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Traveler's Dilemma

An airline loses two suitcases belonging to two different travelers. Both suitcases happen to be identical and contain identical items. An airline manager tasked to settle the claims of both travelers explains that the airline is liable for a maximum of 100 per suitcase (he is unable to find out directly the price of the items), and in order to determine an honest appraised value of the antiques the manager separates both travelers so they can't confer, and asks them to write down the amount of their value at no less than 2 and no larger than 100. He also tells them that if both write down the same number, he will treat that number as the true dollar value of both suitcases and reimburse both travelers that amount. However, if one writes down a smaller number than the other, this smaller number will be taken as the true dollar value, and both travelers will receive that amount along with a bonus: 2 extra will be paid to the traveler who wrote down the lower value and a $2 deduction will be taken from the person who wrote down the higher amount. The challenge is: what strategy should both travelers follow to decide the value they should write down?

One might expect a traveler's optimum choice to be 100; that is, the traveler values the antiques at the airline manager's maximum allowed price. Remarkably, and, to many, counter-intuitively, the traveler's optimum choice (in terms of Nash equilibrium) is in fact 2; that is, the traveler values the antiques at the airline manager's minimum allowed price.

What I don't understand about this is that the maximum that the first traveler is guaranteed to be able to receive even if the other traveler picks the worst possible number, is $98. (He picks 100 and she picks something lower). Therefore, rationally, the traveller should never choose a number lower than 96, because even in the best case scenario, any number lower the than 96 will end with him receiving less money than he otherwise could have if he had picked 100.

So why is the "rational" answer 2? Isn't there a base case type thing that prevents the traveller from rationally choosing a number lower than 96? It seems like the "loop" of cycling downwards as he realizes that she would just choose a number one lower is broken after 96.

I've been trying to understand this for a while, but I just can't seem to grasp why the traveller would choose a number lower than 96. Help?

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  • $\begingroup$ "What I don't understand about this is that the maximum that the first traveler is guaranteed to be able to receive even if the other traveler picks the worst possible number, is $\$98$." No. You get "lower number minus two dollars" if you pick the higher number. So if you say 100, and the other says 2, you get $2-2 = 0$ dollars. $\endgroup$ – Daniel Fischer Oct 20 '14 at 18:53
  • $\begingroup$ Ohhhhh... That explains it. Thank you! Sorry I didn't read more carefully $\endgroup$ – Hermes Oct 20 '14 at 19:05
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As Daniel Fischer has explained in a comment, you have misunderstood the payoffs. If you pick \$98 and the other traveler picks $x<\$98$, then you are paid off with $x-\$2$ and the other traveler is paid off with $x+\$2$. So if you pick \$98 and the other traveler picks \$2, the other traveler receives \$4 and you receive nothing.

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