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I have read about Blackwell's bet paradox on Futility closet. Here is the summary: you are presented with two envelopes, $E_x$ and $E_y$. The envelopes contain a random amount of money, but you don't know anything about the distribution about the money. You open one, check how much money is in there ($x$), and have to choose: take the envelope $E_x$ or $E_y$?

Futility Closet refers to a mathematician called Leonard Wapner: “Unexpectedly, there is something you can do, short of opening the other envelope, to give yourself a better than even chance of getting it right.”

The idea, which seems wrong to me, is as follows: choose a random number $d$. If $d < x$, take $E_x$. If $d > x$, choose $E_y$.

Wapner: “If d falls between x and y then your prediction (as indicated by d) is guaranteed to be correct. Assume this occurs with probability p. If d falls less than both x and y, then your prediction will be correct only in the event your chosen number x is the larger of the two. There is a 50 percent chance of this. Similarly, if d is greater than both numbers, your prediction will be correct only if your chosen number is the smaller of the two. This occurs with a 50 percent probability as well.”

If the probability that $d$ is in $[x,y]$ is greater than zero, then the average success of this method is $\frac{1}{2} + \frac{p}{2}$. This would mean that by observing an unrelated random variable gives us additional information.

I think that this is all wrong, and that the problem lies in choosing a random -integer- number. What does it mean? Like, any integer? In that case, the probability $p$ that $d$ lies between $x$ and $y$ is zero, because both $x$ and $y$ are finite.

If we say that there is a limit on the maximal amount of money, say $M$, or at least we choose d from $1...M$, then the recipe boils down to the trivial advice of choosing $E_y$ if $x < M/2$ and choosing $E_x$ if $x > M/2$.

Do I miss something here?

EDIT

OK, now I begin to see where the apparent paradox comes from. It seemed to me impossible that an unrelated random variable can provide additional information.

However, note that we need to consciously choose a distribution of d. For example, choose the boundaries for a uniform distribution, or $\lambda$ of the Poissionian distribution etc. Clearly, if we are playing for peanuts, and we chose the distribution of d to be uniform on $[10^9, 2\cdot 10^9]$ dollars, $P(d \in (x,y)) = 0$. This last probability will depend first and foremost on our judgement of what can be in the envelopes.

In other words, if the technique works, then the assumption that we do not know what is the distribution of the money in the envelopes (how the amount of money for the envelopes was chosen) is violated. However, if we truly don't know what is in the envelopes, then in the worst case scenario, we don't loose anything by applying it.

EDIT 2

Another thought. Given $x$, let us choose, for drawing $d$, a continuous non-negative distribution such that $P(d < x) = P(d > x)$. We are allowed to do that, am I correct? We proceed as instructed - if $d < x$, we keep the envelope, if $d > x$, we change the envelope. The reasoning does not change, depending how we choose the distribution it can be that $P(d \in [x, y]) > 0$ (or am I mistaken?).

However, given how we chose the distribution, what we now do is equivalent to a coin toss. We toss a coin, and if it is heads, we change envelopes, if it is tails, we stick to the envelope we hold. Where am I wrong?

EDIT 3:

OK, I get it now. If we base the probability function of $d$ on $x$ (e.g., we sample $d$ from an uniform distribution in range $(1, 2 \cdot x)$, then probability $P(d \in (x,y))$ is not independent of $P(\text{correct decision}|d \notin (x,y))$.

So, if $d \in (x,y)$ (with probability $p$), the guess is always correct, as before. If $x$ is the lower number, however, and $d \notin (x,y)$, than $d$ has a higher chance to be lower than $x$ than being higher than $x$, so we are biased towards an incorrect decision. Same reasoning applies when $x$ is the higher of the two numbers.

That means that we have to choose the process of drawing $d$ independently of $x$. In other words, we need to make a guess about parameters of distribution from which $x$ and $y$ are drawn; worst that happens is that we still guess randomly, but best what happens is that our guess was correct -- and then we have an advantage. How this should be better than guessing "x and y will, I think, be at least 1\$, but at most 10\$, so if $x > 5$, we keep it, and if not, we exchange it" I am yet to see.

