# A nifty little parodox. Try and crack it!

Discussion in 'Off Topic [BG]' started by Guss, Dec 3, 2002.

1. ### Guss

Suppose you're on a gameshow where you can choose either of two sealed envelopes, A or B, both containing money. The host doesn't say how much money is in each, but he does let you know that one envelope contains twice as much as the other.

You pick envelope A, open it and see that it contains \$100. The host then makes the following offer: you can either keep the \$100, or you can trade it for whatever is in envelope B.

You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, then, you stand to either win \$100 or to lose \$50. Since you stand to win more than you stand to lose, you should switch.

But just before you tell the host you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. But that can't be right.

What's wrong with your reasoning?

2. ### savagelucy

Apr 27, 2002
uh...i dont mean to sound stupid..but.. whats a paradox?

3. ### BTBbassistjoin us for mankala hour!

Apr 20, 2002
Westlake Village, CA
A paradox is basically a seemingly contradictory statement that may nonetheless be true.

I'm still working in it, give me some time. Good one though.

4. ### NicK tHe GK MaNBanned

a paradox is something thats impossible to solve and hmmmmm well yea u would choose the other envelope no matter which one u get the way i see it either way yer going to win money so why not go for the gold

5. ### pmkellySupporting Member

Nov 28, 2000
Kansas City, MO

P@

6. ### pmkellySupporting Member

Nov 28, 2000
Kansas City, MO

what really is the chance that the game show host will tell you such a thing, true or not? the whole point is to pit people against taking what they have and risking it against either ultimate rewards or giving it all up on a chance...

P@

7. ### Joe NerveSupporting Member

Oct 7, 2000
New York City
Endorsing artist: Musicman basses
ok i must be really bored because i read this 3 times.

you SHOULD switch regardless of which envelope you get, and the odds are with you regardless. why should the odds change if you choose the other envelope????? why does the last supposition invalidate the first argument? am I missing something here .

why can't it be right - why is there a problem with the reasoning????

8. ### jobu3Artist formerly known as Big Joe

Feb 17, 2002
Mountain Top, PA
i would take the 100 dollars and have a nice down payment for a bass or cabinet and be happy!

9. ### jonasp

Aug 7, 2002
Duluth, Minnesota
Agreed! I think it is in human nature to not accept what we get. We aren't happy till we know whats on the other side (or in the other envelope).

I only responded to this thread so people will (hopefully) think i'm smart.

10. ### BrendanSupporting Member

Jun 18, 2000
Austin, TX
Unfortunately, we saw through your not-so-clever ruse.

11. ### rabid_granny

Jan 20, 2002
You're damned if you do,
You're damned if you don't.

You only live once. Go for it!

12. ### Aaron

Jun 2, 2001
Bellingham, WA
I'd take the \$100 for two reasons:

a. There might be a favoritism towards the \$50 because it is a little more of a round number than \$200. How often do you see numbers that aren't rounded to a common number on game shows?
Think of monopoly money 1, 5, 10, 20, 50, 100, 500...

b. the game show owner is probably a cheap bastard

13. ### Bryan R. TylerTalkBass: Usurping My Practice Time Since 2002Staff MemberAdministratorGold Supporting Member

May 3, 2002
Connecticut
The problem with the reasoning is that you cannot come to the same conclusion choosing envelope B. Envelope A has \$100. Therefore, envelope B must contain either \$50 or \$200. And if it has either \$50 or \$200, then the resulting multiplication or division by 2 changes the options.

Envelope A= \$100 so either \$50 or \$200 in B
Envelope B= either \$50 or \$200
If it contains \$50, then you would have to believe that envelope A contains either \$25 or \$100
If it contains \$200, then you would have to believe
that envelope A contains either \$100 or \$400.

Switching them will not result in the same yeilds, so that is why the logic is flawed.

14. ### jokerjkny

Jan 19, 2002
NY / NJ / PHL
dood, you rock.

15. ### John K.Guest

By croikey, I think the man did it.

16. ### Bryan R. TylerTalkBass: Usurping My Practice Time Since 2002Staff MemberAdministratorGold Supporting Member

May 3, 2002
Connecticut
Thanks guys. Have yet to find out if I'm correct though.

