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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 8:44 am

Oh, there's the big difference, P(A>B) is A, not 1-A. I have no idea why I started with 1-A.

So starting from scratch

E(w,A) = 2 * (A) + 0 * (1 - A) = 2 A
E(p,A) = 1 * (P) + 2*(A)*(1-P) = P + 2A - 2AP = 2A + P(1 - 2A)

Ew > Ep if 1-2A < 0, so A > 0.5

if value of winning is 5
E(w,A) = 5A
E(p,A) = P + 5A -5AP
so Ew > Ep if P-5AP<0, 1-5A<0,
A > 0.2

if B has declared peace
E(w,A) = 2 * (A)
E(p,A) = 2A + 1 - 2A = 1
2A > 1
A > 0.5
if B has declared war
E(w,A) = 2A
E(p,A) = 2A
Ew = Ep, so it's irrelevant
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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 9:05 am

Double posting to explain, an expected value is the mean value of a function across all values of its parameters. So it's closely tied to probabilities, in that if a function has n states, each with a probability Pn, then the expected value is sum of probabilities multiplied with values.

it's easily proved too, using the definition of probability as favourable states/total states, because the mean value would be
=sum(value at each state)/total number of states
=sum(sum(value at favourable state))/total number of states
=sum(value*number of states that produce that value)/total number of states
= sum(value *( (number of states that produce that value)/(total number of states))
= sum(value * probability of that value)

So in this case, the function is "amount of money A will get if"
E(p,A) is expected value of peace for A, or "amount of money A will get if it's at peace"
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WE ARE ALL FLOATING IN THE WINDS OF TIME. BUT YOUR CANDLE WILL FLICKER FOR SOME TIME BEFORE IT GOES OUT -- A LITTLE REWARD FOR A LIFE WELL LIVED. FOR I CAN SEE THE BALANCE AND YOU HAVE LEFT THE WORLD MUCH BETTER THAN YOU FOUND IT, AND IF YOU ASK ME, said Death, NOBODY COULD DO ANY BETTER THAN THAT...
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Re: Ask a Mathematician

Postby cmsellers » Sat Dec 10, 2016 9:10 am

So it seems like my intuitions are essentially correct, but you're interpreting the bonus questions differently leading to different results for those questions.

Regarding the $5 trillion dollar pot: it seems like the value of peace should scale with the pot to 2.5, since we're assuming both countries have equal resources. I take it that you're assuming either that one country has $1 trillion and another $4 trillion and we're working out the utility for that country, or else that war becomes a positive-sum game with expected utility of going to war equal to $1.5 trillion?

And I still don't get why you're assuming that if one party has declared peace their armed forces could be anywhere between 0 and 1. Surely if they've declared peace (and we're assuming rational actors), the assumption is that the enemy's armed forces are between 0 and .5?
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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 9:18 am

the 5 trillion pot would not change anything if it changes the value of peace, they'd all cancel out in the same way the word 'trillion' does in the calculation.
and in the second case, why would we assume rational actors? A has no way of knowing B's army size regardless of whether they've declared peace.
A has to make their decision independently with just the info that B has not declared war. i.e. the assumption is that the expected value of their decision is independent of the expected value of B's decision, since they're symmetric either way.
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Re: Ask a Mathematician

Postby cmsellers » Sat Dec 10, 2016 9:22 am

Doesn't Game Theory depend on assuming rational actors?
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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 9:27 am

The person whose actions you're predicting has to be rational. Not all actors. In Prisoner's dilemma terms for example, the implication that B is doing the maths would be to assume B is not informing no matter what, which puts a 0 value on not informing.
A rational A assuming a rational B would inform.
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WE ARE ALL FLOATING IN THE WINDS OF TIME. BUT YOUR CANDLE WILL FLICKER FOR SOME TIME BEFORE IT GOES OUT -- A LITTLE REWARD FOR A LIFE WELL LIVED. FOR I CAN SEE THE BALANCE AND YOU HAVE LEFT THE WORLD MUCH BETTER THAN YOU FOUND IT, AND IF YOU ASK ME, said Death, NOBODY COULD DO ANY BETTER THAN THAT...
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Re: Ask a Mathematician

Postby cmsellers » Sat Dec 10, 2016 9:32 am

In the Prisoner's Dilemma, you do expect the other player to defect, unless you're playing iterated games.

This seems to me like a case where if A were a rational actor, they would assume B is also a rational actor. If B declares peace it's because their army is < .5. Therefore if A's army is > .25 and there is no cost to declaring war, the expected utility of declaring war should outweigh the expected utility of peace.
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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 9:35 am

The best outcome of a prisoner's dilemma is to cooperate and not inform, because then neither of them will get the harsher punishment.
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it's made with the full assumption that B could inform, even if it's not the better option. Same rules apply here. The other party's decision can't be assumed to be your decision.
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Re: Ask a Mathematician

Postby cmsellers » Sat Dec 10, 2016 9:43 am

That's the co-operative solution. In a single game without some method of enforcing it, the optimal strategy is always to inform.

