### Comments (83)

seems pretty easy...

EV (expected value) of heads would then be = 25% of \$3, or \$0.75EV of tails would be = 25% of \$1, or \$0.25 Your EV = \$1

EV of no matches = 50% of \$4, or \$2

So she makes \$1 each game... I'd recommend not playing unless you get a blowjob or something else

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johnyred:
seems pretty easy...

EV (expected value) of heads would then be = 25% of \$3, or \$0.75EV of tails would be = 25% of \$1, or \$0.25 Your EV = \$1

EV of no matches = 50% of \$4, or \$2

So she makes \$1 each game... I'd recommend not playing unless you get a blowjob or something else

This would be make sense if not for the mistake in the last line. It's not "EV of no matches = 50% of \$4, or \$2", it's "EV of no matches = 50% of \$2, or \$1"

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lol basic probability; P(A) = 1/4 A = 3, P(B) = 1/4, B = 1. P(c) = 1/2 C = 2. so b/c these are all independent events, the probability that u get 4 is actually 1/8 and the probability that she gets 4 is 1/4....therefore u say screw u and dont play, lol i think

p.s. i have a rule to run if a beautiful stranger stars talking to me about money....usually ends up with YOU PAY NOW!

P
Best Response

So I'm assuming you can pick your choice before hand? (instead of a random flip?)

If so you can set up a payoff matrix....

``````                       Hot Chick
T                      H
``````

You --------------------------------- | \ -2,2 | T | 1,-1 \ | ---------------------------------- | -2,2 \ 3,-3 | H | \ | -----------------------------------

No Nash Equilibrium exist for this game. It's an outguessing game.

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The reason a lot of you are having trouble is that your assuming what the guy & girl show each other is random. She is actually going to try and pick a certain side of the coin...based on what you will pick & what will maximize ur e.v. Hence the"game theory" addendum

I wont ruin the answer yet and let you guys try some more. I have faith in you WSO (kinda)

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There's no solution.

It would make sense for her to show tails all of the time, but then he would just choose tails also. Therefore, she will choose heads as she believes he believes she will show tails.... This will go on forever.

How many times do we play? I think the first round should be random regardless. Then the next round and so forth becomes a strategy play using previous information somehow. dunno much about game thoery

She could play randomly over a distribution favoring tails and win on average. You should not play.

The first two comments are REALLY off base. IF you assumed that they chose heads vs tails at random, THEN it would be an equal expected value game, as the problem suggests. The first two comments got their logic a little twisted, which is why they didn't get this.

The actual question (the game theory component) is how will they act in response to the difference of \$1 for TT and \$3 for HH. My intuition tells me this game gives an advantage to your opponent, but I'm not positive and it might still be even (you definitely do not have an advantage).

I've done game theory problems like this with excel solver before. For instance, I did one that tried to find the optimal strategy for rock paper scissors (the optimal strategy of course turns out to be that you randomly choose either rock, paper, or scissors with 1/3 probability).

I'm actually pretty interested, so I'm going to try to plug this one into Solver. I would guess the optimal strategy turns out to be something like the girl picks Tails 75% of the time, and you pick heads 75% of the time

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you're spot on, its just generalized rps with unequal payoffs. the stranger optimizes going 3:5 head:tails (which is the point at which you are playing 50:50). Small correction, you have an expected loss of -.125.

definitely no need for excel solver. surprised the match penny game doesn't get more love around here though given the implications for asset pricing bubbles.

agree with booth she should play tails with probability 5/8 if she does, your expected value is negative, and altering your heads/tails distribution in any way whatsoever will not change your expected value

The first two comments were (I'm almost positive) assuming you were flipping the coin, not "showing" it... Why use a coin as the item in question if not for that element of 50/50 chance. Anyway, I agree with whoever said that you should always show heads, so she should go tails, but knowing that, you would go tails and so on. Regardless we can conclude she has the advantage and this game has shitty EV for us.

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You should play heads 3/8 of the time and tails 5/8 of the time thus making her indifferent between heads and tails.

She should play heads 3/8 of the time and tails 5/8 of the time thus making you indifferent between heads and tails.

This is the mixed strategy Nash equilibrium.

With these strategies the expected value is that you will pay her on average .33 per round.

Might be wrong, not sure.

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This would be a great interview problem..good job guys

It's a simultaneous game with no Nash Equilibrium, right?

