I was thinking of this problem.


I believe the solution and I think I understand it, but if I take the following approach I'm completely confused.

Problem 1:

I will offer you the following game. You pay me \$10 and I will flip a fair coin. Heads I give you \$5 and Tails I give you \$20.

The expectation is \$12.5 so you will always play the game.

Problem 2:

I will give you an envelope with \$10, the envelope is open and you can check. I then show you another envelope, closed this time and tell you: This envelope either has \$5 or $20 in it with equal probability. Do you want to swap?

I feel this is exactly the same as problem 1, you forgo \$10 for a \$5 or a \$20, so again you will always switch.

Problem 3:

I do the same as above but close the envelopes. So you don't know there are $10 but some amount X. I tell you the other envelope has double or half. Now if you follow the same logic you want to switch. This is the envelope paradox.

What changed when I closed the envelope??


Some have argued that problem 3 is not the envelope problem and I'm going to try and provide below why I think it is by analysing how each views the game. Also, it gives a better set up for the game.

Providing some clarification for Problem 3:

From the perspective of the person organising the game:

I hold 2 envelopes. In one I put \$10 close it and give it to the player. I then tell him, I have one more envelope that has either double or half the amount of the envelope I just gave you. Do you want to switch? I then proceed to flip a fair coin and Heads I put \$5 in and Tails I put \$20. And hand him the envelope. I then ask him. The envelope you just gave me has twice or half the amount of the envelope you are holding. Do you want to switch?

From the perspective of the player:

I am given an envelope and told there is another envelope that has double or half the amount with equal probability. Do I want to switch. I think sure I have $X$, hence $\frac{1}{2}(\frac{1}{2}X + 2X) > X$ so I want to switch. I get the envelope and all of a sudden I am facing the exact same situation. I want to switch again as the other envelope has either double or half the amount.

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    $\begingroup$ At least for me the key understanding is that I can't just say "I have X, hence (1/2*X + 2X)/2 > X" - the total average chance is 50/50, but for any specific X, the expected chances aren't 50/50 anymore; and the larger X, the lower chance of having 2*X in the other envelope (for positive finite distributions); so integrating over the possible X'es sum(p(X) * (1/2X*f(X) + 2X(1-f(X)) ) = X, where f(X) is the likelihood of the first envelope being larger, given any particular X. $\endgroup$
    – Peteris
    Apr 29, 2014 at 23:30
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    $\begingroup$ In the statement of the paradox, there is nothing that says that an amount X is chosen by the experimenter and then the experimenter randomly decides to either put $X$ or $X/2$ in the other envelope. The fact that you keep conflating the situation you created with the two envelope paradox means you do not understand why it's incorrect for the player to believe there is a 50/50 chance the other envelope is either $X/2$ or $2X$. In the actual two envelope problem, the probability that $2X$ is in the other envelope is either 0 or 1. $\endgroup$
    – jsk
    Apr 30, 2014 at 1:58
  • $\begingroup$ you are right. I do not undestand :( hence the question. I am trying to understand the difference between the problem 3 I have stated and the envelope paradox. I understand that in the paradox there are two envelopes so X and 2X and done, but I don't see how that is different from giving someone an envelope and then flipping a coin to decide to put the other amount in. $\endgroup$
    – evan54
    Apr 30, 2014 at 2:03
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    $\begingroup$ The trick to this is the flawed assumption that either $X/2$ or $2X$ outcomes are equally likely. If $2X$ is in the other envelope, then the expected gain from switching is $2X-X=X$. If $X/2$ is in the other envelope, then the expected gain from switching is $X/2 - X = - X/2$. The player doesn't know which of these situations he is in, but that doesn't mean he should believe there is a 50/50 chance. $\endgroup$
    – jsk
    Apr 30, 2014 at 2:29
  • 1
    $\begingroup$ Let's suppose the envelopes contain $X$ and $2X$. If you end up with $X$, then the probability that $2X$ is in the other envelope is 1 and the probability that $X/2$ is in the other envelope is 0. If you end up with $2X$, then the probability that $2*(2X)=4X$ is in the other envelope is 0 and the probability that $2X / 2=X$ is in the other envelope is 1. $\endgroup$
    – jsk
    Apr 30, 2014 at 2:34

7 Answers 7



The next two sections of this note analyze the "guess which is larger" and "two envelope" problems using standard tools of decision theory (2). This approach, although straightforward, appears to be new. In particular, it identifies a set of decision procedures for the two envelope problem that are demonstrably superior to the “always switch” or “never switch” procedures.