Also, only declaring war if your army > .5 assumes that B is acting at random, and we have no reason to assume that. We don't have those probabilities, but saying "since we don't know exactly how irrational B is so we're holding off until .5" is irrationably cautious. If we assume that B. is irrational to an unknown degree this problem becomes unsolvable. Since we weren't given that information, it makes sense to assume that B. is a rational actor.

Also, I'm going to bed now. Hopefully Carrie will weigh in tomorrow.
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Re: Ask a Mathematician

Postby A Combustible Lemon » Sat Dec 10, 2016 9:49 am

It's not assuming irrationality on B's part, it's allowing for irrationality on B's part. And the point of bringing prisoner's dilemma into this is to point out that rational actors would always act symmetrically, and the symmetric act of not informing has more value and less jail time than the symmetric act of informing. A does not believe B would act symmetrically in calculating his chances. i.e, the only actor we're assuming rationality for is the actor whose actions we're choosing.
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WE ARE ALL FLOATING IN THE WINDS OF TIME. BUT YOUR CANDLE WILL FLICKER FOR SOME TIME BEFORE IT GOES OUT -- A LITTLE REWARD FOR A LIFE WELL LIVED. FOR I CAN SEE THE BALANCE AND YOU HAVE LEFT THE WORLD MUCH BETTER THAN YOU FOUND IT, AND IF YOU ASK ME, said Death, NOBODY COULD DO ANY BETTER THAN THAT...
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Re: Ask a Mathematician

Postby CarrieVS » Sat Dec 10, 2016 12:21 pm

I don't probability well, I'm gonna leave this one to Lemon.
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Re: Ask a Mathematician

Postby DoglovingJim » Sat Dec 10, 2016 3:00 pm

I was talking to a little Asian primary school kid in the pool (being an old white man this should spark alarms, but I know his mother so I guess it's okay). And he asked me what 2+2 was, and I responded 4.

He said "nooooo.... It is fish, you're so stupid Jim". First I thought he must have been drunk and considered splashing him in the face for his response but than I realised that kids don't drink alcohol and I had to be the bigger man, so I allowed it to slide.

Anyway my question is this. Is he crazy? Or is this 2+2= Fish thing something that makes sense for mathematicians?
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Re: Ask a Mathematician

Postby CarrieVS » Sat Dec 10, 2016 5:24 pm

That is not a thing that makes sense. I assume it's some kind of joke: when I was a kid we had 1+1= window.

You draw 1 + 1 with the 1's just straight lines and the + touching each 1 and as high as them, then you draw an = big over the top so that the two bars join the top and bottom of each 1. When I was in primary school this was high comedy.

EDIT: I'd never heard of 2+2= fish but apparently this is it.
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Re: Ask a Mathematician

Postby cmsellers » Sat Dec 10, 2016 7:34 pm

A Combustible Lemon wrote:It's not assuming irrationality on B's part, it's allowing for irrationality on B's part. And the point of bringing prisoner's dilemma into this is to point out that rational actors would always act symmetrically, and the symmetric act of not informing has more value and less jail time than the symmetric act of informing. A does not believe B would act symmetrically in calculating his chances. i.e, the only actor we're assuming rationality for is the actor whose actions we're choosing.

Your interpretation of the Prisoner's Dilemma is wrong. In a single incidence of the Prisoner's Dilemma, rational actors will always inform, unless there's an outside force encouraging them not to. In iterated Prisoners' Dilemmas not informing the first round and a tit-for-tat strategy is thereafter usually the optimal solution, but in a single game you need some outside reason (like trust in your other prisoner) not to inform.
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Re: Ask a Mathematician

Postby cmsellers » Wed Dec 21, 2016 7:14 am

My answer was too simple. And I had to read the real answer three times before I finally understood it. It's a devilishly clever problem.

Two countries, call them A and B, with armies of randomly determined strengths somewhere between 0 (very weak) and 1 (very strong), are deciding whether or not to declare war on each other to win a stash of gold. Each country knows its own strength, but not that of the other. They declare “war” or “peace” simultaneously. If either declares war, they go to war, and the country with the stronger army wins. The optimal strategy? Always declare war!

Why? Suppose the optimal strategy is to declare war when your strength is greater than some number X and that, because each player is facing an identical situation at the beginning of the game, both players play according to this strategy. Here’s where the game theory comes in. If Country B declares war if and only if its strength is greater than X, Country A would do well to declare war whenever its strength is greater than X/2. Why? There are two cases: In those situations where B’s strength (unknown to A) is greater than X, it doesn’t make any difference what A does, since B will declare war and A will be forced to fight. But in those situations where B’s strength is less than X, A profits by lowering its threshold to X/2. In half of those scenarios, it will be stronger than B, win the war, and double its gold cache.

But if A is declaring war when its strength is greater than X/2, then B will do well to declare war when its strength is greater than X/4, but then A would do well to declare war when its strength is greater than X/8, and so on. It’s bloodshed all the way down. They don’t stop until they’re both declaring war in every situation. Put another way, the only equilibrium “threshold” strength is 0. The strengths are always greater than 0, so both countries will always declare war.

I like it, but also hate myself for not getting it.
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