Yeah, I just solved it out by looking at all possible set of strategies, there's no Nash Equilibrium. It basically has to turn into a random draw of heads or tails because neither party has any incentive to go with one versus the other, and there's no point of that because the expected value of the game is 0 in that case.

see http://www.siam.org/pdf/news/331.pdf if you want the answer, apparently Marilyn vos Savant (Guinness record of highest IQ) got it semi-wrong, so don't feel bad if you were way off

Marilyn's answer: Should you play? No. She can win easily. One way: If she shows you twice as many tails as heads, she wins an average of \$1 for every six plays." Marilyn's answer, which has the stranger playing twice as many tails as heads and winning \$1 in six plays, is wrong if the opponent elects to play a pure strategy of tails all the time

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AndyLouis:
see http://www.siam.org/pdf/news/331.pdf if you want the answer, apparently Marilyn vos Savant (Guinness record of highest IQ) got it semi-wrong, so don't feel bad if you were way off
Marilyn's answer: Should you play? No. She can win easily. One way: If she shows you twice as many tails as heads, she wins an average of \$1 for every six plays." Marilyn's answer, which has the stranger playing twice as many tails as heads and winning \$1 in six plays, is wrong if the opponent elects to play a pure strategy of tails all the time

Damn proof owned me.

If the opponent plays randomly, the beautiful stranger makes (3/8)(1/2)(-\$3+\$2)+(5/8)(1/2)(-\$1+\$2) = \$0.125

Andy I lost all respect for Marilyn when she attacked Wile's Fermat's Last Theorem proof. Her argument was really disappointing

solb22:
Andy I lost all respect for Marilyn when she attacked Wile's Fermat's Last Theorem proof. Her argument was really disappointing
Her assertion that Wiles' proof should be rejected for its use of non-Euclidean geometry was especially contested. Specifically, she argued that because "the chain of proof is based in hyperbolic (Lobachevskian) geometry," and because squaring the circle is considered a "famous impossibility" despite being possible in hyperbolic geometry, then "if we reject a hyperbolic method of squaring the circle, we should also reject a hyperbolic proof of Fermat's last theorem."

maybe she just had a brain fart? later she redeems herself...

In a July 1995 addendum to the book, vos Savant retracts the argument, writing that she had viewed the theorem as "an intellectual challenge-'to find a proof with Fermat's tools.'

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The real question is: Is this person a hot/cute/beautiful girl?

is it once you know what one of the coins is (either heads or tails), one of your two possibilities of hh or tt gets canceled out while hers can work either way?

You were not specific enough. CASE 1: you both flip your coin and show it to each other possibilities: H,H: +3 T,T: -1 H,T: -2 T,H: -2 Because each outcome has an equal likelihood, your average net loss per instance is (3 - 1 - 2 - 2) / 4 = -.5 Therefore, you are at a clear disadvantage.

CASE 2: you can choose which side to show, kind of like rock-paper-scissors The girl simply has to show tails each time, and you lose \$1 if you show tails as well, and \$3 if you show heads.

Either way, you'd be a dumbass to agree.. unless you're a magician

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This was a great question. I thought it was a simple expected value question but it actually does involve thinking about best response functions.

There is no pure strategy Nash solution as has been stated above. The mixed strategy Nash equilibrium is 3/8 heads for both players. In that Nash you have an expected value of -0.125 per round.

here's another one (an old favorite) -

rules: non-zero-sum/no perfect information, 2 players, strategies per player: N >> 1, not sequential, # of pure strategy Nash equilibria: 1

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/malus: \$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?

extra credit: What is the Nash equilibrium of the game? How do you think the #'s varied when this game was actually experimented in real life?

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AndyLouis:
here's another one (an old favorite) -

rules: non-zero-sum/no perfect information, 2 players, strategies per player: N >> 1, not sequential, # of pure strategy Nash equilibria: 1

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/malus: \$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?

extra credit: What is the Nash equilibrium of the game? How do you think the #'s varied when this game was actually experimented in real life?

I know this one...I'll give the other monkeys some time. I'm a fan of the Game Theory threads, Andy.

AndyLouis:
here's another one (an old favorite) -

rules: non-zero-sum/no perfect information, 2 players, strategies per player: N >> 1, not sequential, # of pure strategy Nash equilibria: 1

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/malus: \$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?

extra credit: What is the Nash equilibrium of the game? How do you think the #'s varied when this game was actually experimented in real life?