Section 2 introduces (standard) terminology, concepts, and notation. It analyzes all possible decision procedures for the "guess which is larger problem." Readers familiar with this material might like to skip this section. Section 3 applies a similar analysis to the two envelope problem. Section 4, the conclusions, summarizes the key points.

All published analyses of these puzzles assume there is a probability distribution governing the possible states of nature. This assumption, however, is not part of the puzzle statements. The key idea to these analyses is that dropping this (unwarranted) assumption leads to a simple resolution of the apparent paradoxes in these puzzles.


An experimenter is told that different real numbers $x_1$ and $x_2$ are written on two slips of paper. She looks at the number on a randomly chosen slip. Based only on this one observation, she must decide whether it is the smaller or larger of the two numbers.

Simple but open-ended problems like this about probability are notorious for being confusing and counter-intuitive. In particular, there are at least three distinct ways in which probability enters the picture. To clarify this, let's adopt a formal experimental point of view (2).

Begin by specifying a loss function. Our goal will be to minimize its expectation, in a sense to be defined below. A good choice is to make the loss equal to $1$ when the experimenter guesses correctly and $0$ otherwise. The expectation of this loss function is the probability of guessing incorrectly. In general, by assigning various penalties to wrong guesses, a loss function captures the objective of guessing correctly. To be sure, adopting a loss function is as arbitrary as assuming a prior probability distribution on $x_1$ and $x_2$, but it is more natural and fundamental. When we are faced with making a decision, we naturally consider the consequences of being right or wrong. If there are no consequences either way, then why care? We implicitly undertake considerations of potential loss whenever we make a (rational) decision and so we benefit from an explicit consideration of loss, whereas the use of probability to describe the possible values on the slips of paper is unnecessary, artificial, and-—as we shall see—-can prevent us from obtaining useful solutions.

Decision theory models observational results and our analysis of them. It uses three additional mathematical objects: a sample space, a set of “states of nature,” and a decision procedure.

  • The sample space $S$ consists of all possible observations; here it can be identified with $\mathbb{R}$ (the set of real numbers).

  • The states of nature $\Omega$ are the possible probability distributions governing the experimental outcome. (This is the first sense in which we may talk about the “probability” of an event.) In the “guess which is larger” problem, these are the discrete distributions taking values at distinct real numbers $x_1$ and $x_2$ with equal probabilities of $\frac{1}{2}$ at each value. $\Omega$ can be parameterized by $\{\omega = (x_1, x_2) \in \mathbb{R}\times\mathbb{R}\ |\ x_1 \gt x_2\}.$

  • The decision space is the binary set $\Delta = \{\text{smaller}, \text{larger}\}$ of possible decisions.

In these terms, the loss function is a real-valued function defined on $\Omega \times \Delta$. It tells us how “bad” a decision is (the second argument) compared to reality (the first argument).

The most general decision procedure $\delta$ available to the experimenter is a randomized one: its value for any experimental outcome is a probability distribution on $\Delta$. That is, the decision to make upon observing outcome $x$ is not necessarily definite, but rather is to be chosen randomly according to a distribution $\delta(x)$. (This is the second way in which probability may be involved.)

When $\Delta$ has just two elements, any randomized procedure can be identified by the probability it assigns to a prespecified decision, which to be concrete we take to be “larger.”


A physical spinner implements such a binary randomized procedure: the freely-spinning pointer will come to stop in the upper area, corresponding to one decision in $\Delta$, with probability $\delta$, and otherwise will stop in the lower left area with probability $1-\delta(x)$. The spinner is completely determined by specifying the value of $\delta(x)\in[0,1]$.

Thus a decision procedure can be thought of as a function



$${\Pr}_{\delta(x)}(\text{larger}) = \delta^\prime(x)\ \text{ and }\ {\Pr}_{\delta(x)}(\text{smaller})=1-\delta^\prime(x).$$

Conversely, any such function $\delta^\prime$ determines a randomized decision procedure. The randomized decisions include deterministic decisions in the special case where the range of $\delta^\prime$ lies in $\{0,1\}$.

Let us say that the cost of a decision procedure $\delta$ for an outcome $x$ is the expected loss of $\delta(x)$. The expectation is with respect to the probability distribution $\delta(x)$ on the decision space $\Delta$. Each state of nature $\omega$ (which, recall, is a Binomial probability distribution on the sample space $S$) determines the expected cost of any procedure $\delta$; this is the risk of $\delta$ for $\omega$, $\text{Risk}_\delta(\omega)$. Here, the expectation is taken with respect to the state of nature $\omega$.

