1) in my envelope, I'll be in a position where it is in my interests to switch.
(2) I know that I'll either find $25 or $50 or ... or $25 * 2^(n-1) or ... in my envelope
(3) I'm not right now in a position where it is in my best interests to switch.
The paradox arises because we quite naturally infer from the (1)s and (2) to not-(3). Intuitively, we reason if I'll be better off doing A than B whatever happens next, I'm better off doing A than B. What the paradox shows us is that this argument form, as tempting as it is, cannot be generally valid.
Obviously this kind of situation is quite unlike any we're going to find ourselves in anytime soon. But just how much that should matter to theoretical discussions of reasoning about uncertainty is not entirely clear.
Focusing on the alien envelope problem, it seems to me that this is a mathematical question rather than a statistical one. Cleaning it up slightly, let's rephrase the paradox like this.
1. Pick with equal probability one of the integer numbers > 0.
2. Assume you got the finite number x.
3. Pick one more number like in 1.)
4.This number is almost certain to lie in the infinite range > x, and almost certain to NOT lie in the finite range between 0 and x, i.e. this number is almost certain to be bigger than the first number you picked, the x.
The paradox is that you are almost certain to pick a lower number before a higher number, even though you choose randomly.
My point in quoting an astronomical number above was to demonstrate the purely theoretical, non economical, character of the alien envelope problem.
Trying to solve it though, would probably involve attacking point 2.) above. You are of course almost certain of NOT picking the number x, or anything below it, rather you would be almost certain to pick anything above it. This is why you wish to switch envelopes in the alien problem.
More importantly, the almost zero likelihood of picking anything as low as x as the first number, leads us to conclude that the chain of events 1. to 3. almost never will happen.
Therefore, contrary to the claim of the paradox, there is no way to consequently get smaller numbers before larger if you choose them randomly.
I agree the paradox wouldn't be interesting if it was just another puzzle that arose by assuming that it is possible to pick an integer at random in a way that any integer is equally likely.
There's a way to get the alien version of the paradox working without assuming anything quite so incoherent. As Brad pointed out, just use a St Petersburg method to work out what goes in one of the envelopes, and then toss a coin to determine whether twice as much or half as much goes in the other envelope. The results aren't quite as dramatic, but still paradoxical. If my back of envelope calculations are right, then when you see any amount x larger than $12.50, then your expected profit from switching is x/4, and if you see $12.50 you know you'll profit $12.50 from switching. Still, you shouldn't switch.
In this case, as x gets large, the probability of getting an envelope with less than x in it tends to 1 rather quickly. The probability of getting an envelope with less than $1000 in it is around 0.99.
This only resembles a zen riddle if you make it
This problem becomes easier if you cut out
irrelevant information. It states the alien is
all-knowing but that is not analytically useful
because of predestination. Box configuration is
only relevant in setting options.
So someone offers you $10 if you are the kind
of person to take $10, $1,000,000 if you are not,
nothing if you are and claim you're not and
$1,000,010 if you are but take the 10 anyway.
What IS analytically useful is that xhsbr (Read:
he) has a perfect computer model of your mind so
he knows which option you will choose beforehand
and that translates into infallability. For all
intents and purposes he cannot be wrong.
So the alien is infallible? Cut out the
What IS the kind of human who would not
take $10? That's free money. The numbers are
irrelevent. We must assume the kind of person
not to take free money does not exist.
In order for this experiment to be necessary
to prove a hypothesis, we must cut out the
free money factor by understanding that
"kind" to be the kind of human who would
not take $10 instead of a million.
Ratio is also unimportant as more money is
always better than less money and there is
no probability involved.
To summarise: The alien is infallible. All
humans want to maximise profit. 1M is better
than 10 which is better than 0.
So .. $0 ? Not possible since if you don't take
the 10 you get the 1,000,000.
Only by being the kind of cheater who would
try to take both where that would be logically
impossible would you get $10.
I'm a one-boxer.