Mar. 26th, 2005

Weird math

Mar. 26th, 2005 08:21 am
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I discovered this conundrum yesterday in a discussion with coworkers about a brainteaser that one of us half-heard but didn't hear the end of.

Let's say that I give you an envelope and I tell you (or you look and see) it has x dollars in it. Let's say 100. Then, I have another envelope. This envelope contains an amount that has 50% probability of having twice what you have and 50% probability of having half what you have, I tell you. Would you like to switch envelopes? The expected value of the envelope is (0.5)(200) + (0.5)(50) = $125, which is more than what you have, so of course you switch. Actually, even without doing the math you can see the situation is a bit better than double-or-nothing, so you would still say switch. As usual with these puzzles, we're assuming you are risk-neutral and not, for example, in desperate need of $90 to take your beloved pet to the vet.

So, great, we switch. You now have an envelope containing either $50 or $200 and I have an envelope containing $100. If I ask you if you want to switch again you should certainly say no. And yet, here's what baffles me: again it is entirely correct for me to say that the envelope I have has a 50% probability of containing twice what you have and a 50% probability of containing half what you have.

The English is the same. The math is reversed. And so, if I gave you a sealed envelope and told you that sentence, you should really have no idea what to do since you don't know which envelope's contents is known and which isn't.

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