Decision procedures are compared in terms of their risk functions. When the state of nature is truly unknown, $\varepsilon$ and $\delta$ are two procedures, and $\text{Risk}_\varepsilon(\omega)\ge \text{Risk}_\delta(\omega)$ for all $\omega$, then there is no sense in using procedure $\varepsilon$, because procedure $\delta$ is never any worse (and might be better in some cases). Such a procedure $\varepsilon$ is inadmissible; otherwise, it is admissible. Often many admissible procedures exist. We shall consider any of them “good” because none of them can be consistently out-performed by some other procedure.

Note that no prior distribution is introduced on $\Omega$ (a “mixed strategy for $C$” in the terminology of (1)). This is the third way in which probability may be part of the problem setting. Using it makes the present analysis more general than that of (1) and its references, while yet being simpler.

Table 1 evaluates the risk when the true state of nature is given by $\omega=(x_1, x_2).$ Recall that $x_1 \gt x_2.$

Table 1.

$$\matrix { \text{Decision:}& & \text{Larger} & \text{Larger} & \text{Smaller} & \text{Smaller}\\ \text{Outcome} & \text{Probability} & \text{Probability} & \text{Loss} & \text{Probability} & \text{Loss} & \text{Cost} \\ x_1 & 1/2 & \delta^\prime(x_1) & 0 & 1 - \delta^\prime(x_1) & 1 & 1 - \delta^\prime(x_1) \\ x_2 & 1/2 & \delta^\prime(x_2) & 1 & 1 - \delta^\prime(x_2) & 0 & 1 - \delta^\prime(x_2) }$$

$$\text{Risk}(x_1,x_2):\ (1 - \delta^\prime(x_1) + \delta^\prime(x_2))/2.$$

In these terms the “guess which is larger” problem becomes

Given you know nothing about $x_1$ and $x_2$, except that they are distinct, can you find a decision procedure $\delta$ for which the risk $[1 – \delta^\prime(\max(x_1, x_2)) + \delta^\prime(\min(x_1, x_2))]/2$ is surely less than $\frac{1}{2}$?

This statement is equivalent to requiring $\delta^\prime(x)\gt \delta^\prime(y)$ whenever $x \gt y.$ Whence, it is necessary and sufficient for the experimenter's decision procedure to be specified by some strictly increasing function $\delta^\prime: S\to [0, 1].$ This set of procedures includes, but is larger than, all the “mixed strategies $Q$” of 1. There are lots of randomized decision procedures that are better than any unrandomized procedure!


It is encouraging that this straightforward analysis disclosed a large set of solutions to the “guess which is larger” problem, including good ones that have not been identified before. Let us see what the same approach can reveal about the other problem before us, the “two envelope” problem (or “box problem,” as it is sometimes called). This concerns a game played by randomly selecting one of two envelopes, one of which is known to have twice as much money in it as the other. After opening the envelope and observing the amount $x$ of money in it, the player decides whether to keep the money in the unopened envelope (to “switch”) or to keep the money in the opened envelope. One would think that switching and not switching would be equally acceptable strategies, because the player is equally uncertain as to which envelope contains the larger amount. The paradox is that switching seems to be the superior option, because it offers “equally probable” alternatives between payoffs of $2x$ and $x/2,$ whose expected value of $5x/4$ exceeds the value in the opened envelope. Note that both these strategies are deterministic and constant.

In this situation, we may formally write

$$\eqalign{ S &= \{ x\in \mathbb{R}\ |\ x \gt 0\}, \\ \Omega &= \{\text{Discrete distributions supported on }\{\omega, 2\omega\}\ |\ \omega \gt 0 \text{ and }\Pr(\omega) = \frac{1}{2}\}, \text{and}\\ \Delta &= \{\text{Switch}, \text{Do not switch}\}. }$$

As before, any decision procedure $\delta$ can be considered a function from $S$ to $[0, 1],$ this time by associating it with the probability of not switching, which again can be written $\delta^\prime(x)$. The probability of switching must of course be the complementary value $1–\delta^\prime(x).$

The loss, shown in Table 2, is the negative of the game's payoff. It is a function of the true state of nature $\omega$, the outcome $x$ (which can be either $\omega$ or $2\omega$), and the decision, which depends on the outcome.