I was mislead by the pop-sci formulation of the problem in Wapner's book (Unexpected Expectations: The Curiosities of a Mathematical Crystal Ball), which states

"By any means whatsoever, select a random positive integer" (Wapner suggests a geometric distribution -- tossing coins until the first heads come up, repeating the process if $d=x$) "If $d > x$ guess higher and if $d < x$ guess lower. (...) You will guess correctly more than 50 percent of the time because $d$ points correctly more than 50 percent of the time!"

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    $\begingroup$ Very closely related: stats.stackexchange.com/questions/95694 $\endgroup$ – whuber Jun 28 '16 at 12:40
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    $\begingroup$ This is quite different from the two envelopes problem in the sense that: (1) the argument given for switching in the two-envelopes problem is fallacious, the flaw in the argument can be seen by adding a Bayesian prior while (2) the argument given by Wapner for Blackwell's bet is correct. $\endgroup$ – Matthew Gunn Jun 28 '16 at 13:32
  • $\begingroup$ If the amounts of money in the envelopes are arbitrary elements of a set of numbers S, a sufficient and necessary condition for Wapner's strategy to work is for the CDF of the number you choose to be strictly increasing on S. $\endgroup$ – Solomonoff's Secret Jun 28 '16 at 17:16
  • $\begingroup$ OK, I am still missing something - please see my EDIT 2, but it looks to me as if we could just toss a coin and it still should work, according to the reasoning. Where am I wrong? $\endgroup$ – January Jun 29 '16 at 13:11
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This is more widely known as the two-envelope problem. Most commonly the amounts are given as $A$ and $2A$ but it's not required that this be the case.

Some points:

  1. You cannot choose a random integer uniformly*, but the quoted part doesn't seem to require it be uniform. Choose a distribution - it doesn't matter what it is for the argument - as long as it has some probability of exceeding any finite value.

  2. It wouldn't make sense to choose $d$ integer with the quoted decision rule, because money is discrete which means there's a nonzero chance $d=x$ and there's nothing listed for that case. (Or alternatively, to modify the rule to specify what to do when they're equal)

  3. Leaving that aside, you could choose $d$ from some non-negative continuous distribution -- then we don't have to worry about equality.

* (nor can you choose a uniformly random non-negative integer nor a uniformly random positive integer)


If we say that there is a limit on the maximal amount of money, say $M$, or at least we choose $d$ from $1...M$, then the recipe boils down to the trivial advice of choosing $E_y$ if $x<M/2$ and choosing $E_x$ if $x>M/2$

If it turns out that the random distribution from which $x$ is chosen encompasses $M/2$ this should work (give you better than 50-50); if the distribution is stuck in one half it would not.

However, the versions of this game I was first presented with is that the envelope is presented by someone who (possibly) seeks to minimize your income from the game. The strategy of using a distribution to decide whether to switch to the other envelope will still work in that instance.

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  • $\begingroup$ OK, points (1-3) taken. So, I am allowed to choose such a random, non-negative, continuous distribution of $d$ that $P(d < x) = P(d > x)$, correct? But then the decision is based essentially on a coin toss... am I wrong? $\endgroup$ – January Jun 29 '16 at 13:13
  • $\begingroup$ You don't require $P(d<x)=P(d>x)$ at all. You just need some non-zero probability of getting between the two amounts. $\endgroup$ – Glen_b Jun 29 '16 at 14:49
  • $\begingroup$ Yes, but I am allowed to define the density function for $d$ as I wish, right? I do that in order to lead the argument to an absurd conclusion. $\endgroup$ – January Jun 30 '16 at 6:21
  • $\begingroup$ by making your strategy a function of x you're not giving yourself the advantage of making the correct choice when d is between x and y -- you're defining your way out of winning the game. If the link you give claims that such a strategy will work they'd be wrong $\endgroup$ – Glen_b Jun 30 '16 at 7:44
  • $\begingroup$ What, in Wapner's reasoning, forbids me to define the probability function used to derive $d$ as a function of $x$? As long as $P(d \in (x,y)) > 0$, then his reasoning should still work, am I wrong? If I use a continuous, non-negative distribution that includes $x$ (e.g. uniform distribution on $(1, 2 \cdot x)$, then I am guaranteed that this is the case. And I still make the correct decision if $d \in (x,y)$. $\endgroup$ – January Jun 30 '16 at 9:23
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Wapner's argument is correct!

