The necktie paradox

In the spirit of the season, I thought it would be fun to discuss a Christmas related paradox.

The necktie paradox begins as follows (source). Two men are given neckties by their wives as Christmas presents. Over drinks they start arguing about who has the cheaper necktie.

They decide to make a playful wager to settle the matter. They will talk to their wives to find out which necktie is more expensive. The man with the more expensive necktie will be the loser of the bet, and additionally, he has to hand it over his necktie to the other as a prize.

How fair is this wager to either play?

The other necktie is always better?

The first man is thinking about the bet, and he suddenly feels that he has made a good wager.

He reasons as follows: the probability of me winning or losing is an even 50/50 chance. If I lose, then I lose the value of my necktie. If I win, however, then I win more than the value of my necktie.

That is, I can bet something worth x dollars and have a 50 percent chance of winning something valued at more than x dollars. It’s a winning proposition, and I should definitely make this wager!

The second man, also thinking about the bet, uses the same reasoning to arrive at the same conclusion. He feels the bet is a winning proposition for him, and he is happy to take the wager.

This is obviously not possible that both men can be advantaged in this zero-sum game. What is the reason for the necktie paradox?

Resolving the necktie paradox

The necktie paradox is in fact a special case of the two envelopes paradox. Consequently there are a few ways to resolve the paradox. Here are a couple of them.

Resolution 1: correct the logic

One of the problems is the men are reasoning incorrectly, a common downfall when trying to do math over drinks.

The issue is that each is considering the wager in terms of his own tie being the more expensive and less expensive tie at the same time. Clearly, one’s own tie has to be one or the other, either the more expensive or the less expensive.

Therefore, the reasoning should be:

  • If I have the more expensive tie, and I make the bet, I will lose my more expensive tie
  • If I have the less expensive tie, and I make the bet, I will win the more expensive tie

There is a 50/50 chance of being in either of those situations. Hence, there is no advantage to either player, and there is no mystery to the bet.

The game will end with each man winning the more expensive tie half of the time, and hence the bet is fair.

Resolution 2: correct the expectation calculation

There is another way to view the problem. Suppose the more expensive necktie is worth x. What is the expectation of the bet?

To calculate the expectation, you need to know two things.

First, what is the chance you win the bet? This is the chance that your tie is more expensive. It is assumed that neither you nor the other man knows the value. Under such ignorance, it is reasonable to say there is a 50/50 chance that either tie is more expensive. Hence you win with a 50 percent chance and lose with a 50 percent chance.

Second, what is the payoff to the bet? In the case you win, you win the more expensive tie worth x dollars. In the case you lose, you have to give up your more expensive tie worth x dollars.

Putting these two facts together, we get:

nectie_paradox

So we can see from the math the wager has an expected value of zero, which is entirely sensible.

Resolution 3: never bet a gift from your wife

Of course, it should be clear the game is not really a zero-sum game but a negative sum game.

Upon learning the bet, and that their husbands would wager their thoughtful gifts, both wives will be angry. Clearly, there will be no winners, and the only safe bet is to avoid this game entirely. Consider yourself forewarned :)



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  • http://headinside.blogspot.com/ gmsc

    In other words, “A strange game. The only winning move is not to play.” ( http://www.youtube.com/watch?v=NHWjlCaIrQo#t=3m35s )

    Did you happen to catch the new strategy for the two-envelope paradox, announced by some Australian researchers this past August?
    http://www.physorg.com/news169811689.html

  • Benjamin Vitale

    Presh,

    Say, the less expensive necktie has value “y”, and
    the more expensive one has the value y+z with z > 0.

    Assuming that both men have an equal chance of being correct, the expected value in winnings for either man is,

    (0.50)(y + z) – (0.50)(y) = (0.50)z

    Both men are expected to make money if they bet.
    So both men are correct in choosing to bet.

  • Cecil

    Benjamin, that doesn’t make sense. Both parties in a simple wager can’t have an expected gain.

    The key is that when you lose the bet, you lose the more expensive tie – so the expected value in winnings is

    (0.5)(y+z) – (0.50)(y+z) = 0.

  • WOPR

    “The only winning move is not to play”

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    gmsc: Thanks for the links. I will look into the two envelopes math and I hope to do a writeup of this.

    Benjamin: Actually what you wrote is the incorrect reasoning. You either end up with both ties or with no ties. If y > z, the expression is either:

    0.5(z) – 0.5(z) = 0 [payoffs in terms of "profit"]

    Or

    0.5(y+z) – 0.5(y+z) = 0 [payoffs in terms of "revenue"]

    The second expression being what Cecil wrote.

  • Benjamin Vitale

    Presh, Cecil,
    You’re right, I was wrong.





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