Table 2.

$$\matrix{ & \text{Loss}&\text{Loss} &\\ \text{Outcome}(x) & \text{Switch} & \text{Do not switch} & \text{Cost}\\ \omega & -2\omega & -\omega & -\omega[2(1-\delta^\prime(\omega)) + \delta^\prime(\omega)]\\ 2\omega & -\omega & -2\omega & -\omega[1 - \delta^\prime(2\omega) + 2\delta^\prime(2\omega)] }$$

In addition to displaying the loss function, Table 2 also computes the cost of an arbitrary decision procedure $\delta$. Because the game produces the two outcomes with equal probabilities of $\frac{1}{2}$, the risk when $\omega$ is the true state of nature is

$$\eqalign{ \text{Risk}_\delta(\omega) &=-\omega[2(1-\delta^\prime(\omega)) + \delta^\prime(\omega)]/2 + -\omega[1 - \delta^\prime(2\omega) + 2\delta^\prime(2\omega)]/2 \\ &= (-\omega/2)[3 + \delta^\prime(2\omega) - \delta^\prime(\omega)]. }$$

A constant procedure, which means always switching ($\delta^\prime(x)=0$) or always standing pat ($\delta^\prime(x)=1$), will have risk $-3\omega/2$. Any strictly increasing function, or more generally, any function $\delta^\prime$ with range in $[0, 1]$ for which $\delta^\prime(2x) \gt \delta^\prime(x)$ for all positive real $x,$ determines a procedure $\delta$ having a risk function that is always strictly less than $-3\omega/2$ and thus is superior to either constant procedure, regardless of the true state of nature $\omega$! The constant procedures therefore are inadmissible because there exist procedures with risks that are sometimes lower, and never higher, regardless of the state of nature.


Comparing this to the preceding solution of the “guess which is larger” problem shows the close connection between the two. In both cases, an appropriately chosen randomized procedure is demonstrably superior to the “obvious” constant strategies.

These randomized strategies have some notable properties:

  • There are no bad situations for the randomized strategies: no matter how the amount of money in the envelope is chosen, in the long run these strategies will be no worse than a constant strategy.

  • No randomized strategy with limiting values of $0$ and $1$ dominates any of the others: if the expectation for $\delta$ when $(\omega, 2\omega)$ is in the envelopes exceeds the expectation for $\varepsilon$, then there exists some other possible state with $(\eta, 2\eta)$ in the envelopes and the expectation of $\varepsilon$ exceeds that of $\delta$ .

  • The $\delta$ strategies include, as special cases, strategies equivalent to many of the Bayesian strategies. Any strategy that says “switch if $x$ is less than some threshold $T$ and stay otherwise” corresponds to $\delta(x)=1$ when $x \ge T, \delta(x) = 0$ otherwise.

What, then, is the fallacy in the argument that favors always switching? It lies in the implicit assumption that there is any probability distribution at all for the alternatives. Specifically, having observed $x$ in the opened envelope, the intuitive argument for switching is based on the conditional probabilities Prob(Amount in unopened envelope | $x$ was observed), which are probabilities defined on the set of underlying states of nature. But these are not computable from the data. The decision-theoretic framework does not require a probability distribution on $\Omega$ in order to solve the problem, nor does the problem specify one.

This result differs from the ones obtained by (1) and its references in a subtle but important way. The other solutions all assume (even though it is irrelevant) there is a prior probability distribution on $\Omega$ and then show, essentially, that it must be uniform over $S.$ That, in turn, is impossible. However, the solutions to the two-envelope problem given here do not arise as the best decision procedures for some given prior distribution and thereby are overlooked by such an analysis. In the present treatment, it simply does not matter whether a prior probability distribution can exist or not. We might characterize this as a contrast between being uncertain what the envelopes contain (as described by a prior distribution) and being completely ignorant of their contents (so that no prior distribution is relevant).


In the “guess which is larger” problem, a good procedure is to decide randomly that the observed value is the larger of the two, with a probability that increases as the observed value increases. There is no single best procedure. In the “two envelope” problem, a good procedure is again to decide randomly that the observed amount of money is worth keeping (that is, that it is the larger of the two), with a probability that increases as the observed value increases. Again there is no single best procedure. In both cases, if many players used such a procedure and independently played games for a given $\omega$, then (regardless of the value of $\omega$) on the whole they would win more than they lose, because their decision procedures favor selecting the larger amounts.

In both problems, making an additional assumption-—a prior distribution on the states of nature—-that is not part of the problem gives rise to an apparent paradox. By focusing on what is specified in each problem, this assumption is altogether avoided (tempting as it may be to make), allowing the paradoxes to disappear and straightforward solutions to emerge.


(1) D. Samet, I. Samet, and D. Schmeidler, One Observation behind Two-Envelope Puzzles. American Mathematical Monthly 111 (April 2004) 347-351.

