New And Improved Wallet Game

[Jazza]

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

[Jazza]

upon further thought i'm not sure there is a strategy that guarantees a minimum EV of 0, however there are strategies that beat the strategy of picking one number no matter what the number is (as long as the number is finite)

but this doesn't have to halt the contest

[kyro]

call me stupid but i don't see how this is any different from the first game. i choose a bazillion. oh you choose a bazillion and one? damn.

[Jazza]

allrighty i'll bust out with a strategy to beat all these pick a number strategies

my probability distribution function p(x) is:

0 if x<1
1/x^2 if x>=1

according to my calculations, no matter how large a number you choose, this strategy will win with a positive infinite EV

[mostsmooth]

[ QUOTE ]
allrighty i'll bust out with a strategy to beat all these pick a number strategies

my probability distribution function p(x) is:

0 if x<1
1/x^2 if x>=1

according to my calculations, no matter how large a number you choose, this strategy will win with a positive infinite EV

[/ QUOTE ]
i agree with kyro
this game appears equally as stupid as the first game

[kyro]

ok, i still pick a bazillion. you still lose.

[mostsmooth]

[ QUOTE ]
allrighty i'll bust out with a strategy to beat all these pick a number strategies

my probability distribution function p(x) is:

0 if x<1
1/x^2 if x>=1

according to my calculations, no matter how large a number you choose, this strategy will win with a positive infinite EV

[/ QUOTE ]
and wheres the strategy? is it hidden ? (im not that good at math)

[kyro]

[ QUOTE ]
[ QUOTE ]
allrighty i'll bust out with a strategy to beat all these pick a number strategies

my probability distribution function p(x) is:

0 if x<1
1/x^2 if x>=1

according to my calculations, no matter how large a number you choose, this strategy will win with a positive infinite EV

[/ QUOTE ]
and wheres the strategy? is it hidden ? (im not that good at math)

[/ QUOTE ]

I think he's trying to apply a probability distribution to an infinite range which is really quite silly.

[Jazza]

[ QUOTE ]
I think he's trying to apply a probability distribution to an infinite range which is really quite silly.

[/ QUOTE ]

if you're talking about the fact that the probability density function is non-zero on the interval [1,infinity) why is this silly?

and according to my calculations if you picked a bazzilion you would lose to this strategy on average

[kyro]

Fine, humor me. Pick a random number using your method.

[LetYouDown]

This entire conversation is absurd. For any finite number you choose, there are a finite number of ways to win and an infinite number of ways to lose. This isn't even an interesting discussion, it's an exercise in futility.

[kyro]

QUIET LYD, I'm waiting for hero to use his formula to produce a number so he can beat me when I only pick 2,552,109

[UprightCreature]

The distribution Jazza has chosen is a little like being on the casino side of things in the Martingale system. The expectation value of the number Jazza picks is infinite, even though he is picking a finite number every time. If you pick a fixed, but large number, you will win almost every time, just like the Martingale system. However when you lose that one time you will lose on average an infinite amount. This gives you an overall EV of -Infinity.

Jazza's system is guaranteed to beat any system that has a distribution that covers a finite range in a cash game. It is not difficult to come up with a system beats Jazza's system but the distribution must cover a semi-infinite range.

The funny thing is that if you play tournament style, that is a finite number of contests where the winner is determined by the person who has won the most at the end, Jazza's system actually becomes a losing system as long as your large number is sufficiently large compared to the number of contests in the tournament.

Just like poker the strategy that maximizes ev for each contest gets you the money in a cash game, but not necessarily in a tournament.

[mostsmooth]

[ QUOTE ]
The distribution Jazza has chosen is a little like being on the casino side of things in the Martingale system. The expectation value of the number Jazza picks is infinite, even though he is picking a finite number every time. If you pick a fixed, but large number, you will win almost every time, just like the Martingale system. However when you lose that one time you will lose on average an infinite amount. This gives you an overall EV of -Infinity.

Jazza's system is guaranteed to beat any system that has a distribution that covers a finite range in a cash game. It is not difficult to come up with a system beats Jazza's system but the distribution must cover a semi-infinite range.

The funny thing is that if you play tournament style, that is a finite number of contests where the winner is determined by the person who has won the most at the end, Jazza's system actually becomes a losing system as long as your large number is sufficiently large compared to the number of contests in the tournament.

Just like poker the strategy that maximizes ev for each contest gets you the money in a cash game, but not necessarily in a tournament.

[/ QUOTE ]
i still dont see any strategy? is this due to my piss poor math education? is 0 if x<1, 1/x^2 if x>=1 a number picking strategy?

[Siegmund]

[ QUOTE ]

i still dont see any strategy? is this due to my piss poor math education? is 0 if x<1, 1/x^2 if x>=1 a number picking strategy?

[/ QUOTE ]

Yes, and yes.

Sorry, not to be rude, let me spell out a bit more:

The contest organizer asked people to submit their strategy in the form of a probability distribution function. You can use this function and its derivative directly to calculate the expectation of one strategy against another. Or, to run a simulation, you

Generate a random number k in [0,1]
find x such that F(x)=k
and play the game with x as this player's chosen strategy.

For example, with the "1/x^2" strategy, we can rearrange k=1/x^2 as x=sqrt(1/k), and our first five plays of the game might be

k=.9577 -> x = $1.02
k=.9827 -> x = $1.00
k=.8923 -> x = $1.05
k=.3103 -> x = $1.79
k=.1464 -> x = $2.61

[kyro]

Well that's nice and simple. Oh wait, it's for a bounded range. The range he wants is boundless, and I'm still waiting for him to show me that his PD helps him pick a number that is bigger than my googolplex^googolplex a googolplex times.

[UprightCreature]

[ QUOTE ]
For example, with the "1/x^2" strategy, we can rearrange k=1/x^2 as x=sqrt(1/k), and our first five plays of the game might be


[/ QUOTE ]

You skipped some steps here. 1/x^2 is the proability density not the probability, you need to integrate to get the probability. In the case of the random number being k in [0,1] then x=1/k for the 1/x^2 distribution.