I'd go with \$100. In real life I'd rather take the risk of getting nothing instead of a guaranteed \$2 for the chance of getting somewhere upwards of \$90. I don't really care if the opponent chooses \$99 and I get \$97. Thinking along those lines is only going to get me back to \$2. I think the Nash equilibrium assumes complete risk aversion, while in real life it's different...especially for small amounts like this.

For the first one, how did we know that our optimal strategy is the one that made us indifferent whether she showed a head or tails? (Or that her optimal strategy was the one that made her indifferent what we showed?)

For the second one, I'm pretty sure the Nash equilibrium is to list the value at \$2 I think. Having said that, it would go WAY different in real life. If I got this in real life, I'd definitely list it as \$100 (unless I just hated the other guy or something, haha).

The Nash is \$2 in the second one. Prisoner's Dilemma more or less with a nice undercutting competition thrown in. No price pair is stable until you get to \$2; someone can always do better by defecting and going lower.

Trying to figure out if there's any skew here and if we should be charging or paying for it. If you play 100 independant games with \$200 at stake, you should come out with \$0, on average. But if you start with \$4, plunk down \$2 on the table waiting for her to get one head and one tail, and then place a new bet of \$3.50 if you win \$3 or \$1 if you lose \$2, skew starts to have an impact here.

I'd go with 99 in the second one. Most are going to say 100 so I do best at 101, if 98 or lower then I'm simply eating whatever price difference there is plus \$2. No rational person should ever play anything lower than \$98. The Nash has to be \$2 though since you can always Greek the other guy by sticking him \$1 lower until that point. Not sure though I don't know how to do math...

I took a game theory class once. Long time ago. The first question that should be asked is: Are the coins are actually fair? Yeah, I know they're pennies, but still...we were always required to ask that first. If you both have coins that are double sided but opposite (example only heads on one coin, only tails on another) you can never beat the "beautiful stranger". Secondly you need to know in regards to fairness if there is an equal probability to land on either side. This will affect how you calculate the results. Lastly you need to know if the stranger will actual pay if you win. Without the answer to those questions you can't respond with 100% accuracy.

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Edit: looking at the blog, this is not a coin toss game. She gets to pick heads or tails, and so do you. Then you disclose your choices at the same time.

I think this is not really a probability problem but a calculus problem if you both have an unlimited bankroll. She is choosing a probability with which to show tails, and, given the probability she opts for, is it impossible for you to respond with a strategy that does not lose money?

We know she can't show heads more than 2/3 of the time. Otherwise, a very basic strategy for me is just to choose heads all of the time and make money off of her as she pays me \$1 more than twice as often as I pay her \$2.

However, if she shows tails more than 40% of the time, your strategy, in the very worst case, is to always show tails.

So now we look at the spot in between 60% heads and 66.67% heads. Is it impossible for you to come up with a winning- or tying- strategy if she chooses some probability in between that. I think this boils down to a problem of finding a local minima along the y axis of x local maximas.

My TI-89 is out of batteries and my multivariable calculus skills are a little rusty right now, so I am going to go GRE strategies and try it out for 63%:

Let X be the probability I choose heads: Expected Net (X)= \$1 * .63 * X - \$2 *.63 * (1-X) - \$2 * .37 * X + \$3 * .37 (1-X)

Uhoh. This is a first-order polynomial with no local maxima or minima. This is probably a bad sign for us because our best strategy is probably going to be an edge case- choosing heads or tails 100% of the time:

Expected Net(X) = -\$1.26+\$1.89X - \$0.74x+ \$1.11- \$1.11X =-\$0.15 + \$1.15X -\$1.11X =-\$0.15 +\$0.04X

Indeed, if she chooses to show 63% heads, our best response is to respond by showing heads 100% of the time. Even then, we lose \$0.11 per game.

If I were in a great mood for algebra right now, I would go back through that expected net equation and generalize it for Y, her strategy, which probably does have a local maxima between 60% and 63%, and give an exact number. But assuming it's not a coin toss- that she chooses a probability to show heads with, this is not a good game to play for us, because she can pick a probability that we will always lose money with.

I'm not really smart enough to answer this question, and have never taken a game theory class. Here's my intuitive answer.

Game favors her because there's a higher likelihood that she gets \$4 bucks in a shorter period of time – i.e., it takes two games for me to get \$4 bucks half of the time (meaning I'll have to spend more time in a goddamn public library for me to increase my odds of winning), and one game for her to get \$4 bucks half of the time.

That said, I'm still playing because i) she is a beautiful stranger, ii) I'm in a public library so I clearly don't have much else going on (no opportunity cost) and iii) if I were single, I wouldn't really give a shit if I'm out \$20 bucks but up one hottie in my address book.

