New And Improved Wallet Game

[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 [img]/images/graemlins/cool.gif[/img]). 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 [img]/images/graemlins/cool.gif[/img]
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 [img]/images/graemlins/laugh.gif[/img]

[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[img]/images/graemlins/cool.gif[/img](ever heard of being right back where you started from?)

[Jazza]

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

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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]

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

[/ QUOTE ]

that's ok right?

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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:

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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.