Edit: And be carfull mapping [1,Infinity) to [0,1]. 0 was included in your range and should not have been. K should be in (0,1] in the example above.

[UprightCreature]

[ QUOTE ]
Oh wait, it's for a bounded range

[/ QUOTE ]

No its not. As k->0 x->Infinity.

If you pick a constant number C (albeit a large one) your EV is as follows.

You win C whenever he picks a number less than C.
The EV of this is:
Integral from 1 to C of C/x^2 dx.
(C - 1)

When he picks a number greater than C you lose x
The EV of this is:
-Integral from C to infinity of x/x^2 dx.
=-Infinity

The probability of you winning any given round is (1 - 1/C) and the probability of losing is 1/C.

As you can see for large C you usually win, but there is no bounds to how much you can lose on that rare occasion where you lose.

[Siegmund]

[ QUOTE ]

You skipped some steps here. 1/x^2 is the proability density not the probability, you need to integrate to get the probability.


[/ QUOTE ]
jazza's own words were "distribution function" not "density function." Did he mean it? I don't know. But yes, I should have noticed that he would need to put 1-(1/x^2) if he he wanted a valid cdf. (Or, perhaps, 1-1/x, if he intended his answer to be a density.)

[ QUOTE ]

Edit: And be carfull mapping [1,Infinity) to [0,1]. 0 was included in your range and should not have been. K should be in (0,1] in the example above.

[/ QUOTE ]

That is true. Most of the pseudorandom generators in fact give numbers in [0,1) and a simulation ought to have an extra line in the code to catch the chance of 0 being returned.

I confess to having never bothered to include such a line in code I have written - and don't recall ever having a crash because of it. Yet. /images/graemlins/smile.gif

It's an interesting curiosity that the simulation is probably going to return incorrect results - lots of numbers larger than 2^32 have been mentioned as possible strategies, enough to make them always win against built-in RNGs. That's going to be quite a task, finding a RNG capable of returning a number less than 1/googleplex or whatever with positive probability!

[PairTheBoard]

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

[UprightCreature]

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

[/ QUOTE ]

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.


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

[/ QUOTE ]

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]

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

[/ QUOTE ]

I'll take the largest number imaginable +1.

[marv]

[ QUOTE ]
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

[/ QUOTE ]

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]

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

[/ QUOTE ]

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]

[ QUOTE ]
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

[/ QUOTE ]
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]

[ QUOTE ]
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

[/ QUOTE ]

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]

[ QUOTE ]
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

[/ QUOTE ]
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
/images/graemlins/cool.gif

[Jazza]

no worries

like i said before if we suppose that when i pick the larger number that i win $0, then your EV is googlex-1

but now suppose that when you win you get $0, my EV is:

integegral of (how much i win given what number i picked * the chance i picked that number) over the range of one googleplex to infinity

which is:

int(x*(1/x^2),x=googleplex..infinity)

and x*(1/x^2) = 1/x, and the integral of 1/x is ln(x), so the above integral is (in a not formal way):

ln(infinity)-ln(googleplex)

the log (the ln function) of infinity is infinity, and the log of googleplex is finite, hence the result is infinity (stricter mathematicians would demand i should have used limits, but i see that as a formality in this case)

if you are not a fan of integrals, this can be avoided, i will present another similar but different strategy:

i will pick a positive number x, with chance K/x^2

where K happens to be a constant 6/Pi^2 (not that important)

so for instance, there is a K chance that i will pick the number 1, K is about .61, so there is about a 61% chance i pick the number 1

and there is a K/4 chance i pick the number 2

and there is a K/9 chance i pick the number 3, and so on..

so, say that you always pick 1000

the chance that i pick a number less than 1000 is pretty close to 1, so i'll give you an edge here and just call it 1 for now

and if you win $1000 each time this happens, this will contribute almost $1000 to your EV

but, there is a K/1001^2 chance that i pick $1001 and beat you, and if i do i will win $1001 off of you, so this will contribute K/1001^2*$1001 = $K/1001 to my EV

and similarly I might pick $1002 which will contribute $K/1002 to my EV, and again I might pick $1003 which will contribute $K/1003 to my EV, and I keep going and going

if you add up all these contributions to my EV you get:

K*(1/1001+1/1002+1/1003+1/1004+1/1005+1/1006+1/1007+....)

and as it turns out this sum is infinite, and this whole process will work for any number you choose

i hope this makes sense, i'm not very good at explaining things

[mostsmooth]

i dont have much time to discuss right now (bad work day), but 2 things: you stated my number has to be finite, yet you are apparently introducing infinity into your number(im not sure /images/graemlins/cool.gif). 2. when i win, i dont win $1000 (or whatever your number is), i win double my number (for example my number is 999 trillion trillion trillion trillion googleplexes ^999 trillion trillion trillion trillion googleplexes, which from now on i will call "RBN", which again is just an example as id add a lot more trillions and googleplexes to it for the real deal)
so i would win an amount of RBN*2, RBN*(RBN-1) times, and thats a lot of moolah. your random number will win once in RBN tries. although your choice is infinite apparently(is that legal? you said i couldnt sit down and write out a number forever because that would make my number infinite), your final number is finite, just as mine is which means your winnings arent infinite, no?
obviously i dont know anything about math, so humor me /images/graemlins/cool.gif
on a side note, part of your formula is 1/x^2 if x>=1, what if used 1/x^3 if x>=1, would mine beat yours?

[kyro]

[ QUOTE ]
on a side note, part of your formula is 1/x^2 if x>=1, what if used 1/x^3 if x>=1, would mine beat yours?

