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<h2 class="hd hd-2 unit-title">The Two-Envelope Paradox</h2>
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<h3 class="hd hd-2">Introducing the Two-Envelope Paradox</h3>
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<p><span style="font-family: 'book antiqua', palatino;">“Free Money!”, reads a sign near the town square. Next to the sign is small man in a white suit. Next to the man is a table, with two envelopes on top. “This must be a joke!” you think to yourself as you walk by. “Who could possibly be giving out free money?” A few days later, however, a friend informs you that the sign is for real. The small man in the white suit is actually giving out free money. “There must be some kind of catch!” you say. “How does it work?”</span></p>
<p><span style="font-family: 'book antiqua', palatino;">“It’s very simple,” your friend explains. “The man has placed a check in each envelope. We don’t know how much money the checks are made out for—it’s different each day. But we do know about the method that the man uses to fill out the checks. The first check is made out for <span class="math inline">\(n\)</span> dollars, where <span class="math inline">\(n\)</span> is a natural number chosen at random. The second check is made out for <span class="math inline">\(2n\)</span> dollars. Unfortunately, you won’t know which of the two checks has the larger amount and which one has the smaller one.”</span></p>
<p><span style="font-family: 'book antiqua', palatino;">“That’s it?” you ask, incredulous. “That’s it!” your friend replies. “All you need to do is pick one of the envelopes, and you’ll get to keep the check. I played the game last week, and I ended up with a check for $1,834,288. I took it to the bank, and cashed it without a hitch. Now I’m looking to buy a beach house. It’s a shame the man won’t let you play more than once!”</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Unable to help yourself any longer, you decide to pay a visit to the man in the white suit. “Welcome!” he says, with a big smile. “Select one of these envelopes, and its contents will be yours.” After a moment’s hesitation, you reach for the envelope on the left. How wonderful it would be to have your very own beach house! As you are about to open the envelope, the man interjects: “Are you interested in mathematics?”</span></p>
<p><span style="font-family: 'book antiqua', palatino;">“Why, yes. I am,” you say. “Why do you ask?”</span></p>
<p><span style="font-family: 'book antiqua', palatino;">“You’ve selected the envelope on the left, but I’ll give you a chance to change your mind. Think about it for a moment,” the man adds with a smile, “it might be in your interests to do so.” You consider the issue for a moment, and conclude that the man right: you should switch! For you reason as follows:</span></p>
<blockquote>
<p><span style="font-family: 'book antiqua', palatino;">I know my envolope contains a certain amount of money: \(k\) dollars, say. This means that the other envelope must contain either \(2k\) dollars or \(k/2\) dollars. These outcomes ought to have equal probability, since the initial number \(n\) was selected at random. And if the outcomes have equal probability, I should switch. For although it is true that if I switch, I am just as likely to gain money as I am to lose money, what I’ll gain if I gain is more than what I’ll lose if I lose. More specifically, if I gain, I’ll gain k dollars, and if I lose, I’ll lose \(k/2\) dollars. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The same point can be made formally. The expected value of switching is as follows: <span class="math display">\[EV(\text{switch}) = k/2\cdot 0.5 + 2k\cdot 0.5 =5/4 \cdot k\]</span> And, of course, the expected value of staying is \(k\). Since <span class="math inline">\(5/4 \cdot k\)</span> is always larger than <span class="math inline">\(k\), </span>the Principle of Expected Value Maximization entails that you should switch!</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">“All right!” you cry, “I’ll switch!” With a trembling hand, you place the envelope you had been holding back on the table, still unopened, and pick up the other. As you prepare to open it, the man in the white suit interrupts: “Excuse me, but are you sure you don’t want to switch again? It might be in your interests to do so…” It is at that point that the problem dawns on you. The exact same reasoning that led you to switch the first time could be used to argue that you should switch again, and go back to your original selection.</span></p>
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<h2 class="hd hd-2 unit-title">Paradox</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> We are before a paradox. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">We have an apparently valid argument telling us that you should switch <em>regardless of which of the envelopes you’ve selected.</em> And that can’t be the right conclusion, since it would lead you to switch envelopes indefinitely. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Where have we gone wrong?