Conundrum: Two Envelopes
I overheard this once on a train and was never able to figure it out. Maybe someone here can help me.
Imagine I have two envelopes and I tell you truthfully that both contain money and that one envelope contains twice as much money as the other. I offer you your choice of envelope and you choose one of them without opening it.
Now I ask you if you would like to switch envelopes. You chose yours randomly, so it’s a 50/50 chance whether the other envelope contains half as much or twice as much. So, if the amount you now have is x, there’s a 50 percent chance that switching would get you 2x and a 50 percent chance it will get you x/2. You have twice as much to gain as you have to lose, regardless of how much is in your envelope, so it makes sense mathematically to switch envelopes.
But of course, this is ridiculous, since you have no new information about the two envelopes than you had before. Once you’ve made that switch, by the same logic, you should want to switch again. This much seems obvious. So where’s the flaw in the math above?
By the way, I consulted our good friend Wikipedia before posting this, and it was little help. It just mumbled something about Bayesian Decision Theory and said the problem would be easy if I were a mathematician. It then went on to pose a harder problem in which you can look inside one of the envelopes, and an even harder problem that was way over my head at 5:30 am. Thanks, Wikipedia.
April 10th, 2007 at 12:57 pm
This is a more succint explanation – http://en.wikipedia.org/wiki/Monty_Hall_problem
Did you look at this?
April 10th, 2007 at 3:16 pm
The Monty Hall problem is similar, in that it involves making a random choice and then having the option to switch, but it’s not the same situation. Actually, it’s not really a paradox; it’s a question that does have a correct answer, but one that seems counterintuitive to some.
The thing about our paradox is that there’s no way to make any distinction between the two envelopes. So why is it that we can construct a (seemingly) sound mathematical argument that switching is favorable? Where’s the flaw?
April 13th, 2007 at 8:50 am
You know, I think I got it.
It is true that it’s a 50/50 chance between two scenarios: A) the other envelope contains 2x, and B) the other envelope contains x/2.
But x isn’t really a fixed constant once we start doing probabilities. It’s true that there is a fixed amount of money in the envelope after it’s chosen, but without knowing what it is, I have to take both scenarios into account when I figure my odds. And in scenario A, x is only half the amount that it is in scenario B.
Put another way, let n=the smaller amount and 2n=the larger amount. If the envelope I chose is still x, then in scenario A, x=n, but in scenario B, x=2n. So I’m either trading n for 2n, or I’m trading 2n for 2n/2. No more paradox.
Any math people read this blog? Did I get it? At least it solves the paradox for me. Damn, that was driving me crazy.
I have to say that I started to move in the right direction when I was considering the similarities and differences between this problem and the Monty Hall problem. Thanks, DeLisa!
April 13th, 2007 at 8:54 am
This from the Wikipedia:
Okay, that’s the answer then. I didn’t understand what they were talking about the first time I read it, but now it makes sense. Just like Shakespeare makes a lot more sense after you already know what’s going on.