Some comments:

  • Following the described cut-off strategy where we switch envelopes if $x < d$ is at worst useless in ex-ante expectation. With a good choice of $d$, it can be quite useful.
  • If you add a Bayesian prior (i.e. you add beliefs about the initial distribution of money in the envelopes), you can solve for the optimal value of $d$ given your prior beliefs.
  • In certain situations (eg. where the more you observe, the more likely it is you got the big envelope), a cut-off strategy is even optimal.
  • In a more general Bayesian setting, you can do better than a simple cutoff strategy for many priors.

A related but different problem:

As several @Glen_b and @whuber have mentioned, there's a related puzzle known as the Two Envelope Problem where a fallacious argument is given for always switching envelopes and the flaw in the argument can be seen by taking a Bayesian approach and adding prior beliefs over the contents of the two envelopes.

In some sense though, the puzzle described here is rather different. Wapner's argument is correct!

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    $\begingroup$ OK, now I see where the paradox comes from. Or, to be specific, where the additional information flows into the system. By consciously choosing the distribution of d, we use our a priori knowledge about where, more or less, the amounts of money in both envelopes should be. Worst case scenario, our knowledge is useless, but the method guarantees that we will not be at a disadvantage if using it. $\endgroup$ – January Jun 29 '16 at 11:28
  • $\begingroup$ After some thought, I still don't get it -- see EDIT 2. $\endgroup$ – January Jun 29 '16 at 13:14
  • $\begingroup$ Scenario (A) Imagine the small envelope has $10$ and the big envelope has $20$. Let's choose $d$ = 15. $P(x < d) = P(x > d)$. The decision rule would lead you to the correct choice 100% of the time! $\endgroup$ – Matthew Gunn Jun 29 '16 at 13:37
  • $\begingroup$ Now let's examine some Scenario (B). Imagine the small envelope has an odd number of dollars from 1 to 9 (eg. 1 or 3 or 5 or 7 or 9) and the large envelope has 1 dollar more. Choose $d = 5.5$ and then $P(x<d) = P(x > d)$. Here though, your repick if $< 5.5$ decision rule isn't as helpful! It leads to the right decision if $x = 1, 3, 5, 6, 8,$ or $10$ and the wrong decision if $x = 2, 4, 7, 9$. Recall the possible pairs are (1,2), (3, 4), (5, 6), (7, 8) (9, 10)$ The OPTIMAL BAYESIAN thing to do knowing this initial distribution is repick if you see an odd amount of money. $\endgroup$ – Matthew Gunn Jun 29 '16 at 13:48
  • $\begingroup$ We don't know the distribution of $x$ and $y$, so we cannot pick it in a way that you propose it. Once we opened the envelope, we know $x$, but we have no idea that it was randomly chosen from integers 1 to 9, and thus we cannot choose $d$ to be 5.5. As mentioned by @Glen_b above, $d$ must be picked from a non-negative, continuous distribution. $\endgroup$ – January Jun 29 '16 at 13:58
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I was intrigued by this and took the pragmatic approach of playing with it in Excel.

I generated three random numbers for x, y, and d in the range 1-100. I then did the comparison between d and x and between x and y and looked at the result, right or wrong.

I did this 500 times and repeated that several times and regularly got the right answer arounf 330 out of 500, as predicted.

I then increased the range of d to 1-10000 and the correct answer dropped to about 260 for 500 runs.

So yes, the selection of d is dependant on the expected values of x and y.

BoB

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I think the apparent paradox with the Wapner expansion of the equation p + (1-p)/2 is that it assumes that (1-p)/2 >0. For many ranges of d this value is 0.

For example: any d selected from a symmetric distribution centered on the value in the open envelope, gives a probability of wrong 1/2 and correct 1/2.

Any asymmetrically chosen distribution appears to bias the choice the wrong way 1/2 the time.

So is there a way to choose a range and distribution for d such that this equation holds?

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