Follow me on insta @FinancialDemigod
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See my edited solution above. If she shows heads with between a 60% and 66+2/3% probability, there is no strategy for us to respond with in which we break even or make money.

: 'Let's show pennies to each other, either heads or tails. If we both show heads, I pay you \$3. If we both show tails, I pay you \$1. If they don't match, you pay me \$2.' At this point, she is shushed.

s = {hh, tt, th, ht}

hh = \$3 tt = \$1 ht = -\$2 th = -\$2

do you really need game theory to tell you not to play this game based on these payout?

blastoise:
: 'Let's show pennies to each other, either heads or tails. If we both show heads, I pay you \$3. If we both show tails, I pay you \$1. If they don't match, you pay me \$2.' At this point, she is shushed.

s = {hh, tt, th, ht}

hh = \$3 tt = \$1 ht = -\$2 th = -\$2

do you really need game theory to tell you not to play this game based on these payout?

Yes .. . you are making the same mistake as lots of other posters, that this is a game wherein both player flip randomly. Either player can choose the distribution over which they pick H/T. It does not have to be 50/50.
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Boothorbust:
blastoise:
: 'Let's show pennies to each other, either heads or tails. If we both show heads, I pay you \$3. If we both show tails, I pay you \$1. If they don't match, you pay me \$2.' At this point, she is shushed.

s = {hh, tt, th, ht}

hh = \$3 tt = \$1 ht = -\$2 th = -\$2

do you really need game theory to tell you not to play this game based on these payout?

Yes .. . you are making the same mistake as lots of other posters, that this is a game wherein both player flip randomly. Either player can choose the distribution over which they pick H/T. It does not have to be 50/50.

explain how this changes anything

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blastoise:
: 'Let's show pennies to each other, either heads or tails. If we both show heads, I pay you \$3. If we both show tails, I pay you \$1. If they don't match, you pay me \$2.' At this point, she is shushed.

s = {hh, tt, th, ht}

hh = \$3 tt = \$1 ht = -\$2 th = -\$2

do you really need game theory to tell you not to play this game based on these payout?

yes..because this tells you that the e.v. will be zero. but if the girl is smart...it will negative. so yes you need game theory

bump waiting for booth to answer

Because the optimal strategy for each player depends on what they think the other players distribution of h/t is going to be. The stranger can pick a distribution over which your optimal strategy is still negative EV. This is the equilibrium that the game will end in.

EDIT: I have not thought about this super rigorously yet, you have a math degree - tell me why I'm wrong. Not trying to be snarky here.

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Boothorbust:
Because the optimal strategy for each player depends on what they think the other players distribution of h/t is going to be. The stranger can pick a distribution over which your optimal strategy is still negative EV. This is the equilibrium that the game will end in.

ok i as a viewer enter the room watch one exchange / E_1/ i then enter the room again and see another exchange /E_2?

are these independent or dependent?

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blastoise:
Boothorbust:
Because the optimal strategy for each player depends on what they think the other players distribution of h/t is going to be. The stranger can pick a distribution over which your optimal strategy is still negative EV. This is the equilibrium that the game will end in.

ok i as a viewer enter the room watch one exchange / E_1/ i then enter the room again and see another exchange /E_2?

are these independent or dependent?

Each exchange is independent.
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blastoise:
prove it
You contend that you don't need game theory to solve this. Please explain. To know whether or not I can get +ev out of this, I need to know what the h/t distribution of the stranger will be. You need to find the mixed strategy equilibrium to determine this, no?

We agree if there were to multiple trials of this game I can determine probabilities of her selection choice based on past events ?

^ if it's no then the events are random if yes then the events are conditional

check mate.

sb please.

blastoise:
^ if it's no then the events are random if yes then the events are conditional

check mate.

sb please.

Wait - this is wrong or I don't understand what you're saying. Just because you can deduce a probability distribution does not mean events are conditional. i can watch someone roll dice and deduce that the probability of rolling a 3 is 1/18. But each roll is still a random, independent variable.
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Boothorbust:
blastoise:
^ if it's no then the events are random if yes then the events are conditional

check mate.

sb please.