(2) J. Kiefer, Introduction to Statistical Inference. Springer-Verlag, New York, 1987.

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    $\begingroup$ This is a short article I wrote ten years ago but never published. (The new editor of the AMM saw no mathematical interest in it.) I have given talks in which I played the two-envelope game with the audience, using substantial amounts of real money. $\endgroup$
    – whuber
    Apr 29, 2014 at 21:26
  • 1
    $\begingroup$ Very nice write up! Joe Blitzstein talked about the two evelope problem in a Harvard Stat 110 lecture which is available free on youtube if anyone is interested btw. $\endgroup$ May 25, 2014 at 10:05
  • $\begingroup$ @whuber Consider this variant. Suppose I choose two amounts of money such that one is twice as much as the other. Then I flip a fair coin to decide which amount goes in which envelope. Now you pick an envelope at random, and imagine the amount inside it, calling it $x$ (if this step is questionable, consider the case of opening up the envelope and looking at the actual amount - since the reasoning applies no matter what value you see inside, it should apply with a general $x$). Then calculate the expected value of the money in the other envelope as $E = (1/2)(x/2) + (1/2)(2x) = 1.25x > x$... $\endgroup$ Oct 9, 2014 at 3:04
  • $\begingroup$ I guess I don't understand where in that reasoning I "assumed a prior distribution on the states of nature". Did I? Clearly the reasoning cannot be correct, because I cannot justify switching to the other envelope by merely thinking about the first envelope (since the same logic would apply to the second, once I switch once). $\endgroup$ Oct 9, 2014 at 3:05
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    $\begingroup$ @Zubin There is a basic (but interesting) mistake in that analysis. Let $\theta$ be the smaller amount in the two envelopes. Given an observation of $x$, you know that either $\theta=x$ or $\theta=x/2$ and that the likelihood of this observation in either case is $1/2$. In the former case the amount $Y$ in the other envelope is $2x$ and in the latter case it is $x/2$, but in order to assign a valid expectation to $Y$ you must assume there is some probability distribution for $\theta$. Equal likelihood is not equivalent to equal probability. $\endgroup$
    – whuber
    Oct 9, 2014 at 3:48

The issue in general with the two envelope problem is that the problem as presented on wikipedia allows the size of the values in the envelopes to change after the first choice has been made. The problem has been formulized incorrectly.

However, a real world formulation of the problem is this: you have two identical envelopes: $A$ and $B$, where $B=2A$. You can pick either envelope and then are offered to swap.

Case 1: You've picked $A$. If you switch you gain $A$ dollars.

Case 2: You've picked $B$. If you switch you loose $A$ dollars.

This is where the flaw in the two-envelope paradox enters in. While you are looking at loosing half the value or doubling your money, you still don't know the original value of $A$ and the value of $A$ has been fixed. What you are looking at is either $+A$ or $-A$, not $2A$ or $\frac{1}{2}A$.

If we assume that the probability of selecting $A$ or $B$ at each step is equal,. the after the first offered swap, the results can be either:

Case 1: Picked $A$, No swap: Reward $A$

Case 2: Picked $A$, Swapped for $B$: Reward $2A$

Case 3: Picked $B$, No swap: Reward $2A$

Case 4: Picked $B$, Swapped for $A$: Reward $A$

The end result is that half the time you get $A$ and half the time you get $2A$. This will not change no matter how many times you are offered a swap, nor will it change based upon knowing what is in one envelope.

  • $\begingroup$ IMO, the problem says that you cannot lose A no matter what. So, your +A vs -A cannot be appropriate. You either win A or 2A. $\endgroup$ Sep 24, 2016 at 9:38

My interpretation of the question

I am assuming that the setting in problem 3 is as follows: the organizer first selects amount $X$ and puts $X$ in the first envelope. Then, the organizer flips a fair coin and based on that puts either $0.5X$ or $2X$ to the second envelope. The player knows all this, but not $X$ nor the result of the coin-flip. The organizer gives the player the first envelope (closed) and asks if the player wants to switch. The questioner argues 1. that the player wants to switch because the switching increases expectation (correct) and 2. that after switching, the same reasoning symmetrically holds and the player wants to switch back (incorrect). I also assume the player is a rational risk-neutral Bayesian agent that puts a probability distribution over $X$ and maximizes expected amount of money earned.

Note that if the we player did not know about the coin-flip procedure, there might be no reason in the first place to argue that the probabilities are 0.5 for the second envelope to be higher/lower.