[/ QUOTE ]

I don't think yours would qualify as a probability distribution, however. I think if you toyed with a range though, you could have a better EV than Jazza. I could be wrong though.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
on a side note, part of your formula is 1/x^2 if x>=1, what if used 1/x^3 if x>=1, would mine beat yours?

[/ QUOTE ]

I don't think yours would qualify as a probability distribution, however. I think if you toyed with a range though, you could have a better EV than Jazza. I could be wrong though.

[/ QUOTE ]
i thought it was a density function?

[kyro]

Jazza, a few things I'd like cleared up if you wouldn't mind. (This really does interest me now)

1. You say your PDF is P(x) = 0 if x<1
1/x^2 if x >= 1

This doesn't make sense. P(1) = 1/1^2 = 1. Do you mean P(X > 1) = 0 if x < 1
1/x^2 if x >=1?

Also, here's something to think about. Let's call the number you derive from your formula 'X'

I will use the same formula you come up with. Then, I'll set Y = X + 1. My EV should be greater than yours, correct?

[kyro]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
on a side note, part of your formula is 1/x^2 if x>=1, what if used 1/x^3 if x>=1, would mine beat yours?

[/ QUOTE ]

I don't think yours would qualify as a probability distribution, however. I think if you toyed with a range though, you could have a better EV than Jazza. I could be wrong though.

[/ QUOTE ]
i thought it was a density function?

[/ QUOTE ]

It is. I got it confused. There's a reason it doesn't make sense as a distribution function /images/graemlins/laugh.gif

[mostsmooth]

[ QUOTE ]
Jazza, a few things I'd like cleared up if you wouldn't mind. (This really does interest me now)

1. You say your PDF is P(x) = 0 if x<1
1/x^2 if x >= 1

This doesn't make sense. P(1) = 1/1^2 = 1. Do you mean P(X > 1) = 0 if x < 1
1/x^2 if x >=1?

Also, here's something to think about. Let's call the number you derive from your formula 'X'

I will use the same formula you come up with. Then, I'll set Y = X + 1. My EV should be greater than yours, correct?

[/ QUOTE ]
if that works im gonna set mine to be Y=X*RBN/images/graemlins/cool.gif(ever heard of being right back where you started from?)

[Jazza]

[ QUOTE ]
2. when i win, i dont win $1000 (or whatever your number is), i win double my number

[/ QUOTE ]

i'm not sure i follow here, if i put $1 in my wallet and you put $1000 in your wallet, you win $1000, not $2000, agreed?

[ QUOTE ]
you stated my number has to be finite, yet you are apparently introducing infinity into your number

[/ QUOTE ]

yeah it's not as you'd expect, but my strategy always produces a finite number, yet the average is infinite

take my discrete distribution:

[ QUOTE ]
i will pick a positive number x, with chance K/x^2

where K happens to be a constant 6/Pi^2 (not that important)

so for instance, there is a K chance that i will pick the number 1, K is about .61, so there is about a 61% chance i pick the number 1

and there is a K/4 chance i pick the number 2

and there is a K/9 chance i pick the number 3, and so on..

[/ QUOTE ]

if you work out the average amount that i choose, it would be:

$1 * the probability i choose $1 +
$2 * the probability i choose $2 +
$3 * the probability i choose $3 +
and so on..

which is:

$1 * K/1^2 +
$2 * K/2^2 +
$3 * K/3^2 +
and so on

which is:
K * (1/1 + 1/2 + 1/3 + 1/4....)

which oddly enough is infinite

this is sort of the whole 'trick', to make it so that your average amount of money you choose is infinite

[ QUOTE ]
on a side note, part of your formula is 1/x^2 if x>=1, what if used 1/x^3 if x>=1, would mine beat yours?

[/ QUOTE ]

as it turns out no, the average of your distribution is actually finite, so it doesn't do to well

however 2/x^1.5 for x>=1 and 0 for x<0

would beat my other probability density distribution

[Jazza]

[ QUOTE ]
This doesn't make sense. P(1) = 1/1^2 = 1

[/ QUOTE ]

that's ok right?

[ QUOTE ]
Do you mean P(X > 1) = 0 if x < 1
1/x^2 if x >=1?

[/ QUOTE ]

i don't think i follow here

[ QUOTE ]
I will use the same formula you come up with. Then, I'll set Y = X + 1. My EV should be greater than yours, correct?

[/ QUOTE ]

so you're saying your PDF is:

0 if x<2
1/(1-x)^2 if x>=2

according to my calculations this new distribution beats my original distribution, this new one has an EV of +$1 against the old one

although i don't know if this just happens to be the case this time, or that this will always be a way to find a better strategy

i think pairtheboard said what you are saying:

[ QUOTE ]
Take your probablity P for example.

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

P* beats your P.


[/ QUOTE ]

it seems to be the case, but i'm waiting for him to prove it

[kyro]

For some reason I keep confusing density and distribution. Ignore the majority of that post.

However, for any formula you come up with, I can come up with a larger number on average.

One more thing. The winner wins the amount that THEY put in the wallet. So let's say your formula does come up with a number that is larger than the one i picked. Guess what the very next number I choose is? Now that I think about it, your strategy is terrible for this particular game. Basically, the winner wins whatever amount he chose, so the loser's amount is irrelevant. Why would you knowingly have a number that had a 1/bajillion chance of winning, if you know that each time you're going to be paying out a bajillion? This would work better I think if the winner won the loser's amount.

[Jazza]

[ QUOTE ]
However, for any formula you come up with, I can come up with a larger number on average.


[/ QUOTE ]

maybe so, but you have to know my formula first

[ QUOTE ]
Why would you knowingly have a number that had a 1/bajillion chance of winning, if you know that each time you're going to be paying out a bajillion?

[/ QUOTE ]

because when i win (even thought it's only 1 out of every bajillion times) i will be winning on average an infinite amount

[ QUOTE ]
This would work better I think if the winner won the loser's amount.

