# Two questions on the Traveler's Dilemma (statistics):

**URL:** <https://boards.straightdope.com/t/two-questions-on-the-travelers-dilemma-statistics/638748>\
**Category:** Factual Questions\
**Created:** [October 24, 2012, 6:13pm UTC](https://boards.straightdope.com/t/two-questions-on-the-travelers-dilemma-statistics/638748 "2012-10-24T18:13:55Z")\
**Posts on this page:** 1\
**Page:** 6

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**Author:** ![Kraydak](https://avatars.discourse-cdn.com/v4/letter/k/5daacb/32.png) [@Kraydak](https://boards.straightdope.com/u/Kraydak)\
**Post date:** [January 29, 2013, 1:46pm UTC](https://boards.straightdope.com/t/two-questions-on-the-travelers-dilemma-statistics/638748/101 "2013-01-29T13:46:12Z")

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> [@Hellestal](#):
>
> **Kraydak** is mistaken.
> 
> The Nash equilibrium is $2. **Indistinguishable** is a mathematician. **Chronos** is a physicist. I personally won the most gold stars in my fifth grade class. We’re heavyweights on this topic. Or hell, if you don’t trust us, you can read the words from [the person who actually devised the Traveller’s Dilemma](http://www.scientificamerican.com/article.cfm?id=the-travelers--dilemma). He says the same thing. He says two dollars is the “logical” choice.
> 
> But you have to understand, words like “logical” or “rational” are essentially meaningless in this context. They’re just another way of saying “Nash equilibrium”. They are not the ultimate expression of optimal play. The entire point of the Traveler’s Dilemma is that it’s one striking case where Nash doesn’t always work.
> 
> There are some other things in the post that look a bit like [evidential decision making](http://en.wikipedia.org/wiki/Evidential_decision_theory), but that’s another kettle of fish. The fact remains that $2 is the Nash equilibrium.

And I’m an astrophysicist who, mea culpa, originally got the definition of the Nash equilibrium wrong, and therefore thought it actually applied to the TD. Shrug. One can argue that $2 is the Nash equilibrium to the TD (I’d argue it isn’t, but we can run with it). However, unless you manage to come up with a way that players A and B can submit different bids (remember, players A and B are perfect rational players playing a symmetric game), then they must submit the same bid…

The Nash equilibrium argument claims that player A will go: if player B bids $2, then my best bid is $2, so I’ll bid $2 because the same is true for player B. But that is just a statement that all bids like ($4, $2) give player A a smaller payoff than ($2, $2). We know that all bids ($X, $2) for X =/= 2 are forbidden by symmetry. So the Nash equilibrium statement is that: Player A will submit a bid of $2 because his payoff in the ($2,$2) case is better than than in a large set of impossible cases that the system can never see. A rational player doesn’t give weight to impossible scenarios, so a rational player will ignore the Nash equilibrium because it isn’t applicable to symmetric games.

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