17. ### Fretless12ver

Nov 28, 2002
Nampa, Idaho USA
You have a 50% chance of picking either A or B. It is irrelevant which one you pick to start. This is because if you pick A, then B could be either half or double the amount of A. But if you pick B, then A could be either half or double the amount B. Same argument either way. Let's say you pick A which contains \$100. Envelope B, which represents 50% of your original choice, has a 50% chance of being half the amount and a 50% chance of being double. The 50% probability of getting \$50 gives an expected value of \$25. And the 50% probability of getting \$200 gives an expected value of \$100, exactly what you already have. Stick with envelope A!

18. ### geshel

Oct 2, 2001
Seattle
I got it! I got it!

The original logic is flawed because we don't really know what the odds are that any envelope can have \$100 in it, is. If we assume the prize money is unlimited, we get the feeling that any value is equally likely (and, therefore, the odds are 50/50 that the other envelope is double or half).

But, first let's start with a simplified case, where there's a cap on the prize money. Let's say the Lesser envelope contains between \$1 and \$N. The Greater envelope contains, therefore, the corresponding double amount ranging from \$2 to \$(2*N).

So, if you open the first envelope and it's greater than \$N, you already know the answer: it's the Greater envelope. But, let's look at the case for now where the value is between \$1 and \$N.

What are the odds that you opened the Lesser envelope, given that it contained \$X? Well, using Bayes' Rule: The probability that you picked the lesser envelope, GIVEN that the one you picked had X (X<=N) dollars in it, is:

(A. the probability that you picked the Lesser envelope in the first place) * (B. the probability that the Lesser envelope contained X dollars) / (C. the probability of finding X dollars in either envelope)

A. This is 1/2 - since you picked randomly

B. This is 1/N : it can contain any value from 1 to N inclusive, so odds that it contains X is 1/N (for example, pick a number from 1 to 10, inclusive. The odds you picked 5 are 1/10) ***

C. This is 1/2 * 1/N + 1/2 * 1/(2N) : the probability you pick the Lesser envelope times the probability it contains X dollars, plus the probability you pick the other envelope times the probability it contains X dollars

So we have (1/2 * 1/N) / (1/2 * (1/N + 1/(2N))) which reduces to (1/N) / (3/(2N)) and further to 2/3. So, the odds are 2/3 you picked the smaller envelope. Remember though this is for the cases where X<=N.

For these cases, the expected value if you switch to the other envelope is 2/3*2X + 1/3*X/2, or 9/6*X.

For the cases where X>N, the expected value if you switch is X/2 (easy, since we know you picked the Greater envelope, no matter what the other one has half of what you have).

So, since odds are even as two whether or not X>N, the overall expected value if you switch is 1/2 * 9/6 * X + 1/2 * X/2 = X. This is exactly what you'd expect if you don't switch (again, easy: you stick with what you got, X).

Since this answer has no dependency on N, we can increase N as high as we want without changing the result. Odds are even if you switch, just like they should be.

I realize that this was rather drawn out and probably won't make a lot of sense to someone who hasn't had a probability class. I'll try to figure out a way to explain it in layman's terms.

I guess, suffice to say it this way: the two envelopes aren't equal, since the realm of possible amounts in one of them is twice what it is in the other one.

*** this is not really right for a continuous case, where X can be \$4.248294 or something. But it still works out: odds of finding X in the Lesser envelope (X<=N) are still half those of finding X in the Greater envelope). So if it's p/N, then the other probability is p/(2N)

19. ### geshel

Oct 2, 2001
Seattle
You missed a step: the overall expected value for switching is the sum over all cases, or \$125. This is what is said in the original problem, though in normal language: you can bet \$50, on a 50/50 chance, but with 3:1 odds (you'd win \$150 if the other envelope has \$200). This is a good bet, hence you feel like switching. But, as I've tried to demonstrate above, the odds aren't really 50/50 because the probability curves for the two envelopes aren't identical (the Greater envelope has only half the chance of containing any particular dollar amount that the Lesser envelope does).

20. ### j b scott

Nov 14, 2002
Melbourne, Australia
geshel, maybe you should play some sport or something...