Nice question and nice response Chris.

I'd like to attempt a different way to formulate the problem, if I may, and then attempt a clear illustration of how to resolve it. The interesting thing for me here is that the 'payoff' is actually negative, and so as framed I will attempt a purely economic solution.

Suppose Alice and Bob each have a Necktie, and these cost x_a and x_b. It is more likely that the more expensive one will be judged to be more 'beautiful', and we assign a probability distribution function to this such that the probability that Alice's hat will be preferred is f_a (x_a - x_b), where f_a is some monotonic function from 0 to 1 where f_a(0) = 0.5, g_a(-\infinity) = 0 and f_a(\infinity) = 1.

The expected payoff for Alice is now

$ = -f_a x_a + (1-f) x_b

Alice's strategy as regards fitting out her hat then follows from taking derivatives, i.e.

d$/dx_a = -f - (x_a + x_b) df/dx_a

As f, df/dx_a, x_a and x_b are all greater than 0 (unless the hat was a liability!), the expectation value of payoff is strictly negative, and increases with increasing x_a. Hence the only rational strategy for Alice is to minimise the cost of her hat, which of course increases the probability of losing.

I think that should be right. Note that it is quite an interesting function of the payoff here, and I think that modifying f could give a more interesting more general solution.

Interesting question as ever Derek

As stated, I think the dilemma is unsolvable, because there is clearly an even symmetrical probability of one half on both sides.

But isn't that a theoretical, ideal case - a case in which an infinity of 'beautifulness statuses' are all evenly probable?

The only way to perhaps solve it would likely be to further specify 'better looking' : you could build a table of grades of better-lookingness (thus defining a finite number of better-looking statuses, rather than an infinity of possibilities) and assign a probability weighting to each grade. Then you would have a clearly specified definition of beauty status, with a probability weighting attached to each grade.

Then you would have to build in profile properties pertaining to the judge: does he or she prefer warm or cold colors, etc. Then you'd have a real-world case, easily solvable. The ideal case though as simply stated above is interestingly most likely unsolvable.

@Chris. The above problem has all the elements you need. You can remove any apparent subjectivity by replacing the two neckties with two values X and Y. If Alice has X and Bob has Y, then Alice wins if X>Y and Bob wins if Y>X. You then have to make the rule that they do not see each other's values before they enter the game.

Kraitchik's way of stating the game in terms of neckties simply is a more compact way to say all that.

Nice question and nice response Chris.

I'd like to attempt a different way to formulate the problem, if I may, and then attempt a clear illustration of how to resolve it. The interesting thing for me here is that the 'payoff' is actually negative, and so as framed I will attempt a purely economic solution.

Suppose Alice and Bob each have a Necktie, and these cost x_a and x_b. It is more likely that the more expensive one will be judged to be more 'beautiful', and we assign a probability distribution function to this such that the probability that Alice's hat will be preferred is f_a (x_a - x_b), where f_a is some monotonic function from 0 to 1 where f_a(0) = 0.5, g_a(-\infinity) = 0 and f_a(\infinity) = 1.

The expected payoff for Alice is now

$ = -f_a x_a + (1-f) x_b

Alice's strategy as regards fitting out her hat then follows from taking derivatives, i.e.

d$/dx_a = -f - (x_a + x_b) df/dx_a

As f, df/dx_a, x_a and x_b are all greater than 0 (unless the hat was a liability!), the expectation value of payoff is strictly negative, and increases with increasing x_a. Hence the only rational strategy for Alice is to minimise the cost of her hat, which of course increases the probability of losing.

I think that should be right. Note that it is quite an interesting function of the payoff here, and I think that modifying f could give a more interesting more general solution.

Hi Derek,

You entered your clarification as I was writing mine ;)

The paradox is apparent.

Alice think that the game is advantageous since the maximal loss is one necktie and the potential gain is two neckties including one judged superior. It is wrong.

If Alice has the superior necktie, the payoff is one nicktie. If Alice has not the superior nicktie, the loss is one necktie. Without the judgment, the expected gain is zero.

@Feraud - I believe you have the payoff matrix the wrong way around. The best necktie is awarded to the _other_ player. Hence if Alice has the superior necktie, the payoff is negative one necktie, and if Alice has the worse necktie, the gain is one (superior) necktie

Oops, you are right. I read too fast, "the judge takes the better necktie and awards it to the other player.", and not "the best necktie wins..."

Our solution to the problem can be found here:

https://www.researchgate.net/publication/228664715_The_two-envelope_problem_revisited

...but I was wondering if others have different insights into it.