[/ QUOTE ]

according to my calculations in this game the simple pick a number strategy dominates all other types of strategies, so in my opinion it's less interesting

[kyro]

[ QUOTE ]


because when i win (even thought it's only 1 out of every bajillion times) i will be winning on average an infinite amount

[/ QUOTE ]

I don't see this...

It has to be a finite number. And when you win the finite number from me, I'm just going to bet your finite number plus some and win it back. What am I missing.

[Jazza]

this is what i posted for mostmooth:

[ QUOTE ]
i will pick a positive number x, with chance K/x^2

where K happens to be a constant 6/Pi^2 (not that important)

so for instance, there is a K chance that i will pick the number 1, K is about .61, so there is about a 61% chance i pick the number 1

and there is a K/4 chance i pick the number 2

and there is a K/9 chance i pick the number 3, and so on..

so, say that you always pick 1000

the chance that i pick a number less than 1000 is pretty close to 1, so i'll give you an edge here and just call it 1 for now

and if you win $1000 each time this happens, this will contribute almost $1000 to your EV

but, there is a K/1001^2 chance that i pick $1001 and beat you, and if i do i will win $1001 off of you, so this will contribute K/1001^2*$1001 = $K/1001 to my EV

and similarly I might pick $1002 which will contribute $K/1002 to my EV, and again I might pick $1003 which will contribute $K/1003 to my EV, and I keep going and going

if you add up all these contributions to my EV you get:

K*(1/1001+1/1002+1/1003+1/1004+1/1005+1/1006+1/1007+....)

and as it turns out this sum is infinite, and this whole process will work for any number you choose


[/ QUOTE ]

[kyro]

I must be dumb. If it appears I still don't get it after this, I'll let it go...

I pick $1000 because it's my favorite number. 999 times, you pick a number under $1000. So I win 999,000. On the 1000 try, you pick a number, and by some luck, you hit 1,000,000 and beat my 1,000. Now you are up 1,000. So for the next number, I pick 1 trillion, and now I just erased any edge you had over me. I can do this all day too. Any time you pick a number that is larger than mine, why can I not just pick a larger number than you just picked the very next time and win my money back? I have control over every number that I pick. Yes, I understand your distribution, and I realize the math behind it. But it works against a person who picks a constant number every time.

[Jazza]

there's a chance that even though you keep upping your number, i keep winning every time

but i'm actually not sure if overall i am winning on average, as you are no longer just picking one number over and over again, but i'm pretty sure i'm still + infinity EV

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
2. when i win, i dont win $1000 (or whatever your number is), i win double my number

[/ QUOTE ]

i'm not sure i follow here, if i put $1 in my wallet and you put $1000 in your wallet, you win $1000, not $2000, agreed?

[/ QUOTE ]
i dont agree, you said the winner gets double of what he puts in the wallet.(which may or may not affect any of the rest of the discussion)

[Jazza]

srry, i didn't mean getting the double the money, i meant just winning $1000 in this case

[mostsmooth]

[ QUOTE ]
srry, i didn't mean getting the double the money, i meant just winning $1000 in this case

[/ QUOTE ]
then arent we back to the original game that you admitted was stupid /images/graemlins/confused.gif

[kyro]

Yeah, and I knew that was coming. I'm sure you know what my next argument is so I'll just not say it. Your EV may be positive, but you'll have to play this game an infinite times to show it.

[Jazza]

in the original game, if you picked $1000, and i picked $1, you would only win $1

[mostsmooth]

[ QUOTE ]
there's a chance that even though you keep upping your number, i keep winning every time

but i'm actually not sure if overall i am winning on average, as you are no longer just picking one number over and over again, but i'm pretty sure i'm still + infinity EV

[/ QUOTE ]
fwiw, i wouldnt be picking the same number every time either, especially if i lost. i didnt think you would assume that we were

[Jazza]

[ QUOTE ]
Yeah, and I knew that was coming. I'm sure you know what my next argument is so I'll just not say it. Your EV may be positive, but you'll have to play this game an infinite times to show it.

[/ QUOTE ]

yeah i reckon that's true

[mostsmooth]

[ QUOTE ]
in the original game, if you picked $1000, and i picked $1, you would only win $1

[/ QUOTE ]
and in this game if you pick $1 and i pick RBN, i get RBN*2, correct?

[Jazza]

nah, i pay you just RBN

[mostsmooth]

[ QUOTE ]
nah, i pay you just RBN

[/ QUOTE ]
lol, then you need to change the rules again, because the rules now are different from that. and again, i dont know if it matters, but if you were to pay me double my bet, i think it does? nah, probably doesnt matter /images/graemlins/frown.gif

[Jazza]

well, everything i've said so far refers to the 'only recieving RBN' rule, the only thing i need to change is my bad wording in my OP

but yeah i'm pretty sure it doesn't matter much anyway

[mostsmooth]

[ QUOTE ]
well, everything i've said so far refers to the 'only recieving RBN' rule, the only thing i need to change is my bad wording in my OP

but yeah i'm pretty sure it doesn't matter much anyway

[/ QUOTE ]
2 more things:
1.my plan is to come up with RBN, then every trial the new RBN will be the old RBN^oldRBN=new RBN. this plan has got to win!!
2.is infinity-1 an acceptable amount to put in my wallet?

[kyro]

[ QUOTE ]
[ QUOTE ]
well, everything i've said so far refers to the 'only recieving RBN' rule, the only thing i need to change is my bad wording in my OP

but yeah i'm pretty sure it doesn't matter much anyway

[/ QUOTE ]
2 more things:
1.my plan is to come up with RBN, then every trial the new RBN will be the old RBN^oldRBN=new RBN. this plan has got to win!!
2.is infinity-1 an acceptable amount to put in my wallet?

[/ QUOTE ]

2. infinity-1 = infinity

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
well, everything i've said so far refers to the 'only recieving RBN' rule, the only thing i need to change is my bad wording in my OP

but yeah i'm pretty sure it doesn't matter much anyway

[/ QUOTE ]
2 more things:
1.my plan is to come up with RBN, then every trial the new RBN will be the old RBN^oldRBN=new RBN. this plan has got to win!!
2.is infinity-1 an acceptable amount to put in my wallet?