</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> The first time I heard about the Two-Envelope Paradox, I was unimpressed. It seemed to me that the setup relied on a problematic assumption: the assumption that the man in the white suit is able to pick a natural number at random, and do so in such a way that different numbers always have the same probability of being selected. (We relied on this assumption above in assigning equal probability to the outcome in which you get [mathjaxinline]2k[/mathjaxinline] dollars by switching and that outcome in which you get [mathjaxinline]k/2[/mathjaxinline] dollars by switching.) As we saw in Lecture 6.4, however, there can be no such thing as a <em>uniform</em> probability distribution over the natural numbers, on pain of violating Countable Additivity—a principle which I very much hoped to keep in place. No uniform probability distribution, no paradox! Or so I thought. . .</span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> I have since come to see that I was wrong. It is a mistake to think that the paradox requires a uniform probability distribution over the natural numbers. As it turns out, there are non-uniform probability distributions which yield the result that switching envelopes always has a higher expected value than staying put. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Here is an example, due to Oxford philosopher John Broome. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Take a die, and toss it until it lands One or Two. If the die first lands One or Two on the [mathjaxinline]k[/mathjaxinline]th toss, place [mathjaxinline]2^{k-1}[/mathjaxinline] in the first envelope, and twice that amount in the second. As you’ll be asked to prove below, this setup entails that the expected value of switching is always greater than the expected value of staying put.</span></p>
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<span style="font-family: 'book antiqua', palatino;">What is the probability that the man fills the first envelope with <span class="math inline">\(2^n\)</span>, and the second envelope with twice that amount?</span>
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<span style="font-family: 'book antiqua', palatino;">If your envelope contains <span class="math inline">\(\$1\)</span>, you should definitely switch. Suppose your envelope contains <span class="math inline">\(2^{k}\)</span> dollars for some <span class="math inline">\(k&gt;0\)</span>. Then the expected value of not switching is $<span class="math inline">\(2^{k}\)</span>. What is the expected value of switching?</span>
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<h2 class="hd hd-2 unit-title">A way out?</h2>
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<p><span style="font-family: 'book antiqua', palatino;">My favorite answer to the Two-Envelope Paradox is due to Broome, and to New York University philosopher David Chalmers. One starts by asking a simple question: in a Two-Envelope setup such as Broome’s, what is the expected value of a given envelope? As I’ll ask you to verify below, the answer is that each envelope has <em>infinite</em> expected value.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">As it turns out, this is no accident: one can prove that a two-envelope setup can only lead to trouble when the expected value of the envelopes is infinite. But, as Broome and Chalmers point out, we have independent reasons for thinking that decisions involving options of infinite expected value are problematic.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Consider, for example, the <strong>St. Petersburg Paradox</strong>. A coin is tossed until it lands Heads. If it lands Heads on the <span class="math inline">\(n\)</span>th toss you are given <span class="math inline">\(\$2^n\)</span>. How much should you pay for the privilege of playing such a game? The expected value of playing is infinite: <span class="math display">\[EV(\text{play}) = \frac{1}{2^1} \cdot 2^1 + \frac{1}{2^2} \cdot 2^2 + \dots = 1 + 1 + \ldots = \infty\]</span> So (assuming you value only money) the Principle of Value Maximization entails that you should be willing to pay <em>any finite amount of money</em> for the privilege of playing. And this is surely wrong: it is surely reasonable to balk at the prospect of paying, e.g. a million dollars for a chance to play.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">I’m inclined to think that Broome and Chalmers are right to think that decisions involving options of infinite expected value are problematic, and that this helps answer the Two-Envelope Paradox. I must confess, however, that I do not feel ready to let the matter rest. Like the St. Petersburg Paradox, the Two-Envelope Paradox suggests that our decision theoretic tools fall short when we try to reason with infinite expected value. But I’m not sure I have a satisfactory way of theorizing about such cases. If I were to assign the Two-Envelope Paradox a “paradoxicality grade” of the kind we used in Lecture 3, I would choose a number somewhere between 6 and 8. It is clear that the paradox brings out some important limitations of decision-theory, but I do not feel able to gauge just how important they are.</span></p>
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Problem 1
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<span style="font-family: 'book antiqua', palatino;">Suppose that an envelope is filled using Broom&#8217;s method: a die is tossed until it lands One or Two. If the die first lands One or Two on the <span class="math inline">\(k\)</span>th toss, the envelope is filled with <span class="math inline">\(2^{k-1}\)</span> dollars. What is the expected value of the envelope?</span>
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<span class="math inline">\(2^{k-1}+ k^{2}\)</span>
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<span class="math inline">\(\infty\)</span>
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