Article The two-envelope problem revisited

The stated necktie version of the problem has a simpler way to avoid the paradox. Assuming that the players are allowed to choose which necktie they enter into the contest, the minimax solution is to choose a necktie with value zero. The situation is indeed symmetric and the payoff is zero.

If you argue that a necktie has at least scrap value, I would argue that some neckties are so bad that they are not worth the cost of disposal. By continuity, some would have net value 0. Allowing neckties with negative value does not change the minimax solution becuase presenting a necktie with negative value runs the risk of the other player offering a necktie with a higher value that is still negative. The minimax solution is still zero.

The interesting mathematical questions come from the advanced set theory considerations of each player trying to choose the smallest number that is still greater than zero. Limiting the player strategies to the open interval that does not include zero is a purely mathematical abstraction. Although it results in interesting mathematics, it does not reflect real-world economics or utility. Even if the value of the real world objects (whether they be neckties or something else) is bounded away from zero, the set of values or utilities would surely be a closed set, so the optimum strategy would be to present a necktie of minimum value if the minimum is greater than zero. That still gives a minimax solution with a net value of zero.

The game theoretic question, as opposed to the mathematical paradox, gets more interesting if we take into account the fact that, for things like neckties, "the beauty is in the eye of the beholder." That is, the two players will have different utility functions for the neckties in a given set. This produces a non-zero-sum game that could be generalized to a model for a barter economy.

nice Q！

Hi @James Baker,

I was wondering about these solutions you mention - and started thinking about this game as being similar to a Misere bid in a game like 500. Again, it is not clear that this anything other than a contrived problem - but are you aware of an analysis of such problems in a multi-player setting?

Hi Andrew,

No, I am not aware of a multi-player analysis. However, like you say, the paradoxical versions of this set up are highly contrived. In what realistic circumstances would the lower-valued necktie be rewarded?

I think that the more interesting multi-person analysis would be of my second version, in which we allow for the fact that a given necktie may be worth more to a different player than to the one now possessing it. Then, appropriate trades among the players would be like a multi-person, heavenly version of the Long Spoons parable. Obviously, there may be barter exchanges that only work with three or more simulataneous participants. I think that has value as an example for real-world business deals, but I am not sure that it involves any interesting theoretical mathematics.

Hi James,

I like your suggestion (and also enjoyed looking up the parable of the Long Spoons!).

So I suppose the problem you are discussing would play out something like this. There are m players (labelled by i), each with n neckties (labelled by j). Each player values the neckties with their personal preference as x_{ij}. The idea then is to determine a strategy such that each player maximises their own perception of the value of the neckties they have after trading., so the payoff for a given player is $_i = sum_j x_{ij}.

Such a model actually should become a simple microcosm for a realistic market-type model as even the 'value' of a necktie to a given player might be low, a rational player would still attempt to acquire the necktie if the 'price' is less than what another player would be likely to pay for it.

I suppose the interesting questions that might come out from this (if indeed it hasn't been solved to death already) are things like: what is the optimal strategy? How long does it take to converge based on trading implementations.

The latter two are probably the most interesting questions as any scenario would have to take into account not only the players own evaluation of the value of the necktie (their personal value), but also the 'market expectation' of it, as a player would be advised to trade a necktie if the trade price were such that they could sell it later on.

This is not a game or paradox at all. It is the process of using a "judge" to rank two items. The higher ranked item is "awarded"(passed) by the judge to the person whose tie was judged in second (last) place.

The term "awarded" is ambiguous since the necktie is "taken" by the judge but no legal title of transfer is involved so the "loser" has among his/her options the arrest of the judge or a civil suit against the new possessor of the necktie judged as best in game.

Wise players more interested in physics then histrionics, lawyers and judges (of the legal sort) would likely restore the tie to its original owner and go have a glass of wine with the judge's fee.

Vic is correct that it is not a (realistic) game. However, I think he misunderstood the since in which it is a paradox. I will address that issue because my earlier answers may have confused Vic and other readers on the issue of the paradox.

In my previous answers, I deliberately set aside the paradox, which I think is a purely mathematical paradox and that the real world game has alternate interpretations and alternate solutions that do not involve the paradox.

However, as reflected in the reference in Derek's original post and in some of the answers, there is a real and well-known mathematical paradox related to the artificial, unrealistic situation described by the abstract version of the game. In its purest form, each player is trying to pick the smallest number that is strictly greater than zero. My earlier answers avoided this situation by allowing neckties to have zero (or possibly negative) values, which I believe is the case in the real world.

