When it comes to the game in this article, I still think you make no sense. However, I now see that eliminating dominated strategies is not necessarily rational.

For the traveler’s dilemma with a choice between $180 and $300 and a p=$60, choosing 299 does not completely dominate choosing 300. In fact, for most choices (119 out of 121 possible choices) of the oponent, 300 will be a dollar better than 299. For the other 2 choices 299 is better by $59. But 300 does completely dominate 180.

For the modified version of the dilemma, 299 does dominate 300. And by iterating one down per step, 180 does become the supposed rational strategy (when one knows the opponent is also rational).

For the steal or split game, you don’t need to resort to iteration. There are only 2 choices, and stealing always dominates splitting (stealing will either double the winnings, or have no effect), so stealing seems the obvious rational strategy.

The problem with eliminating the dominated strategies in these 2 cases is that you end up with the following situation. 2 “rational” people playing one another will both lose, while 2 “irrational” people will both win. That seems like a strange definition of “rational”.

But I’m afraid micheal webster doesn’t seem to make sense. It’s true that in the situation that Presh gives there is no incentive to pick 2/3 of 10 if it is known that all others will pick 10, but that isn’t known. If any one drops below 10, then all those who stay lose. So, if you don’t know what every one else is picking, there is rational incentive to drop by 2/3.

Anything above 13 is dominated by 13 ( it only takes 7 iterations to get to 0, not 20). To decide to pick a number under 13 you don’t need to make any assumption about the other participants. To pick 0 you need to know that all participants are “rational”, and that all participants know that everyone involved is “rational” (the quotes are because it must be assumed that the desire to win the prize is rational, and any contrary motivation is not – it’s not rational to dislike smoothies or to want to avoid time with the professor).

As for the traveler’s dilemma, if I understand correctly, it is poorly described. 2 people aren’t offered between $180 and $300, but rather they get to pick any value between $180 and $300, and they are offered their bid + or – some predefined P. + to the low bid and – to the high bid.

No where does it mention what happens if the bids are equal, but I’m guessing that each would get their bid with no gain or loss.

This is not a 0 sum game, it is a positive sum game with the sum being anywhere from $360 to $600. I’d rather share $600 over $360. With a P of $300, this is like the steal of share game (although now there are 120 levels of attempting to steal – and if you both steal, you don’t go away with nothing, you just each have $120 less).

If P is less than $60, it makes absolutely no sense to chose $180 as the maximum you could get is $240, and you can garantee yourself at least $240 by choosing $300 (so $300 dominates $180). So, if you thought that $180 was dominant by iteration, you stopped the iteration too soon. And it is no surprise that people are picking between $295 and $300.

Has anyone averaged the winnings of those bidding high versus those bidding $180? That is the test of “rationality” where social aspects are considered.

If both are rational, and both know they are rational, and they will have future dealings with one another, they will both choose $300. Rational people know that cooperation benefits all in the long run, and it is hard to get cooperation if you are known to stab people in the back.

In your modified traveler’s dilemma, there is more incentive to go to $180. But a rational person who trusts his partner will still choose $300. However, it will take a smaller amount of doubt in the oponent to switch to $180 unless p is quite small.

As for a persons values and desires affecting what is rational (I think that is what you are trying to say with your “choice function”), I wholly agree.

The group size in this example was 50, assuming all rational members who all are using game theory, the member can only pick between 0 and 1.
Assume everyone picks 0, in this case everyone is the winner.
In case everyone picks 1, average would be 1, 2/3rd of which would be rounded to 1 again, hence as expected, 1 should win.
However, consider that some students picked 0 and others picked 1.
Now when only 2 unbiased choices are possible, knowledge of probability dictates that with a large enough group the set would be approximately equally divided.
Hence, assume for a moment, that a 25 pick 1 and 25 pick 0. This would mean that the average would be 25/50 = 0.5. The 2/3 of 0.5 would be would be less than 0.5 which would be rounded to 0. Hence 0 would be the winning bet.
Even if there are more than 25 people who pick 1 as long as the 2/3rd of the average remains below .5, the number would round off to 0.
Going backwards, as long as the average remains below 0.5*3/2 (i.e. 0.75) all picking 0 would win. Hence in a group of 50, until 38 people pick 1, those picking 0 would win. (this is applicable to all groups with more than 3 students)

Hence anyone with the knowledge of probability along with game theory should pick 0.