Why there is no paradox

Your problem 3 (as interpreted in my answer) is not the envelope paradox. Let the $Z$ be a Bernoulli random variable with $P(Z=1)=0.5$. Define the amount $Y$ in the 2nd envelope so that $Z=1$ implies $Y=2X$ and $Z=0$ implies $Y=0.5X$. In the scenario here, $X$ is selected without knowledge of the result of the coin-flip and thus $Z$ and $X$ are independent, which implies $E(Y\mid X) = 1.25X$. \begin{equation} E(Y) = E(E(Y\mid X)) = E(1.25X) = 1.25E(X) \end{equation} Thus, if if X>0 (or at least $E(X)>0$), the player will prefer to switch to envelope 2. However, there is nothing paradoxical about the fact that if you offer me a good deal (envelope 1) and an opportunity to switch to a better deal (envelope 2), I will want to switch to the better deal.

To invoke the paradox, you would have to make the situation symmetric, so that you could argue that I also want to switch from envelope 2 to envelope 1. Only this would be the paradox: that I would want to keep switching forever. In the question, you argue that the situation indeed is symmetric, however, there is no justification provided. The situation is not symmetric: the second envelope contains the amount that was picked as a function of a coin-flip and the amount in the first envelope, while the amount in the first envelope was not picked as a function of a coin-flip and the amount in the second envelope. Hence, the argument for switching back from the second envelope is not valid.

Example with small number of possibilities

Let us assume that (the player's belief is that) $X=10$ or $X=40$ with equal probabilities, and work out the computations case by case. In this case, the possibilities for $(X,Y)$ are $\{(10,5),(10,20),(40,20),(40,80)\}$, each of which has probability $1/4$. First, we look at the player's reasoning when holding the first envelope.

  1. If my envelope contains $10$, the second envelope contains either $5$ or $20$ with equal probabilities, thus by switching I gain on average $0.5\times(-5) + 0.5\times10 = 2.5$.
  2. If my envelope contains $40$, the second envelope contains either $20$ or $80$ with equal probabilities, thus by switching I gain on average $0.5\times(-20) + 0.5\times(40) = 10$.

Taking the average over these, the expected gain of switching is $0.5\times2.5 + 0.5\times10 = 6.25$, so the player switches. Now, let us make similar case-by-case analysis of switching back:

  1. If my envelope contains $5$, the old envelope with probability 1 contains $10$, and I gain $5$ by switching.
  2. If my envelope contains $20$, the old envelope contains $10$ or $40$ with equal probabilities, and by switching I gain $0.5\times(-10) + 0.5\times20 = 5$.
  3. If my envelope contains $80$, the old envelope with probability 1 contains $40$ and I lose $40$ by switching.

Now, the expected value, i.e. probability-weighted average, of gain by switching back is $0.25\times5+0.5\times5+0.25\times(-40) = -6.25$. So, switching back exactly cancels the expected utility gain.

Another example with a continuum of possibilities

You might object to my previous example by claiming that I maybe cleverly selected the distribution over $X$ so that in the $Y=80$ case the player knows that he is losing. Let us now consider a case where $X$ has a continuous unbounded distribution: $X \sim \textrm{Exp}(1)$, $Z$ independent of $X$ as previously, and $Y$ as a function of $X$ and $Z$ as previously. The expected gain of switching from $X$ to $Y$ is again $E(0.25X) = 0.25E(X) = 0.25$. For the back-switch, we first compute the conditional probability $P(X=0.5Y \mid Y=y)$ using Bayes' theorem: \begin{equation} P(X=0.5Y \mid Y=y) = P(Z=1 \mid Y=y) = \frac{p(Y=y\mid Z=1)P(Z=1)}{p(Y=y)} = \frac{p(2X=y)P(Z=1)}{p(Y=y)} = \frac{0.25e^{-0.5y}}{p(Y=y)} \end{equation} and similarly $P(X=2Y \mid Y=y) = \frac{e^{-2y}}{p(Y=y)}$, wherefore the conditional expected gain of switching back to the first envelope is \begin{equation} E(X-Y \mid Y=y) = \frac{-0.125y e^{-0.5y} + ye^{-2y}}{p(Y=y)}, \end{equation} and taking the expectation over $Y$, this becomes \begin{equation} E(X-Y) = \int_0^\infty \frac{-0.125y e^{-0.5y} + ye^{-2y}}{p(Y=y)}p(Y=y) dy = -0.25, \end{equation} which cancels out the expected gain of the first switch.