However, in the abstract situation, each player is trying to pick a number (or select a necktie from an infinite number of neckties) that is smaller than the number picked by the other player. Pause for a moment and think about what you would use as a strategy.

If you try to think of a strategy that works no matter what the other player does, you may tie yourself into a mental knot, (-; not to be confused with a necktie knot ;-).

A minimax strategy has to work even if the other player knows your strategy. Is there a minimax strategy for this "game"? John von Neumann proved that any finite, zero-sum two-player game has a minimax solution. That is the fundamental theorem of game theory.

It is obvious that picking any particular number, no matter how small, is a losing strategy if known to the other player. The other player could always choose a lower number, for example, half whatever number you pick.

What about a mixed strategy that randomly choses among a set of numbers?

It is not as obvious, not a little reflection shows that a random selection among any finite set of numbers doesn't work either. The other player could use the same probability distribution but substitute one-half the value of each of your numbers.

However, the situation is more complicated with infinite sets. In advanced set theory, mathematicians enjoy designing infinite sets with strange (and paradoxical) properties. For example, you can chop a ball into a finite number of pieces and then reassmble the piecies into two balls identical to the original ball (the Banach Tarski theorem, which is also called a "paradox"). The mathematics is way beyond me, but it is a fun game for set theorists. For example, it depends on which set of axioms you use for set theory.

I just realized that I failed to mention an important point.

The "paradox" IS NOT that game description sets up a somewhat unrealistic situation in which each player is, in a sense, betting against himself. That would not rise to the level of a "paradox" even if people never behaved that way. But, in fact, people do behave that way sometimes. When I was a freshman in college, a group of students had a bet in which the winner would be the one who got the lowest grade in the freshman physics course. If the contest had been to get the lowest grade without actually flunking, and if there were a continuum of possible grades, then that would have been an instance of this paradox, which would be quite separate from the irrationality of actually trying to win the contest.

Yes, the game is contrived, artificial and unrealistic, but that is not the paradox. The mathematical paradox comes from trying to find a minimax solution to this contrived game. Using neckties instead of just numbers tries to give it a real-world setting, but the paradox comes from accepting the abstraction.

Despite the description James Baker and others want to interpret this as a mathematically simple set of choices for two sentient players. Further the ties are to be given a scalar value which may be zero or negative, which the judge does no more then process as higher or lower then the other player.

However the choice being made is for "better looking", and the judge can regard lower value as better looking or higher value as better looking. the players cannot know this in advance of the JUDGEMENT much like the state of the polarization of a photon before that polar state is read.

So either player has a 50% chance of winning of winning no matter what is chosen or what strategy used.

Our universe for this problem is the one created by Derek Abbott in his description above. Indeed on further reading we see that this universe has the state "better looking", and the state "superior" having four possible outcomes. So a player only has a 25% chance of winning or losing with two NULL Judgements where neither wins or loses.

Curiouser and curiouser - must be all those bunnies in Australia. {: )

Don't know what the Aussie bunnies have to say on the matter, but I think that James' interpretation regarding 'value' is the only one that is really consistent, or at least open to mathematical interpretation.

However, even if we assume that we will always have a 50% chance of winning because the judge is post-modern in their interpretation of the best necktie (or should we say 'random'?) then the optimal strategy for each player is still the same: present the necktie with the lowest cost to you (as personally defined), as you might lose your tie.

Actually, I must confess that I have presented three separate interpretations, depending on how we measure the von Neumann utility value of a necktie and the assumptions we make about the possibility of a value of zero . However, I accept Andrew's statement about consistency, because I have tried to make each interpretation consistent within its own point of view.

Furthermore, any game theoretic analysis of the situation must look at the minimax solution if the game is zero-sum or the cooperative or equilibium solutions if it is non-zero-sum.

Allowing the judge's judgement to be different from the players' is consistent with the players' having different utility functions from each other, hence a non-zero sum game, which is interesting as part of a model for a barter economy, but is not a paradox. In a realistic situation with players' utiitity functions differing, the players will be better off bartering directly, if possbile, rather than relying on a judge whose judgement may be unrelated to their utility functions.

Perhaps James Baker you should state the problem in a manner that unambiguously lays out the possible structure precisely.

Then we can see if we want to engage your version to a solution.