[/ QUOTE ]

2. infinity-1 = infinity

[/ QUOTE ]
thats what i was afraid of

[Jazza]

[ QUOTE ]
1.my plan is to come up with RBN, then every trial the new RBN will be the old RBN^oldRBN=new RBN. this plan has got to win!!


[/ QUOTE ]

according to my calculations nope

although after 10 rounds you have an absurdly high chance of winning, you will only be winning a finite amount, and i still have a non-zero finite chance to win an infinite amount on average

same goes after 100 rounds, 1000 rounds and so on

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
1.my plan is to come up with RBN, then every trial the new RBN will be the old RBN^oldRBN=new RBN. this plan has got to win!!


[/ QUOTE ]

according to my calculations nope

although after 10 rounds you have an absurdly high chance of winning, you will only be winning a finite amount, and i still have a non-zero finite chance to win an infinite amount on average

same goes after 100 rounds, 1000 rounds and so on

[/ QUOTE ]
ok my final plan, if i win the old RBN^oldRBN=new RBN, if i lose then old RBN^your winning number= new RBN.i know your gonna say the same thing as before, but oh well. but then again with this new plan, if i happen to lose a trial due to youre random number being insanely higher than my number, i would only be one trial away from being ahead of your winnings. the only way you come out ahead is if youre random number somehow gets bigger everytime, and that seems like it wouldnt happen.
i realize how infinity dominates finite numbers, and youre plan is cheating. of course all this is moot because my original plan was to never stop writing my first number. /images/graemlins/cool.gif
btw, if your little trick truly makes your selection range infinite, then my chances of winning any trial ever arent really absurdly high, they are absurdly low, no? my chances of winning a trial are RBN/infinity, right?
and lastly, you keep saying i can only win a finite amount, and you win an infinite amount on average, but how can you be winning an infinite amount? we put the money in the walets, we both have finite amounts. if you win, you win a finite amount just as i do, the next trial i win more than you just did.
lastly, everything above probably contradicts each other /images/graemlins/confused.gif
now my head hurts

[pzhon]

It looks like many pairs of strategies have undefined equity against each other. Conditioned on winning, the average win is infinite. Conditioned on losing, the average loss is infinite.

[PairTheBoard]

Another poster said this about your last wallet game and sounded like he knew what he was talking about. I don't think you've fundamentally changed the game. He said something like it's a game theoretic unbounded problem and thus has no Nash Equilibrium.

Yes I didn't prove that with P*[a,b]=P[a-1,b-1], P* beats P because I don't want to go to the touble of figuring out how to do it. But I'd bet a dollar it's true. P* simply picks larger numbers more often than P does. It should be as intuitively clear as seeing that picking 506 beats 505.

Neither your 1/x^2 type X pdf and my 0 on [0,2], 1/(1-x)^2 pdf for Y have finite expectations, so

Z = {2Y if Y>x, -2X if Y<X} may not either. However if that's the case I think it's just a matter than Y beats X infinitely badly.

PairTheBoard

[Jazza]

I agree pretty much with most you have said, except

[ QUOTE ]
I don't think you've fundamentally changed the game

[/ QUOTE ]

this may be true, but i have complicated it enough for a contest to be possible

[ QUOTE ]
Neither your 1/x^2 type X pdf and my 0 on [0,2], 1/(1-x)^2 pdf for Y have finite expectations, so

Z = {2Y if Y>x, -2X if Y<X} may not either. However if that's the case I think it's just a matter than Y beats X infinitely badly.



[/ QUOTE ]

according to my calculations, agaisnt X, Y has an EV of exactly $1 per round

[Jazza]

[ QUOTE ]
It looks like many pairs of strategies have undefined equity against each other. Conditioned on winning, the average win is infinite. Conditioned on losing, the average loss is infinite.

[/ QUOTE ]

yeah, that's why i changed the contest to getting 2 points for a win, 1 for a tie, and 0 for a loss, instead of adding up EV's

[PairTheBoard]

[ QUOTE ]
I agree pretty much with most you have said, except

[ QUOTE ]
I don't think you've fundamentally changed the game

[/ QUOTE ]

this may be true, but i have complicated it enough for a contest to be possible

[ QUOTE ]
Neither your 1/x^2 type X pdf and my 0 on [0,2], 1/(1-x)^2 pdf for Y have finite expectations, so

Z = {2Y if Y>x, -2X if Y<X} may not either. However if that's the case I think it's just a matter than Y beats X infinitely badly.



[/ QUOTE ]

according to my calculations, agaisnt X, Y has an EV of exactly $1 per round

[/ QUOTE ]

I would like to see those calculations. If you're right it would prove my point. But upon further reflection I think that for

Z = {2Y if Y>X , -2X if Y<X }

The pdf is going to be of the following kind (in principle not exactly)

0.5/(x^2) For x<-1 or x>1
0 For -1<x<1

I guess the weights would actually be For x<-2 or x>2 and there might be more weight on the right than the left, but still have both expectation integrals on the left and right infinite as in the example above. If that's the case I'm not sure it would make sense to say Y is beating X even though there's more probabilty weight on the right.

I think the guy to listen to is marv in this post

http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2477191&page=0&view=c ollapsed&sb=5&o=14&vc=1

He seems to know what he's talking about. I think what he says is probably right.

PairTheBoard

[pzhon]

[ QUOTE ]
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.

[/ QUOTE ]
It sounds like it would be better to require the strategies to have mean 1. However, the lack of compactness might still be a problem. Perhaps increasingly good strategies would approximate a delta function.

[Jazza]

[ QUOTE ]
Perhaps increasingly good strategies would approximate a delta function.