General solution

The situation seen in the two examples must always occur: you cannot construct a probability distribution for $X,Z,Y$ with these conditions: $X$ is not a.s. 0, $Z$ is Bernoulli with $P(Z=1)=0.5$, $Z$ is independent of $X$, $Y=2X$ when $Z=1$ and $0.5X$ otherwise and also $Y,Z$ are independent. This is explained in the Wikipedia article under heading 'Proposed resolutions to the alternative interpretation': such a condition would imply that the probability that the smaller envelope has amount between $2^n,2^{n+1}$ ($P(2^n<=\min(X,Y)<2^{n+1})$ with my notation) would be a constant over all natural numbers $n$, which is impossible for a proper probability distribution.

Note that there is another version of the paradox where the probabilities need not be 0.5, but the expectation of other envelope conditional on the amount in this envelope is still always higher. Probability distributions satisfying this type of condition exist (e.g., let the amounts in the envelopes be independent half-Cauchy), but as the Wikipedia article explains, they require infinite mean. I think this part is rather unrelated to your question, but for completeness wanted to mention this.

  • $\begingroup$ I edited my question trying to explain why I think it is similar to the envelope paradox and you would want to switch forever. $\endgroup$
    – evan54
    Apr 29, 2014 at 22:51
  • $\begingroup$ @evan54 I rewrote my answer to contain my interpretation of the setting problem 3, more explanation about why the situation is not symmetric, examples etc. $\endgroup$ Apr 30, 2014 at 14:46
  • $\begingroup$ I think I'm close to getting it. I think that once there is a coin flip and envelope 2 contains half/double the amount in your hand you are basically in the situation of the envelope paradox BUT the way you got there guarantees you that you are better off switching. Does that make sense? $\endgroup$
    – evan54
    Apr 30, 2014 at 17:07
  • $\begingroup$ also, if it does, is there a way to make it more formal? I may ponder on it more.. $\endgroup$
    – evan54
    Apr 30, 2014 at 17:08
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    $\begingroup$ @evan54 Not sure. The whole point of the paradox is that it is a situation in which there is no advantage to switching. Thus, anything you change to the setup of the problem that results in it being advantageous to switch, at least initially, must therefore not be equivalent to the setup of the two envelope paradox. Note that in your setup, it only makes sense to switch the very first time. After you switch the first time, you expect to lose by switching back. The flawed logic in the paradox comes into play if you attempt to argue that you should switch back. $\endgroup$
    – jsk
    Apr 30, 2014 at 18:33

Problem 1: Agreed, play the game. The key here is that you know the actual probabilities of winning 5 vs 20 since the outcome is dependent upon the flip of a fair coin.

Problem 2: The problem is the same as problem 1 because you are told that there is an equal probability that either 5 or 20 is in the other envelope.

Problem 3: The difference in problem 3 is that telling me the other envelope has either $X/2$ or $2X$ in it does not mean that I should assume that the two possibilities are equally likely for all possible values of $X$. Doing so implies an improper prior on the possible values of $X$. See the Bayesian resolution to the paradox.

  • $\begingroup$ I see we interpret problem 3 slightly differently. I assumed OP specifically constructs the setting in problem 3 so that the 2nd envelope has probabilities 0.5/0.5. This is clearly possible without improper distributions, but then the possibilities for envelope 1 are not equally likely given the amount in the second envelope. $\endgroup$ Apr 29, 2014 at 20:58
  • $\begingroup$ Agreed, if OP meant that you are told that the other envelope either has $X/2$ or $2X$ with equal probabilities, then problem 3 would not be equivalent to the 2 envelope paradox. $\endgroup$
    – jsk
    Apr 29, 2014 at 21:09
  • $\begingroup$ yes that was my thinking, that in problem 3 there is equal probability between X/2 and 2X. So you hold 3 envelopes give him the 10 and then flip a coin to see if you give him the 20 or 5 (they are closed) if he decides to switch $\endgroup$
    – evan54
    Apr 29, 2014 at 22:21
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    $\begingroup$ @evan54 - if you make the random flip after you choose which envelope to give me, then it's equivalent to problem 1; if you choose both amounts of money, and then make a random flip on which envelope you give me, then it's the situation described above; they're different situations. $\endgroup$
    – Peteris
    Apr 29, 2014 at 22:41
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    $\begingroup$ @evan54 - the optimal player's decision depends on how you made those envelopes. If you don't tell the player how you did that (only that 50/50 sentence), then the optimal strategy depends on player's assumptions on how likely you are to do it one way or another - the first envelope you prepared is less valuable than the second envelope you prepared; if they were fairly shuffled (and unopened) then it doesn't matter what the player chooses; if the player thinks that you likely (>50%) initially gave him the first envelope, then player should switch and stick with that. $\endgroup$
    – Peteris
    Apr 29, 2014 at 23:12

This is a potential explanation that I have. I think it is wrong but I'm not sure. I will post it to be voted on and commented on. Hopefully someone will offer a better explanation.

