Let's discuss Mathematics

[MCdread]

Ahh yeah, I like that MCdread - I think that's the clearest intuition for me. It has the added benefit of making it a lot easier to tweak the odds on my casino scam

Yeah. Mathematically, and for any C_min and C_max between 0 and 1, you just have to calculate what the mean of y = C^2 would be for a large number of trials (so long as you sample from uniform distributions of course).

 

[dutchfire]

Given any distribution p(x), x in [0,1], with unit volume, you just calculate the integral p(x) x^2 dx, x from 0 to 1, right?

edit:
and if players a,b and c are given by distributions p_a, p_b, p_c, then you calculate:

P(C wins) = integral dz dy dx p_a(x) p_b(y) p_c(z)

with boundaries x,y in [0,z] and z in [0,1]

 

[MCdread]

Yeah, that's the integration you need to do for Mise's scenario 1.

For the general case, where you sample C in range [C_min, C_max] (with 0 < C_min < 1; 0 < C_max < 1), you do the same thing and integrate z in [C_min, C_max].
But with p_c(z) corrected by a factor that reflects "player C's loss of freedom" (to use non-rigorous language).
Integrating 1/(C_max - C_min)*p_c(z), where p_c(z) has unit volume, should work, I think.

EDIT: Actually, I'm overcomplicating things. Because the p.d.f. for C equals zero outside the sampling interval anyway, and dutchfire defines it as having an area of 1, his formula will work just fine for any case.

 

[Harv]

Bump for Paradigm Shifter...

From memory, this is called the hailstones problem, or series, or something.

Take any integer x

If x is even, halve it
If x is odd, then do 3x+1

rinse & repeat.

The conjecture is that it will always reach the loop 1-->4-->2-->1, but that's yet to be proven.

Prove that no matter which number you start from, you'll always reach a number divisible by 4.

Bump. You did mean positive integer, correct?

Start with -11
(-11)x3+1 = -32
(-32)/2 = -16
(-16)/2 = -8
(-8)/2 = -4
(-4)/2 = -2
(-2)/2 = -1
(-1)x3+1 = -2
(-2)/2 = -1 and so on

Nobody has proved this? I did not read all the posts here, but I did read the Wikipedia link provided by Paradigm Shifter.

EDIT: The Wikipedia link does specify a positive integer. My mathematical notation is rusty, and typing out a proof in mathematical notation on a forum like this is a bit awkward.

Second EDIT:
It looks like in fact we eventually do get to a number divisible by 4.
If we express an odd number as A*2^N-1, (where A is an odd number and N is an integer greater than zero) then:
a. After (2N-1) iterations, we will arrive at a number divisible by 4.
b. This number divisible by 4 will be 2*(A*3^N-1).

 

[PlutonianEmpire]

How do I use this calculator: http://www.1728.org/ellipse.htm ? I'm entering numbers, but nothing's working.

At all.

Nothing shows up, zilch.

 

[Samson]

How do I use this calculator: http://www.1728.org/ellipse.htm ? I'm entering numbers, but nothing's working.

At all.

Nothing shows up, zilch.

What are you trying to do? I follow the instructions : "click on the "Eccentricity to yx" button, enter .8, click "Calculate"" and it shows " Y/X Ratio =" " 0.6".

 

[PlutonianEmpire]

EDIT: Turns out it refuses to work offline. It's an online-only calculator, apparently. (Those) Bastards.

 

[sanabas]

Nobody has proved this? I did not read all the posts here, but I did read the Wikipedia link provided by Paradigm Shifter.

EDIT: The Wikipedia link does specify a positive integer. My mathematical notation is rusty, and typing out a proof in mathematical notation on a forum like this is a bit awkward.

Second EDIT:
It looks like in fact we eventually do get to a number divisible by 4.
If we express an odd number as A*2^N-1, (where A is an odd number and N is an integer greater than zero) then:
a. After (2N-1) iterations, we will arrive at a number divisible by 4.
b. This number divisible by 4 will be 2*(A*3^N-1).