Wait - this is wrong or I don't understand what you're saying. Just because you can deduce a probability distribution does not mean events are conditional. i can watch someone roll dice and deduce that the probability of rolling a 3 is 1/18. But each roll is still a random, independent variable.

all i asked was is it possible for me to determine how she will respond next round based on the previous rounds

it's a yes or no question XD

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I think you can determine her optimal strategy before the game begins. She will play in a manner that makes you indifferent between heads and tails with p = probability that she plays heads. So she is playing pH and (1-p)T. You will be indifferent when

3p-2(1-p)= -2p + 1(1-p) Solving for p yields 3/8. She will play heads with probability 3/8, and you lose 12.5 cents on average each game. Or a dollar every 8 games.

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It all depends, do you show your coins at the same time? Or can she pick after you show?

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This is the ultimate paradox.

If we can't model what choices she will make based on previous events, then the pay out of the pay out of the union of events might approach zero or approach infinity. Though, after ~6 events and if given a known amount of money it is extremely easy to say who will win.

^ If you want I can explain to you why it is a known problem in probability theory

If we can model what choices she will make based on previous events. then the pay out probability is conditional hence not random.Which again, after a few rounds can ironically be made clear who will win and who will lose.

If it's the first case then, no I wouldn't play.

If it's the second then, no I wouldn't play.

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Blastoise do you know what game theory is? Haha.

Game theory problems are problems in which which you can look at the incentives provided to the game's players to determine their actions. You assume that both your opponent and you are inifinitely knowledgeable, and you find out how you and your opponent would act in this situation. If there is an optimum specific way the two of you would act then this is called the Nash Equilibrium, and is the 'solution' to the problem.

This problem sounded like a game theory problem so we treated it as such - meaning we assumed you are your opponent have infinite knowledge. Also, this problem has a mixed strategy Nash Equilibrium - an assumption here is that each player recognizes the mixed strategy that their opponent is playing (which will happen in the long run if they play the game millions of times).

I'm not following what you're saying, but it sounds like you're doing something where the players are NOT infinitely knowledgeable, and have not been able to recognize their opponents mixed strategy Nash equilibrium

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apm412:
Blastoise do you know what game theory is? Haha.

Game theory problems are problems in which which you can look at the incentives provided to the game's players to determine their actions. You assume that both your opponent and you are inifinitely knowledgeable, and you find out how you and your opponent would act in this situation. If there is an optimum specific way the two of you would act then this is called the Nash Equilibrium, and is the 'solution' to the problem.

This problem sounded like a game theory problem so we treated it as such - meaning we assumed you are your opponent have infinite knowledge. Also, this problem has a mixed strategy Nash Equilibrium - an assumption here is that each player recognizes the mixed strategy that their opponent is playing (which will happen in the long run if they play the game millions of times).

I'm not following what you're saying, but it sounds like you're doing something where the players are NOT infinitely knowledgeable, and have not been able to recognize their opponents mixed strategy Nash equilibrium

fine if they are infinity knowledgeable about what choices then all the choices are conditional hence can be after X amount of plays it will be clear who will win

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im not trying to mean but it sounds like game theory is a load of bullshit

a lot of hand waving in your argument

blastoise:
a lot of hand waving in your argument

blastoise you're only look at this from the point of view from the guy. yes-you do have prior knowledge of past events...but you are missing two key points. One- SO DOES SHE! This negates your ability to "predict her next move." Because she can predict yours..and then you can predict her prediction of yours...this goes on infinitely until you reach equilibrium.

More important is that you don't know when the game will end. All she needs to do is set up her distribution in the long run. You might think-WOW she showed 10 tails in a row...she needs to show heads to get back to the nash equilibrium. But for al you know-she's going to play another 20 heads and then 50 heads in the next 80 rounds.

Simple one sentence answer: You cannot use conditional probability based on the knowledge of nash equilibrium without knowing exactly how long you two will be playing.

Check mate to you, sir.

(I can show you a proof offline, but I doubt you're really that interested).

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solb22:
blastoise:
a lot of hand waving in your argument

blastoise you're only look at this from the point of view from the guy. yes-you do have prior knowledge of past events...but you are missing two key points. One- SO DOES SHE! This negates your ability to "predict her next move." Because she can predict yours..and then you can predict her prediction of yours...this goes on infinitely until you reach equilibrium.

More important is that you don't know when the game will end. All she needs to do is set up her distribution in the long run. You might think-WOW she showed 10 tails in a row...she needs to show heads to get back to the nash equilibrium. But for al you know-she's going to play another 20 heads and then 50 heads in the next 80 rounds.

Simple one sentence answer: You cannot use conditional probability based on the knowledge of nash equilibrium without knowing exactly how long you two will be playing.

Check mate to you, sir.