Maybe the "necktie paradox" never made it into the annals of game theory because it is no "paradox" at all (at least not in the sense of violating perfect rationality). Here's the perhaps most simple answer why: Set up a simultaneous game in which A and B decide whether to take part (y) or keep out (n). If both choose y, then the payoff of A is qX+(1-q)0=qX, while B's expected payoff is (1-q)Y, where q is the ex ante unknown probability for A to be selected by the judge, and X is A's utility from prevailing (Y is B's utility of prevailing). If one player (or both) chooses n, then both players receive 1 (the utility of keeping the own nectie in the absence of a judicial selection). From the description of the game it follows that X,Y>2 (prevailing is better than just having two neckties).

If qX>1 and (1-q)Y>1, then the game obviously has two Nash equilibria; (n,n) and (y,y), as y is weakly dominant for both players. For q=1/2 (as both players have no prior information about the judicial decision), the conditions translate to X>2 and Y>2, which is consistent with the story. So, given the assumptions made in the story, it is fully rational for both players to play (y,y), i.e., to mutually take part in the game.

The other onterpretation of the term "paradox" was like "why do both players find it beneficial to take part"? In terms of economics: Why does mututal participation create a rent? This is less miraculous as it seems, as it happens all the time in markets (in voluntary exchange, the fact that one party takes an advantage does not exactly exclude the possibility that the other party also takes an advantage). In this game, value is created (and allocated) IF AND ONLY IF the judge makes a decision. If he does so, the total value of the necties becomes either X or Y, which are both greater than 2. If the judge is not called for (i.e., if one player chooses n), then this value is not created, and the joint payoff of the two players is just 2, which is lower. So asking for the judge's judgment is a Kaldor-Hicks-improvement, and for q=1/2 it is even a Pareto-improvement.

There is a quite similar paradox. I think the answer to this is quite close to the problem presented:

A wealthy father passes on two envelopes each containing a check to his children, say Mike and Susan. The only information he gives is that one envelope contains exact the double value of the other. But neither Susan nor Mike know which envelope is more valuable.

They are free to interchange the envelope if both of them agree (not knowing the content of the other envelope). Let us assume both are risk averse. Paradox: Both could increase their expected value by switching. (What's wrong with this argument?)

Erwin, I'm not a trained game theorist, but there's no paradox here because it's equally true that both could also *decrease* their expected value by switching. The chance (risk) of loss exactly balances the chance (hope?) of gain.

Actually @Mark Mandel, there really is a paradox here. The clue is in the phrase, one is double the other. So it doesn't matter whether either player knows the value of their envelope or not. From the view point of Mike, with his value of M (say), then there is assumed to be an equal chance that his envelope is the larger of the two.

The outcomes are: if his was the lower value envelope, then switching gives him 2M

Conversely, if his was the higher value, then switching yields M/2

These outcomes are equally likely, hence the overall expectation value due to switching is

0.5*2M + 0.5*M/2 = 5M/4 > M.

But as correctly pointed out, Susan is in the same position, hence her expectation value is also 5S/5 (where S is the value of her envelope). Clearly they cannot both gain by switching!

In the versions I have heard, this seems to be a paradox only in the limit of infinite resources (in which case the paradox is probably correct). If we assume a total finite amount of money to share, then the probability that a given envelope will be the largest value envelope varies as a function of the amount of money in the envelope (so an envelope cannot contain more than 2/3 of the total pool of money to be shared, and if it has more than 1/3 of the total pool, it must be the largest envelope - assuming the father might choose not to divide all of the money between Mark and Susan). Again, these arguments only hold in the limit that the total funds are finite.

Note there is a related paradox (St Petersburg paradox) explicitly to do with gambling and infinite doubling sequences.

@Andrew Greentree, thanks for this explanation, but I don't see that it addresses my point. You worked out the values and concluded, "Clearly they cannot both gain by switching!" But I never said they could.

• @Erwin Amman said (of his version):

— Paradox: Both could increase their expected value by switching. (What's wrong with this argument?)

• I said:

— both could also *decrease* their expected value by switching. The chance (risk) of loss exactly balances the chance (hope?) of gain.

I think the flaw in both your reasoning is *the assumption that they already have some of the money under consideration*. But they don't, not until at least one of them has made their choice and opened their envelope (or the allotment is judged to have been decided for some other reason). Even the one with the smaller amount will enjoy a net gain *with respect to their present holdings*. They are both going to receive a bequest of some size, whether they trade envelopes or not, and each of them will wind up richer than they were before.

* not Mark; I'm not in this game! :-)

Mike's* net expectable gain across the possibilities is

(2*M) / 2 + (M/2) / 2 = (5/4) * M

The same is true for Susan, of course. So total expectable gain across the possible outcomes *is the same for both players*, and there is no paradox. Maybe there's a formal sense of "paradox" here that I'm not getting, but this is how I see it.