All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The Grass is Always Greener II (Posted on 2005-08-11)
In an earlier puzzle, you were handed two envelopes, one of which contained twice as much money as the other. After opening one, you were given the chance to swap. At first glance, it appeared that the your chance of getting more money could only increase each time the envelopes were swapped, but clearly this was nonsense: since there is no probability distribution which allows all real numbers to have the same probability, some values would have to have been more likely than others.

Suppose instead envelopes contain the non-negative integer sums 2n and 2n+1 with probability q(1 − q)n for some fixed q < 1/2

Now of course if the envelope you open contains a 1, you know the other must contain 2, so you ought to swap.

But you can do even better than this. Suppose you open an envelope and find an amount of money 2k

What would the expected value of the second envelope be?

 No Solution Yet Submitted by Sam Rating: 3.8000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 No Subject | Comment 1 of 7

Given that the first envelope (e1) has 2k, the expectation for the second envelope is

E(e2 | e1=2k)
= P(e2=2k-1 | e1=2k) * 2k-1 + P(e2=2k+1 | e1=2k) * 2k+1

simplifying, we get

E = [P(2k-1, 2k) * 2k-1 + P(2k, 2k+1) * 2k+1] /
[P(2k-1, 2k) + P(2k, 2k+1)]

= [2k-1q(1-q)k-1 + 2k+1q(1-q)k] / [q(1-q)k-1 + q(1-q)k]

= 2k-1 * [1 + 4(1-q)] / [1 + (1-q)]

= 2k-1 * [5 - 4q] / [2 - q]

expanding the fraction, we get

E = 2k + 2k-1* [1 - 2q] / [2 - q]

since q < 1/2, 1 - 2q and 2 - q are both positive. Thus,

E > 2k = e1.

This may seem like a paradox, but just because the odds are in your favor doesn't mean you'll always win. In fact, we know

2 - q > 1 - 2q,

so E(e2 | e1 = 2k) is between 2k-1 and 2k+1, as expected.

Also, note that before either envelope is opened, the expectation for the lesser of the two envelopes is

E(elow) = 20q(1-q)0 + 21q(1-q)1 + 22q(1-q)2 + ...

= q * [(2-2q))0 + (2-2q))1 + (2-2q))2 + ...]

and since 2 - 2q > 1, this sum diverges, and the expectation is infinite. This is still not a paradox, but I would sure like to play this game using real money ;).

Edited on August 12, 2005, 12:05 am

Edited on August 12, 2005, 12:06 am
 Posted by Josh70679 on 2005-08-11 23:56:08

 Search: Search body:
Forums (0)