[/ QUOTE ]

huh? you mean like pick one number all the time?

i spent the majority of the thread explaining why this is no good

[Jazza]

well, i'll show you my method, and you can tell me if it is valid (i think it's ok, but i've never taken a stats course so i could be wrong)

p(x) is my probability density function
p(y) is your's

my EV =

int( int(x*p(x)*p(y),y=-n..x) + int(-y*p(x)*p(y),y=x..n), x=-n..n)

then take the limit as n approaches infinity

[Jazza]

[ QUOTE ]
the only way you come out ahead is if youre random number somehow gets bigger everytime, and that seems like it wouldnt happen.


[/ QUOTE ]

there's a chance this will happen

[ QUOTE ]
btw, if your little trick truly makes your selection range infinite, then my chances of winning any trial ever arent really absurdly high, they are absurdly low, no? my chances of winning a trial are RBN/infinity, right?

[/ QUOTE ]

if you pick a high number you win most of the time, but when i win, my average is infinite, making up for all those other losses i take

[ QUOTE ]
and lastly, you keep saying i can only win a finite amount, and you win an infinite amount on average, but how can you be winning an infinite amount? we put the money in the walets, we both have finite amounts. if you win, you win a finite amount just as i do, the next trial i win more than you just did.

[/ QUOTE ]

yeah this is what you'd expect, but the maths says otherwise, when you calculate the average number i choose, it's infinite

[pzhon]

[ QUOTE ]
[ QUOTE ]
Perhaps increasingly good strategies would approximate a delta function.

[/ QUOTE ]

huh? you mean like pick one number all the time?


[/ QUOTE ]
No. "Delta function" can mean more than one thing related to this problem. What I meant was something like choosing n 1/n of the time, and 0 the rest. Perhaps as n increases, the strategy improves in some context (such as against each other).

If you identify the space of outcomes with [0,1], you can graph the strategies, and they approximate a delta function.

[pzhon]

[ QUOTE ]

It sounds like it would be better to require the strategies to have mean 1. However, the lack of compactness might still be a problem. Perhaps increasingly good strategies would approximate a delta function.

[/ QUOTE ]
A way to fix this might be that if you win with x, you get paid sqrt(x).

[Jazza]

interesting idea, but i still don't get why you're trying to 'fix' this, can't we still have a good contest with what i've set up?

[pzhon]

[ QUOTE ]
interesting idea, but i still don't get why you're trying to 'fix' this, can't we still have a good contest with what i've set up?

[/ QUOTE ]
As far as I can tell, no. Under your rules, you can't compare strategies with infinite means. Every strategy with an infinite mean will beat every strategy with a finite mean.

[Jazza]

i agree that every strategy with infinite mean beats a strategy with finite mean, by as far as i can see you can compare strategies with infinite means

this is what i posted in response to pairtheboard just a little while ago:

[ QUOTE ]
p(x) is my probability density function
p(y) is your's

my EV =

int( int(x*p(x)*p(y),y=-n..x) + int(-y*p(x)*p(y),y=x..n), x=-n..n)

then take the limit as n approaches infinity


[/ QUOTE ]

i reckon this is valid, but i'm not 100% sure, what do you think?

[pzhon]

[ QUOTE ]

[ QUOTE ]
p(x) is my probability density function
p(y) is your's

my EV =

int( int(x*p(x)*p(y),y=-n..x) + int(-y*p(x)*p(y),y=x..n), x=-n..n)

then take the limit as n approaches infinity


[/ QUOTE ]

i reckon this is valid, but i'm not 100% sure, what do you think?

[/ QUOTE ]
The expected value does not exist in the usual sense because the expected win, conditional on winning, is infinite, and so is the expected loss, conditional on losing. You are trying to extend the definition of expected value.

In some situations, you may be able to pair up some of the losses with some of the wins, and come up with an answer. However, it is possible that you will get a different answer than another method of pairing up wins and losses, and it is easy for the limit not to exist.

It is like you have an absolutely divergent sequence, but the terms are not presented in a particular order. You impose an order on the terms, and then the resulting series might be conditionally convergent, or might not. However, even if your order produces a conditionally convergent series, another order would produce another answer. Actually, the terms may not even go to 0.

Under your rules, there is never any benefit to producing a lower number rather than a higher number, right? However, it is possible that by raising the numbers generated by a strategy, you would change from winning to losing against another strategy! This is analogous to the following: You can take the sequence 1-1/2+1/3-1/4+1/5..., which is conditionally convergent to a positive number, and double all of the positive terms but insert them deeper in the series to get a series that converges to a negative number.

So, basically, this scoring method for comparing strategies with infinite means may produce no answer, and the answers it does produce are not meaningful.

[Jazza]

[ QUOTE ]
The expected value does not exist in the usual sense because the expected win, conditional on winning, is infinite, and so is the expected loss, conditional on losing. You are trying to extend the definition of expected value.

[/ QUOTE ]

oh, i wasn't aware you couldn't have an infinite EV

[ QUOTE ]
However, it is possible that you will get a different answer than another method of pairing up wins and losses, and it is easy for the limit not to exist.


[/ QUOTE ]

my intuition tells me that any method that is symetric will produce the same answer, and the limit will always exist (i count the limit equalling infinity as the limit existing)

[ QUOTE ]
Under your rules, there is never any benefit to producing a lower number rather than a higher number, right?

[/ QUOTE ]

right

[ QUOTE ]
This is analogous to the following: You can take the sequence 1-1/2+1/3-1/4+1/5..., which is conditionally convergent to a positive number, and double all of the positive terms but insert them deeper in the series to get a series that converges to a negative number.

[/ QUOTE ]

well this is weird, now you are just confusing me...

i see your point, which is why this is confusing, could you argue then that ln(2) is negative??

[ QUOTE ]
So, basically, this scoring method for comparing strategies with infinite means may produce no answer, and the answers it does produce are not meaningful.