(I can show you a proof offline, but I doubt you're really that interested).

Then the events are random since they are not conditional.

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I mean I think it's kind of bullshit-ish too I guess. It's meant as a predictive to tool for modeling how smart opponents (organizations) behave in real life competitions. Example: In the Cold War, the Nash equilibrium strategy was to not nuke your opponent, as long as your opponent could promise "Mutually Assured Destruction."

For this problem, the argument is something like "I assume my opponent is going to play an optimum strategy, in which she assumes that I will also play the optimum strategy." Then you see if those strategies exist .An optimum strategy would be one in which you minimize the degree to which your opponent can take advantage of you once they discover your strategy.

My understanding is there are only a few examples in which game-theory even resembles the real world, but it's still interesting.

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I also still don't understand what you're saying haha, so I don't know if you're argument has some important insight or anything haha

And you are putting words in my mouth and telling me why I am wrong GG.

solb22:
They are neither random nor conditional.

formal proof that this is possible?

then i should be able to model your strategy and hers

hence it's conditional on the previous events....

There's a massive lack of communication going on here.

Blastoise, explain what you mean by events being random? Do you mean each time we show a penny the side if it we show is a random variable based on some probability distribution? And that each time we show the penny is an independent event? Then YES, they are random.

The optimal strategy is a 'random' one in which randomly decide to show heads with 3/8 probability and tails with 5/8 probability (or something like that, I've forgotten what the actual solution is by this point).

It would be retarded for your strategy to be 'conditoned' on what the previous events were, because to your knowledge, the previous events will shed no light on what your opponent will play next. In Rock, paper, scissors, people pretend there are 'mind games' and that you can predict what your opponent will play based on past events, but this is just a psychological perception that is not consistent with how the game really works.

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apm412:
There's a massive lack of communication going on here.

Blastoise, explain what you mean by events being random? Do you mean each time we show a penny the side if it we show is a random variable based on some probability distribution? And that each time we show the penny is an independent event? Then YES, they are random.

The optimal strategy is a 'random' one in which randomly decide to show heads with 3/8 probability and tails with 5/8 probability (or something like that, I've forgotten what the actual solution is by this point).

It would be retarded for your strategy to be 'conditoned' on what the previous events were, because to your knowledge, the previous events will shed no light on what your opponent will play next. In Rock, paper, scissors, people pretend there are 'mind games' and that you can predict what your opponent will play based on past events, but this is just a psychological perception that is not consistent with how the game really works.

One person playing the optimal strategy is conditional on the other person doing the same. If she is bimbo and starts throwing heads all the time I'll do the same.

If you make the assumption that both players choose an optimal strategy then each player's strategy is conditional on that assumption.

apm412:
There's a massive lack of communication going on here.

Blastoise, explain what you mean by events being random? Do you mean each time we show a penny the side if it we show is a random variable based on some probability distribution? And that each time we show the penny is an independent event? Then YES, they are random.

The optimal strategy is a 'random' one in which randomly decide to show heads with 3/8 probability and tails with 5/8 probability (or something like that, I've forgotten what the actual solution is by this point).

It would be retarded for your strategy to be 'conditoned' on what the previous events were, because to your knowledge, the previous events will shed no light on what your opponent will play next. In Rock, paper, scissors, people pretend there are 'mind games' and that you can predict what your opponent will play based on past events, but this is just a psychological perception that is not consistent with how the game really works.

I am claiming if she and you always make perfect choices than the events are random. How is this wrong? I don't understand this shit, but it sounds like a load of crap, no offense.

• 2

Hi guys easy one its mixed strategy nash. Every game has a mixed nash E.

Take away: dont play the game as the stranger will randomize H 0.375 and T 0.625 and the payoff for you is \$- 1/8 for her its \$1/8, and there is nothing you can do about it as its a perfect nash ;-)

``````        Stranger
0.375     0.625       -0
H       T
``````

You H 0.374999999 3 -3 -2 2 - 1/8
T 0.625000001 -2 2 1 -1 - 1/8
0 1/8 1/8

Edit: Wow sorry foor that fucked up table, but recalc it, its true :)

• 1

guys for all the good things to say about blastoise, he likes to troll math threads - don't take him seriously.

Like on the first showing of the coins how can you tell me that that event isn't random?

In the first game there is mixed strategy equilibrium which means that players assign probabilities in their choices by equating the expected values under the different strategies of the opponent. In the second game the equilibrium is at 0 because each player has an incentive to undercut the other players' number.

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