The necktie paradox

There's discussion on the 'necktie paradox' on this blog, where I'm a regular visitor.

I don't agree with the perspectives of those who have responded on that blog.

In the wikipedia, The necktie paradox states that each stands to either win or lose an expensive tie, each at 50% probability, so the game has no advantage to either man.

The problem reads:

Two men are each given a necktie by their respective wives as a Christmas present. Over drinks they start arguing over who has the more expensive necktie, and agree to have a wager over it. They will consult their wives and find out which necktie is the more expensive. The terms of the bet are that the man with the more expensive necktie has to give it to the other as the prize.

The first man reasons as follows: the probability of me winning or losing is 50:50. If I lose, then I lose the value of my necktie. If I win, then I win more than the value of my necktie. In other words, I can bet x and have a 50% chance of winning more than x. Therefore it is definitely in my interest to make the wager. The second man can consider the wager in exactly the same way; therefore, paradoxically, it seems both men have the advantage in the bet.

Is there a problem here?


But, I look at this problem differently:

Say, the cheaper necktie has value y, and
the other one has the value (y + z) with z > 0.

Let's assume 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.

Is this correct?
 
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Say, the cheaper necktie has value y, and
the other one has the value (y + z) with z > 0.

Let's assume 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.

Is this correct?
No, since you assume that z > 0, you have that y + z > y and will thus win by a probability 100% (not 50%)

The correct way of calculating expected gain is:

for player with necktie A (worth A dollars)
P(B>A)*B - P(A>B)*A = 0.5*(B-A)

for player with necktie B (worth B dollars)
P(A>B)*A - P(B>A)*B = 0.5*(A-B)

but since A>B or B>A with equal probability, both have equal chance of winning / losing the difference
 
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If you keep reading that wikipedia article, it answers your question.

The paradox is solved because the men's reasoning is flawed: each is considering his tie to be both the more expensive tie and the less expensive tie at the same time, while it can only be one or the other. The two outcomes of the game each should consider are:
* If I have the more expensive tie, and I make the bet, I lose my more expensive tie.
* If I have the less expensive tie, and I make the bet, I win a more expensive tie.
Thus each stands to either win or lose an expensive tie, each at 50% probability, so the game has no advantage to either man.
 
If you keep reading that wikipedia article, it answers your question.
I read the whole wikipedia article, but i wasn't convinced... i needed to do the math, and see what others may say.

I thought since we have 2 neckties of different $ values, the cheaper one is y and the other is (y+z), with z > 0, etc.

Now, I see that my reasoning was flawed.

I understand this:

No, since you assume that z > 0, you have that y + z > y and will thus win by a probability 100% (not 50%)

The correct way of calculating expected gain is:

for player with necktie A (worth A dollars)
P(B>A)*B - P(A>B)*A = 0.5*(B-A)

for player with necktie B (worth B dollars)
P(A>B)*A - P(B>A)*B = 0.5*(A-B)

but since A>B or B>A with equal probability, both have equal chance of winning / losing the difference
 
525
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Say, the cheaper necktie has value y, and
the other one has the value (y + z) with z > 0.

Let's assume 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
Only the more expensive tie is won or lost, so the expected winnings are zero:

(0.50)(y + z) – (0.50)(y+z) = 0
 
What do you, guys, think of the variation of the "necktie paradox": http://wapedia.mobi/en/Two_envelopes_problem [Broken]

the expected value of the money in the other envelope is:
(0.50) *1/2 * x + (0.50) * 2x = 5x/4

Or

the expected value in either of the envelopes 1.5 C, C being the lower of the 2 amounts

Read about the new strategy for the two-envelope paradox:

http://www.physorg.com/news169811689.html
 
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CRGreathouse

Science Advisor
Homework Helper
2,771
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In the paradox, the essential flaw is calculating relative gain rather than absolute gain. A $100 tie is 10 times more expensive than a $10 tie, but you can't say you expect to gain 900% or lose 90%, since the basis for the two is different. Better to say you stand to gain or lose $90 either way.
 
Thanks to all of you for the feedback and help.

Imagine if we had 3 people exchanging 3 neckties A1, A2 and A3

Would you help me create an interesting problem on this basis?
 

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