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 The grass is always greener... (Posted on 2004-05-06)
You are told there are two envelopes. One contains twice as much money as the other one. You pick one but are allowed to change your mind after picking it. (You are equally likely to pick the one with less money as the one with more money.)

To figure out how much on average the other envelope should contain, one might average x/2 and 2x because one is equally likely to pick one as picking the other. Since this comes out to 5x/4, one might always change his or her mind. But wouldn't this end up with one never making up his or her mind?

 No Solution Yet Submitted by Gamer Rating: 3.8824 (17 votes)

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 Averages Can Be Misleading | Comment 2 of 46 |

As stated, the problem arrives at a false but nevertheless clever conclusion by using inconsistent definitions of the X variable to cause deliberate confusion.  One should not average X/2 and 2X.  Rather one should average X and X/2 (if we wish to define X as the value of the larger dollar amount) or X and 2X (if we wish to define X as the smaller amount).  When we use these two equations the average value of the two sums is always the midpoint (or average) of the sums in each envelope.  In the first case (with X being the larger amount) the average is 3X/4 and in the second case (with X being the smaller amount) the average is 3X/2.   Since one envelope is twice the other, the totals under the two scenarios are the same.

Gordon S.

 Posted by Gordon Steel on 2004-05-06 15:25:12

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