New And Improved Wallet Game

[UprightCreature]

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jazza's own words were "distribution function" not "density function." Did he mean it? I don't know.

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True I read it too fast and read what I wanted to. I assume he meant a probability density because with x = 1/sqrt(k) his average value would be $2 and he would lose anyone who chose a constant value > $2.


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It's an interesting curiosity that the simulation is probably going to return incorrect results

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These kind of problems tend to be very bad for monte carlo simulations. So much of his ev is from values of k that are extremeley small, in fact an infinite amount, that you literally miss all his EV when you try to simulate it. If you somehow did have a perfect non discrete RNG you would still need to run a number of trials >> than the large constant your oponent chose. With the numbers being discussed that would take a really long time. Closed form pen and paper solutions are better here.

[mannika]

If you lose, the amount of money you lose is INDEPENDANT of what you have in your wallet. However, if you win, the amount of money you win is equal to the amount of money in your wallet. Because of this, you should always be picking the largest number imaginable. Regardless of what number your opponent picks, ALL THAT MATTERS is that you pick a number greater than his. You cannot mitigate the risk you face by picking a lower number. There is absolutely no upside to picking any number lower than the highest number imaginable.

Edit: I see what you're saying about your distribution beating any "pick a number strategy". But your strategy is not optimal either, because as another poster mentioned, shifting it to the right by even a small amount results in a dominant portfolio. There is no finite answer to this problem.

[TomCollins]

Just because you don't understand advanced mathematics doesn't mean he is an idiot.

[TomCollins]

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Because of this, you should always be picking the largest number imaginable.

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I'll take the largest number imaginable +1.

[marv]

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there was a post a little while ago about a game called wallet, which got me thinking about the game and the best way to play it

so here are the rules: you and an opponent choose how much money to put in your wallet, when you both have done this you put your wallets on the table, and the person who's wallet has the most money gets his money doubled by the loser (this was different in my other goofed up thread)

you both have infinite bankrolls, but have to choose a finite non-zero amount of money

so what strategy can you use to guarantee a minimum EV of 0?

also, i will create a contest for those who wish to enter:

everyone PM's me their prefered strategy (in the form of a probability distribution function), and after say 3 days i will see who's strategy is best overall, by having a round robin tournament between all participents

2 points for a win, 1 for a tie, and 0 for a loss

most total points wins

one thing to note is that the best strategy for the contest may not be the strategy that guarantees a minimum EV of 0

and btw i have tried to work out the answer (to avoid further embarrisment

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The set of strategies - distribution functions on [0,infinity) - doesn't have the right structure to make this an interesting game from an 'maximise EV' perspective.

Unless both player's strategies have finite mean, the profit/loss function doesn't have a well defined expectation, so you can't evaluate one against the other.

Even if you restrict attention to strategies with finite mean, this still isn't enough for a Nash equilibrium since any strategy with finite mean is beaten by an oppo who just chooses a sufficiently large fixed number.

Marv

[kyro]

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Just because you don't understand advanced mathematics doesn't mean he is an idiot.

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Your reading comprehension is about as solid as my understanding of advanced mathematics. If you could quote me where I even hinted that Jazza was an idiot, I'll retract every statement I've made in the thread and type out a full apology. Until then, you might try to choose your words more carefully.

I have no doubt Jazza is far more knowledgeable in advanced mathematics than I am. All I've asked is that he prove me wrong, which after reading a couple posts by other people, I'm starting to get the impression that I am. So far, all I know is that there's a probability distribution that supposedly will pick a number that will be larger than any number I can come up with on my own which should bring his EV to 0...(or positive, one of the two.) How does he do this? Hell if I know, I haven't been able to get anyone to explain it to me to a point where I completely understand it.

[mostsmooth]

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So far, all I know is that there's a probability distribution that supposedly will pick a number that will be larger than any number I can come up with on my own

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i dont think weve seen this yet, have we? further, i believe somebody has already stated that it is unlikely to get a result that is larger than a googleplex, which is much smaller than the numbers we were throwing around.
of course i dont know what they are talking about, but i understand the concept. it just doesnt seem to be better than writing a really big number.

[Jazza]

the reason my probability density function (which i incorrectly called a distribution function in one of my posts) will beat a googleplex is this:

the chance i will pick a number less than a googleplex is 1-1/googleplex, and if this happens you win a googleplex off of me, so supposing for a minute i win $0 if i pick a larger number, your EV will be:

googleplex*(1-1/googleplex) = googleplex-1

not bad, but the thing is i will pick a larger number than a googleplex one googleplexth of the time (1/googleplex), and when i do, i will win on average an infinite amount of money

and since when you win, you only win a finite amount of money, overall i win an infinite amount of money off of you on average

this whole reasoning can be used for any number you choose, as long as it's finite

if you want i can show you the integrals and stuff to work out the numbers, but it's not that exciting

[Jazza]

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Just as any Delta Distribution - ie. Picking a Number - can be beat by moving the Delta Distribution 1 to the right - ie. Picking the Number +1, so can ANY distribution. Take your probablity P for example.

On any interval [a,b] define P*[a,b] = P[a-1,b-1].

P* beats your P.

PairTheBoard

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i'm not saying you're wrong, but i don't think you proved what you said in this post, please elaborate

assuming what you say is true (which is at least some of the time) then it seems there is no Nash Equilibrium, but this doesn't have to stop the contest, because this may only matter if two people happen to have the same type of probability density function

[mostsmooth]

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the reason my probability density function (which i incorrectly called a distribution function in one of my posts) will beat a googleplex is this:

the chance i will pick a number less than a googleplex is 1-1/googleplex, and if this happens you win a googleplex off of me, so supposing for a minute i win $0 if i pick a larger number, your EV will be:

googleplex*(1-1/googleplex) = googleplex-1

not bad, but the thing is i will pick a larger number than a googleplex one googleplexth of the time (1/googleplex), and when i do, i will win on average an infinite amount of money

and since when you win, you only win a finite amount of money, overall i win an infinite amount of money off of you on average

this whole reasoning can be used for any number you choose, as long as it's finite

if you want i can show you the integrals and stuff to work out the numbers, but it's not that exciting

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how is the amount you win infinite if you happen to pick a larger number?
show me the numbers, i always like to learn stuff
and i still think my really big number beats your plan
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