[/ QUOTE ]

i'm not convinced this is true given that the method is symetric, the quickest way to convince this is true is if you give me an example

[pzhon]

[ QUOTE ]
[ QUOTE ]
The expected value does not exist in the usual sense because the expected win, conditional on winning, is infinite, and so is the expected loss, conditional on losing. You are trying to extend the definition of expected value.

[/ QUOTE ]

oh, i wasn't aware you couldn't have an infinite EV

[/ QUOTE ]
It's not a problem to accept an EV of +infinity. That is standard, just like you can say a limit may be +infinity. It is a problem if you try to subtract infinity from infinity, which happens here.

A Cauchy distribution (c/(1+r^2)) is symmetric about the origin. It has no expected value. The law of large numbers does not work for a Cauchy distribution.

[ QUOTE ]
[ QUOTE ]
However, it is possible that you will get a different answer than another method of pairing up wins and losses, and it is easy for the limit not to exist.


[/ QUOTE ]

my intuition tells me that any method that is symetric will produce the same answer, and the limit will always exist (i count the limit equalling infinity as the limit existing)

[/ QUOTE ]
Your method is well-defined. However, it is not the only way you could try to cancel wins with losses. Instead of pairing things up by the size of the random variable, you could use the percentile: Take the value up to the 90th percentile, then 99th percentile, then 99.9th, etc. This is also well-defined, and it would not necessarily produce the same score, or the same winner.

Your limit does not always exist, even in the extended real line with +-infinity. There can be finite oscillations. There can be unbounded oscillations.

[ QUOTE ]
[ QUOTE ]
So, basically, this scoring method for comparing strategies with infinite means may produce no answer, and the answers it does produce are not meaningful.

[/ QUOTE ]

i'm not convinced this is true given that the method is symetric, the quickest way to convince this is true is if you give me an example

[/ QUOTE ]
You can build your own series by using strategies including the following building blocks: With probability c_n/2^n, take on the number 2^n. Choose your own c_n (and c_n' for a second strategy), and your limit can resemble partial sums of any series with bounded terms. (The terms don't have to go to 0.) You can also get wildly divergent series with unbounded terms by using 2^2^n with probability c_n/2^2^n, so the total probability can be less than 1 even though the c_n may be unbounded.

[PairTheBoard]

[ QUOTE ]
well, i'll show you my method, and you can tell me if it is valid (i think it's ok, but i've never taken a stats course so i could be wrong)

p(x) is my probability density function
p(y) is your's

my EV =

int( int(x*p(x)*p(y),y=-n..x) + int(-y*p(x)*p(y),y=x..n), x=-n..n)

then take the limit as n approaches infinity

[/ QUOTE ]

I'm having trouble with your notation and not sure what your doing. You need to find the expectation of the Random Variable Z where

Z = {2Y if Y>X, -2X if Y<X} where X,Y are the random varibles for P and P* respectively. When you say, let p(x) be your pdf and p(y) be mine you are fouling the notation right off the bat. You can say, let p(x) be your pdf and let p*(y) be mine and then put p* in terms of p. That would be ok.

But it looks like the main fallicy you are working under is integrating from -n to n and letting n go to infinity. That's not the way to integrate over the real line. You need to integrate from -infinity to zero and from zero to infinity seperately and add the results. If both are +infinity or -infinity you can say the entire integral is +infinity or -infinity respectively. But if one integral is -infinity and the other +infinity then the integral is simply undefined as I suspect it will be if you can work out the pdf for Z.

Your on the right track with the double integral though.

PairTheBoard

[Jazza]

my bad, let me clarify:

well, i'll show you my method, and you can tell me if it is valid (i think it's ok, but i've never taken a stats course so i could be wrong)

p(x) is my probability density function
q(y) is your's

my EV =

int( int(x*p(x)*q(y),y=0..x) + int(-y*p(x)*q(y),y=x..n), x=0..n)

then take the limit as n approaches infinity

i will declare that a negative number is not allowed to be chosen, to keep things simple

[Jazza]

just to clarify, when i said symetric, i meant that in the formula to calculate the EV, if you switch the two functions (my pdf and your pdf) you get the same answer

for instance, for my method of calculation to be what i called symetric:

the limit as n approaches infinity of:
int( int(x*p(x)*q(y),y=-n..x) + int(-y*p(x)*q(y),y=x..n), x=-n..n)

should equal:

the limit as n approaches infinity of:
int( int(x*q(x)*p(y),y=-n..x) + int(-y*q(x)*p(y),y=x..n), x=-n..n)

where p(x) is my pdf and q(x) is your's

also, if you don't mind let's just discuss function that always pick non-negative numbers, to keep things simple

now, forgive me but let's back up to something simple, i'm still stuck on ln(2)

according to what you're saying you could argue that ln(2) is negative, by ordering the terms in 1-1/2+1/3-1/4... appropriatly

so why is this not true?

[PairTheBoard]

[ QUOTE ]
my bad, let me clarify:

well, i'll show you my method, and you can tell me if it is valid (i think it's ok, but i've never taken a stats course so i could be wrong)

p(x) is my probability density function
q(y) is your's

my EV =

int( int(x*p(x)*q(y),y=0..x) + int(-y*p(x)*q(y),y=x..n), x=0..n)

then take the limit as n approaches infinity

i will declare that a negative number is not allowed to be chosen, to keep things simple

[/ QUOTE ]

ok. The x and -y should be 2x and -2y if the winner's number gets doubled but that's not really important. I understand your notation. What you're doing looks reasonable and is correctly handling the joint pdf. But here's the problem. The outcome of the game is a Random Variable, say Z, that takes on both positive and negative values. It is the Expectation of that random variable that we are interested in. You are not computing the expectation of Z. You are computing the average value for the game when both X and Y are less than n, then letting n go to infinity. This is like taking the pdf in my example, call it f;

f = 0 on [-1,1]
f(w) = .5/w^2 for w^2>1

computing it's average value for the absoulute value of W less than n, then letting n go to infinity and declaring W to have zero expectation when in fact it's expectation is undefined.

