A man comes up to you on the street that you've never met. He generously offers to play a game with you. He says that he's going to take two envelopes and fill them with money. One of the envelopes is going to contain twice as much money as the other, but he doesn't tell you what either amount will be. He will then give you one of the two envelopes randomly with a 50/50 probability of getting either. He will let you look in it, and decide if you want to keep that money or if you want to switch envelopes and be forced to keep the amount in the envelope.
Let's say that you open the envelop and see $100. Should you stay and be happy with your envelope, or should you switch and potentially get $200 or end up with $50.
Well, half the time you have the higher envelope and half the time you have the lower envelope, so your expected value for switching is:
E = (1/2) * $200 + (1/2) * $50 = $125
Your expected value for switching is more than the $100 that you would always get by staying. So, based on the math, you should switch envelopes. Easy problem, right?
But, if you think about it, the situation makes absolutely no sense. Nothing was special about $100. For any amount X, the expected value of switching is 1.25*X, so we should conclude that we should always switch when we are given the envelope. Thus, we can mathematically switch envelopes even before we look. The paradox lies in the fact that there is a natural symmetry between the envelopes. Since we no nothing special about either one, how could it possibly be preferable to switch? They are both equally likely to be the bigger one, so we should really break even by switching. It should gain us no advantage. But the math is simple and clear. So, what's going on here.
This is known as the two-envelope paradox. It's pretty interesting, I think, and it really stumped me for a while when I was first thinking about it. The solution is somewhat non-trivial but is also enlightening. So, I encourage you to think about the problem and to see if you can figure out what's wrong here. I'll post my take on the solution later.
Incidentally, Wolfram Alpha is now online, and I'm playing with it a bit.
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The man holds 2 envelops, with values either x or 2x. If you pick x, then by switching you will get 2x dollars. If you don't you will get x, NOT x/2 as the problem suggests.
ReplyDeleteE[switching]($) = P(picked x) * 2x + P(picked 2x) * x = 3/2 * x
E[staying]($) = P(picked x) * x + P(picked 2x) * 2x = 3/2 * x
Both are the arithmetic mean of the money in the envelops, there's no problem!
It's misleading to say that the envelopes should break-even. The answer of switching makes perfect sense, you have a 50-50 shot at DOUBLING your money, and the downside is you only lose half. You'd only expect the envelopes to even out if the downside was that you'd lose everything, but either way, you're going away with money. The confusion comes from assuming the positive and negative effect cancel.
ReplyDeleteSo yes, 50% of the time I'll make the wrong choice and end up with 50 instead of 100, but on the other end I have a 50% shot at getting 200. so you should ALWAYS switch, even (as you said) without bothering to look into the envelope the first time.
the problem makes more sense when you look at it this way, instead of two envelopes with an unknown sum, the guy has 3 ($50, 100, 200), he hands you the middle envelope ($100) and says you can keep it, or trade for one of the 2 other envelopes, ($200 or $50) it becomes much clearer in this case that you should always switch, because you're expected value of switching is the same as you stated before.
E=(1/2)*50 + (1/2)*200=125.
Now, if he said instead of $50, you get nothing, then it would even out and not make a difference.
summary: automatically switching is the correct choice
wait....god damn you herbie, I've been lying in bed trying to sleep for the last 2 hours but instead this problem won't leave my head, even after i thought i had it figured out...
ReplyDeleteI think i'm only just understanding the real paradox part of it, which occurs when you shift how you declare your unknowns.
For instance, the above explanation assumed that you declare your envelope X every time. and no matter how many times you simulated the exercise you always started with X, and the other envelope was either 0.5 or 2 times X, then the simulation plays out as above.
Yet the problem could be thought of this way: There are two envelopes, X and (1/2)X, and if you did the exchange with the man 1000 times, half the time you'd start with X, the other half you'd start with (1/2)X, although you wouldn't know which you had at the time. In this view, switching envelopes would be meaningless, because you'd always come out to an average (expected value) of .75*X
So, in conclusion, the cat exists in both it's alive and dead states simultaneously until you look in the box and it retroactively chooses one.
Its almost 3AM and i'm thinking about envelopes with money in them...FML