So the only thing that changed between problem 2 and problem 3 is that the amount became in the envelope you hold became random. If you allow that amount to be negative so there might be a bill there instead of money then it makes perfect sense. The extra information you get when you open the envelope is whether it's a bill or money hence you care to switch in one case while in the other you don't.

If however you are told the bill is not a possibility then the problem remains. (of course do you assign a probability that they lie?)

  • $\begingroup$ Introducing the possibility of negative amounts is an interesting observation, but not needed for resolving the issue in your question. See my answer. $\endgroup$ Apr 29, 2014 at 20:24
  • $\begingroup$ It is not necessary to assume the amount in the envelope is random: it suffices that it is unknown. Assuming randomness adduces information--however little it might be--that was not given in the problem! $\endgroup$
    – whuber
    Apr 29, 2014 at 20:24
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    $\begingroup$ The biggest difference between 2 and 3 is that being told the other amount is either $X/2$ or $2X$ is not the same as being told that the two possibilities are equally likely. Assuming the two amounts are equally likely is not the same as being told the two amounts are equally likely. $\endgroup$
    – jsk
    Apr 29, 2014 at 20:42


I believe that they way you have broken out the problem is completely correct. You need to distinguish the "Coin Flip" scenario, from the situation where the money is added to the envelope before the envelope is chosen

Not distinguishing those scenarios lies at the root of many people's confusion.

Problem 1

If you are flipping a coin to decide if either double your money or lose half, always play the game. Instead of double or nothing, it is double or lose some.

Problem 2

This is exactly the same as the coin flip scenario. The only difference is that the person picking the envelope flipped before giving you the first envelope. Note You Did Not Choose an Envelope!!!! You were given one envelope, and then given the choice to switch This is a subtle but important difference over problem 3, which affects the distribution of the priors

Problem 3

This is the classical setup to the two envelope problem. Here you are given the choice between the two envelopes. The most important points to realize are

  • There is a maximum amount of money that can be in the any envelope. Because the person running the game has finite resources, or a finite amount they are willing to invest
  • If you call the maximum money that could be in the envelope M, you are not equally likely to get any number between 0 and M. If you assume a random amount of money between 0 and M was put in the first envelope, and half of that for the second (or double, the math still works) If you open an envelope, you are 3 times as likely to see something less than M/2 than above M/2. (This is because half the time both envelopes will have less than M/2, and the other half the time 1 envelope will)
  • Since there is not an even distribution, the 50% of the time you double, 50% of the time you cut in half doesn't apply
  • When you work out the actual probabilities, you find the expected value of the first envelope is M/2, and the EV of the second envelope, switching or not is also M/2

Interestingly, if you can make some guess as to what the maximum money in the envelope can be, or if you can play the game multiple times, then you can benefit by switching, whenever you open an envelope less than M/2. I have simulated this two envelope problem here and find that if you have this outside information, on average you can do 1.25 as well as just always switching or never switching.


Problem 2A: 100 note cards are in an opaque jar. "\$10" is written on one side of each card; the opposite side has either "\$5" or "\$20" written on it. You get to pick a card and look at one side only. You then get to choose one side (the revealed, or the hidden), and you win the amount on that side.

If you see "\$5," you know you should choose the hidden side and will win \$10. If you see "\$20," you know you should choose the revealed side and will win \$20. But if you see "\$10," I have not given you enough information calculate an expectation for the hidden side. Had I said there were an equal number of {\$5,\$10} cards as {\$10,\$20} cards, the expectation would be \$12.50. But you can't find the expectation from $only$ the fact - which was still true - that you had equal chances to reveal the higher, or lower, value on the card. You need to know how many of each kind of card there were.

Problem 3A: The same jar is used, but this time the cards all have different, and unknown, values written on them. The only thing that is the same, is that on each card one side is twice the value of the other.

Pick a card, and a side, but don't look at it. There is a 50% chance that it is the higher side, or the lower side. One possible solution is that the card is either {X/2,X} or {X,2X} with 50% probability, where X is your side. But we saw above that the the probability of choosing high or low is not the same thing as these two $different$ cards being equally likely to be in the jar.

What changed between your Problem 2 and Problem 3, is that you made these two probabilities the same in Problem 2 by saying "This envelope either has \$5 or \$20 in it with equal probability." With unknown values, that can't be true in Problem 3.


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