Nobody has proved you'll always reach 1.

Proving you always reach a number divisible by 4 is easier, and you have come up with the proof I was hinting at with:

I think it's merely something interesting that I came up with the first time I met this series. But I couldn't see how it helped me to prove the series. I'll drop an even bigger hint.

Any odd number can be written as (p.2^n) - 1, where p is odd. Correct? What happens when you iterate from that?

 

[Atticus]

What do you mean by p.2^n ?

 

[sanabas]

p * 2^n

Was using . for multiply

 

[Atticus]

Ok. I thought for a while it doesn't work, but some more thought made it obvious that it does.

 

[Harv]

Nobody has proved you'll always reach 1.

Proving you always reach a number divisible by 4 is easier, and you have come up with the proof I was hinting at with:

Any odd number can be written as (p.2^n) - 1, where p is odd. Correct? What happens when you iterate from that?

I found where you said that. Actually, I did not give the proof above. So here it is.

First we define the iterative function f, so that:
f[n+1] = f[n]/2 if f[n] is even
f[n+1] = 3*f[n]+1 if f[n] is odd

It follows that if f[n] is odd, then
f[n+2] = 1.5*f[n]+0.5 and
f[n+2] = 1.5*(f[n]+1)-1

If f[n+2] is odd, then
f[n+4] = 1.5*(f[n+2]+1)-1 = 1.5^2*(f[n]+1)-1

If we express an odd number f[n] = A*2^N-1 where A is an odd number, then
f[n+2N] = 1.5^N*(A*2^N) - 1 = A*3^N - 1 which is an even number.

It follows that
f[n+2N-1] = 2*(A*3^N-1) which is divisible by 4.

I am sorry about the notations which may be difficult to follow. The square brackets following the f as in f[n] is intended as a subscript.

 

[nc-1701]

<snip>
It follows that if f[n] is odd, then
f[n+2] = 1.5*f[n]+0.5 and
f[n+2] = 1.5*(f[n]+1)-1

Not true...
Take n=7
f[9]=28
But 1.5f[7] +0.5 = 1.5*22+0.5=33.5

 

[sanabas]

Not true...
Take n=7
f[9]=28
But 1.5f[7] +0.5 = 1.5*22+0.5=33.5

You've misinterpreted something in there.

If f[n] = 7

then f[n+1] = 3 * 7 + 1 = 22

then f[n+2] = 22/2 = 11 = 1.5 * 7 + 0.5

 

[nc-1701]

You've misinterpreted something in there.

If f[n] = 7

then f[n+1] = 3 * 7 + 1 = 22

then f[n+2] = 22/2 = 11 = 1.5 * 7 + 0.5

<facepalm>

 

[Mise]

This is cool: Snooker break simulation

 

[dutchfire]

Nice.
Someone in my class in uni did a similar project once, though a bit simpler IIRC.

 

[Mise]

Stats question: if I have a model that fits data points going back to 2010, and another that fits the same data but only to 2013, is there any statistic that will say "model A is better than model B because, even though the fit is worse over the respective periods, it fits more data" or alternatively "model B is better than model A because, even though the fit is worse over the whole period, model B fits recent data better than model A"? Basically I can either fit 2013 data really really well, or 2010-present data "kind of okay". I just want a number that will tell me which is better.

I mean so far I've just weighted the square differences by recency before calculating the R^2, which I'm hoping is good enough if I just want to compare the two models, but I'd rather use some sort of standard statistical measure instead.

 

[azzaman333]

I can't think of any statistical measure that would compare something like that. I can't remember even looking at a statistic that might rate a lot of ok errors better than a small set of really low errors.

 

[nerdfighter13]

a=b
a+a=a+b
2a=a+b
2a-2b=a+b - 2b
2(a-b)=a+b -2b
2(a-b)=a-b
2=1

Saw this in a youtube comment. Is this sound reasoning or not?