## The Two Envelope Problem |

**Chris Hardin**

Union College

October 17, 2011

5 pm

Bailey Hall 207

Refreshments will be served in Bailey Hall 204 at 4:45

Suppose that $x$ dollars are placed in an envelope, and $2x$ dollars are placed in another. The envelopes are shuffled and then labeled A and B. You will get to keep one of the envelopes, and you can peek inside one of them before deciding. Suppose you open envelope A and see $\$20$. Should you keep A or switch to B? Naively, there is a 50% chance that envelope B has $\$10$, and a 50% chance that it has $\$40$, so on average B has $\$10\cdot 0.5 + \$40\cdot 0.5 = \$25$. (There's a problem with this naive analysis, but it can be fixed.) So, you should switch, right? For any amount you see in A, you would come to a similar conclusion, so you should switch to B without even bothering to look in A, right? On the other hand, the envelopes were shuffled, so how could always picking B be a better idea than, say, always picking A? This is the \emph{two envelope problem}. We will look at a couple of common but wrong resolutions to this paradox, and one resolution that we argue is correct.

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Union College Math Department Home PageComments to: math@union.edu Created automatically on: Sat Mar 17 08:29:19 EDT 2018 |