Jason_t's flight home...

[El Diablo]

CMI,

Hmmm, that sure sounds like you're now saying that given the problem AS STATED, the solution is 100, not 2. And given the 100k iteration problem AS STATED, the solution is always pick 100.

[Razor]

[ QUOTE ]
2 is only the correct answer to your initial problem if we are 100% certain that we will NEVER play this game against this opponent ever again. Otherwise, 100 is the correct answer.

[/ QUOTE ]

so basically the definition of 'rational' we are using is that being GUARRANTEED SOMETHING no matter how small is ALWAYS better than RISKING getting NOTHING no matter what the size of the RISK or REWARD?

[DMBFan23]

[ QUOTE ]
[ QUOTE ]

if it was 100,000 rounds I would probably enter with tit-for-tat. if it was 3 rounds which strategy would you enter?

[/ QUOTE ]

I'm not sure. Not a lot of time for retaliation and such.

If it was four or five rounds I'd be a bit more comfortable with tit-for-tat but would still have some doubt.

Six to ten rounds would probably be enough for tit-for-tat to do well.

More than 20 rounds and I'd be very comfortable with a tit-for-tat style strategy.

Beyond 100 there would be no doubt at all that a tit-for-tat like strategy would be reasonable and that an unresponsive defecting strategy would be terrible.

There's no hard and fast line. A lot of things in life are like that. It's known as the Paradox of the Heap.

The trouble with the backward induction argument is that it makes no distinction between 1 round, 3 rounds, and 100,000 rounds. It appears logically unassailable when you say it fast but that doens't mean there isn't a subtle fallacy contained therein.

[/ QUOTE ]

yeah, in the rationality paper you linked (with the surprise exam) I believe they discuss this - how humans operate under the assumption that past results impact future decisions. also, not sure if they mention this or now, but humans can't really conceptualize 100,000 moves, so it wouldn't surprise me if many treat it as an infinite scenario. it makes it pretty interesting IMO.

[El Diablo]

Razor,

Let me restate my "final" statement, changing is to can.

2 can only be the correct answer to your initial problem if we are 100% certain that we will NEVER play this game against this opponent ever again. Otherwise, 100 is the correct answer.

[CallMeIshmael]

[ QUOTE ]
Razor,

Let me restate my "final" statement, changing is to can.

2 can only be the correct answer to your initial problem if we are 100% certain that we will NEVER play this game against this opponent ever again. Otherwise, 100 is the correct answer.

[/ QUOTE ]


Obviously Im not a fan of continuing this thread, but I just throught of something, and this is actually wrong.

The problem with inifintely repeated games, is, there is an assumption of the probability of playing again.


For example, tit-for-tat needs to have at least a 2/3 chance of playing again, or it wont work (2/3 is arbitrary, and depends on the payoffs)

Since the probability of playing again in the OP is SOOOO low, Im certain that your conclusion cannot be corrct.

(if the prob. were higher, OR the payoffs were different, your conclusion might have merit...as it the problem stands, it does not)

EDIT: ie. if the prob of playing again is 1 in a million, and the cost of defecting now on later rounds is 98, you only lose 98/milllion + 98/million^2....

This doenst sum to a big enough number to matter

[disjunction]

[ QUOTE ]
2 can only be the correct answer to your initial problem if we are 100% certain that we will NEVER play this game against this opponent ever again. Otherwise, 100 is the correct answer.

[/ QUOTE ]

You mean the problem where jason_t buys the same vase as John Nash for the same price and runs into an airport dude who wants to play a weird math game?

It is NEAR 100% probable that this game will never be played again. I don't think it needs to be absolutely 100% probable, just probable enough to negate the expected value difference from the next time we play.

(except I'm not sure if introducing EV to the conversation is appropriate, but it shouldn't have to be 100%)

[CallMeIshmael]

[ QUOTE ]
You mean the problem where jason_t buys the same vase as John Nash for the same price and runs into an airport dude who wants to play a weird math game?

[/ QUOTE ]

Technically speaking, this game COULD have ramifications on a number of future interations between jason/nash, not JUST this game if it were played again.

But, overall, I agree with your post.

[CallMeIshmael]

[ QUOTE ]
CMI,

Hmmm, that sure sounds like you're now saying that given the problem AS STATED, the solution is 100, not 2. And given the 100k iteration problem AS STATED, the solution is always pick 100.

[/ QUOTE ]

I've already retracted that statement...

If you make the game such that there is a significant probability they will play again, then we'll talk.

Since as the OP was stated there is no reason to believe it, I have found a pretty clear hole in your logic

[Razor]

El Diablo,

2 seems to me to be bad answer even given your stipulation. However, I don't know enough about this nonsense, nor are my thoughts clear enough to make a useful argument.

Any definition of rational that results in 2 being the correct answer for the original problem seems useless.

[Phaedrus]

[ QUOTE ]
yeah, in the rationality paper you linked (with the surprise exam) I believe they discuss this

[/ QUOTE ]

That's a good point. Someone way back in this thread mentioned that the problem reminded them of the surprise exam (aka "pop quiz" or "unexpected hanging" problem).

It's much easier to see that there is a fallacy in backward induction in that problem. (Though putting your finger on exactly where the fallacy lies is extremely difficult).

I'd be interested in knowing whether CMI thinks there is a backwards induction fallacy in the reasoning in the pop quiz problem. And if so, why it isn't also a fallacy in the vase problem.