A harmful big riddle...

College Majors Engineering Majors
1

<p>Let’s say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but once you open it you’re offered the possibility to take the other envelope instead.</p>

<p>Suppose you open one envelope and find in it the amount of money A. You can reason as follows:</p>

<li>The other envelope may contain either ½A or 2A.</li>
<li>The probability that A is the larger amount is ½, and that it’s the smaller is also ½.</li>
<li>If A is the larger amount the other envelope contains ½A.</li>
<li>If A is the smaller amount the other envelope contains 2A.</li>
<li>Thus, the other envelope contains ½A with probability ½ and 2A with probability ½.</li>
<li>So the expected value of the money in the other envelope is ½(½A) + ½ (2A) = 1¼A.</li>
<li>This is greater than A, so you gain by swapping.</li>
<li>Hence, you should swap whatever you see in the first envelope.</li>
</ol>

<p>Or you can reason:</p>

<li>The situation is symmetric</li>
<li>Hence, the expected value of switching is 0</li>
</ol>

<p>Obviously, these two conclusions can’t both be true. Should you switch or not?</p>

2

<p>I’m helping out all the people who don’t care.</p>

<p><a href=“http://en.wikipedia.org/wiki/Two_envelopes_problem[/url]”>http://en.wikipedia.org/wiki/Two_envelopes_problem&lt;/a&gt;&lt;/p&gt;

3

<p>Schrodinger’s Bank of America?</p>

4

<p>Mr Payne, you ruined it :)</p>

5

<p>Well I read the page cited above and haven’t the foggiest what anyone there is talking about. The way I see it is this: the probability/value equation is wrong because you are incorrectly defining the events. What is really going on is this: call the envelope with the smaller amount A and the one with the higher B. What you are trying to determine is the probability that you will end up with envelope B when you randomly choose one of them first. In other words what is the probability that you will choose the envelopes in the exact order A then B. When choosing, you can get A then switch to B or take B then switch to A. Since there are two possible results, AB or BA, each has a 50% chance of occuring. Thus the probability that B will be the second one chosen is exactly 50% and thus your chances that you will improve when you switch is exactly 50%, meaning the situation is symmetric.</p>

6

<p>I agree with drusba. Calculation of expectation value is correct but irrelevant.</p>

7

<p>“Since there are two possible results, AB or BA, each has a 50% chance of occuring. Thus the probability that B will be the second one chosen is exactly 50% and thus your chances that you will improve when you switch is exactly 50%, meaning the situation is symmetric.”</p>

<p>Yes, but when you improve you gain the entire value of your envelope and when you lose you only lose half the value of your envelope. Just because the probability distribution is 50:50 doesn’t mean it’s symmetric.</p>

8

<p>You swap until you have felt both envelopes. You then take the one that’s heavier. :)</p>

9

<p>…which was full of pennies! ;)</p>

10

<p>you should switch. you will win 2/3 of the time. (or was that the 3 doors problem? yeah, oops, i think i was thinking of 3 envelopes) </p>

<p>i remember doing something like this with a monte carlo simulation. (sp)</p>

11

<p>Isn’t this similar to the Prisoner’s Dilemna? Wouldn’t it be wise to switch then?</p>

12

<p>This is stupid. Once you switch on an even interval, you’re back with what you started.</p>

13

<p>The way I see it, you open up the first envelope, and you know how much money you have. The second envelope can either be one half OR twice as much money…meaning it is an unknown. You will on average gain 25%, but you could in fact gain 100% of the time or lose 100% of the time, and even though your chances of one is higher than the other, there is a discrete probability of both happening. You will not gain anything because on an infinite amount of tries of you with 2 envelopes where you go throug this thought process, you will have an infinite amount of 1/2A envelops and 2A envelopes, meaning overall, your net gain is zero. Negative infinite+positive infinity = zero, and you WILL approach both ends in time, no matter how long it may take(assuming negative infinite comes from money lost in the switch from 1A to 1/2A).</p>

<p>This is just my view on it, I dunno if its flawed or not.</p>

14

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