We can use your setup for the Expectation of Z which done properly will be:

Lim n to infinity (int( int(2x*p(x)*q(y),y=0..x), x=0..n)}}

+

Lim m to infinity (int(int(-2y*p(x)*q(y),y=x..n), x=0..m))))

I'll leave it to you to figure out but I'm betting both summands are infinite, one positive and one negative, making the Expectation of Z undefined.

PairTheBoard

[pzhon]

[ QUOTE ]

according to what you're saying you could argue that ln(2) is negative, by ordering the terms in 1-1/2+1/3-1/4... appropriatly

[/ QUOTE ]
No. If you rearrange the terms of a conditionally convergent series to get another conditionally convergent series, you have no guarantee that you will get the same sum. In fact, every conditionally convergent series can be rearranged to diverge, or to converge to 0.

ln 2 is not defined to be the sum of those terms in any order. The sum in that particular order is ln 2.

[PairTheBoard]

hi pzhon,

I somehow missed your posts up till now. I've been saying basically the same thing as you but I've been wondering. My probablity theory is pretty rusty. We know that if the expectation exists then the law of large numbers hold. But is the converse true? It seems to me there are weaker conditions than the existence of the expected value which still give stochastic convergence for the Averaging Statistic. For example for the random variable X with pdf:

p(x) = 0 on [-1,1]
p(x) = .5/x^2 for x^2>1

For X1...Xn independent and identically distributed as X, is it possible that

(X1+X2+...+Xn)/n

might still stochastically converge to zero as n goes to infinity?

I seem to vaguely recall something about Tightness Conditions which are weaker than existence of expected value but which still provide stochastic convergence for certain kinds of statistics. Thanks,

PairTheBoard

[pzhon]

[ QUOTE ]
We know that if the expectation exists then the law of large numbers hold. But is the converse true?

[/ QUOTE ]
Interesting question. I doubt it.

Edit: My guess was wrong. A proof of the converse can be found here (http://www.math.hmc.edu/faculty/krieger/m157llnconverse.pdf).

[ QUOTE ]
For example for the random variable X with pdf:

p(x) = 0 on [-1,1]
p(x) = .5/x^2 for x^2>1

For X1...Xn independent and identically distributed as X, is it possible that

(X1+X2+...+Xn)/n

might still stochastically converge to zero as n goes to infinity?

[/ QUOTE ]
That is too similar to a Cauchy distribution. In fact, you can create Cauchy random variables Yi so that Xi-Yi has an expected value (and in fact, is bounded). The averages behave like the averages of Cauchy random variables: They converge to a Cauchy random variable rather than a constant.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
We know that if the expectation exists then the law of large numbers hold. But is the converse true?

[/ QUOTE ]
Interesting question. I doubt it.

Edit: My guess was wrong. A proof of the converse can be found here (http://www.math.hmc.edu/faculty/krieger/m157llnconverse.pdf).

[ QUOTE ]
For example for the random variable X with pdf:

p(x) = 0 on [-1,1]
p(x) = .5/x^2 for x^2>1

For X1...Xn independent and identically distributed as X, is it possible that

(X1+X2+...+Xn)/n

might still stochastically converge to zero as n goes to infinity?

[/ QUOTE ]
That is too similar to a Cauchy distribution. In fact, you can create Cauchy random variables Yi so that Xi-Yi has an expected value (and in fact, is bounded). The averages behave like the averages of Cauchy random variables: They converge to a Cauchy random variable rather than a constant.

[/ QUOTE ]

Thanks. So in my example,

Sn = (X1+...+Xn)/n

Sn converges to a Cauchy Random Variable? That's interesting.

PairTheBoard

[Jazza]

[ QUOTE ]
ln 2 is not defined to be the sum of those terms in any order. The sum in that particular order is ln 2.

[/ QUOTE ]

this is what i thought, which is why i think just because you can rearange the sum and get a different EV, it doesn't necesarilly mean the intial sum is not the actuall EV

anyways, let me ask you this

if two players use the same strategy, the 1/x^2 strategy i've been going on about, what's the EV of each player? undefined, or 0? and if it's 0, how do you prove this?

[pzhon]

[ QUOTE ]

just because you can rearange the sum and get a different EV, it doesn't necesarilly mean the intial sum is not the actuall EV

[/ QUOTE ]
Do you not believe me when I say the EV is not defined? Are you looking for a reference to confirm what I have been telling you?

You are trying to extend the expected value. Once again, there are extensions, but the one you have proposed has many pathological properties, and doesn't cover many strategies.

[Jazza]

[ QUOTE ]
Do you not believe me when I say the EV is not defined?

[/ QUOTE ]

i'm undecided

anyways, i am curious as to what the EV is when both opponents use the same strategy

[pzhon]

[ QUOTE ]
[ QUOTE ]
Do you not believe me when I say the EV is not defined?

[/ QUOTE ]

i'm undecided

[/ QUOTE ]
If you aren't going to believe a professional mathematician telling you basic things about mathematics, then I've wasted my time. Good luck finding a source you believe.

[Jazza]

well i apoligize if i made you feel bad, but could you answer one last question of mine:

[ QUOTE ]
if two players use the same strategy, the 1/x^2 strategy i've been going on about, what's the EV of each player? undefined, or 0? and if it's 0, how do you prove this?

[/ QUOTE ]

[PairTheBoard]

It will be undefined Jazza. I think the way to convince yourself of this is to run a simulation and look at the average results. If the expectation were zero you would see the average results jumping around closer and closer to zero the more trials you do. Run a simulation and you will not see this. You will see the average Jumping All over the place and the longer you run the simulation the more you will see extreme outliers for the average popping up like you would pulling samples from an unbounded distribution somewhat similiar to the 1/x^2 pdf itself.

PairTheBoard