New And Improved Wallet Game

[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. [img]/images/graemlins/smile.gif[/img]

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