Something I've been thinking about

[Rainclouds]

Ringmaster is spot on and he has answered all questions which can be asked about this situation, imo.

To summarize once again:
1) our equity vs villains range is 50% in all situations (we all agreed on that)
2) therefore, our chance of winning the pot is 50% (ORLY)
3) if you lose the pot, you lose all, if you win the pot, you win the whole pot (ORLY)
4) therefore, no matter what the distribution of villains range is, we have 50% chance of winning the whole pot, and 50% chance of losing the whole pot (YA RLY).
5) variance is the mean square deviation of values from their arithmetic mean
6) the arithmetic mean is 0.5
7) if we win, we win 1. The deviation from the arithmetic mean is 0.5. The square deviation is 0.5*0.5=0.25
8) if we lose, we win 0. The deviation from the arithmetic mean is -0.5. The square deviation is -0.5*-0.5=0.25
9) The variance is the mean of 0.25 and 0.25, which is 0.25 (ORLY)
10) From (1) to (9) follows that the variance is the same for all situations.

It's mathematically proven so there's not really anything to argue about. And now I must go to sleep.

[DaycareInferno]

i can smell the money now.

[Rainclouds]

Oh yeah, I'd like to add one thing:

Some posters are trying to calculate the variance of the distribution of villains hand.
The only thing that matters to the poker player though is the variance of the outcomes of the hands.

If we speak about variance in the cheese thread or in bbv, we are always addressing the variance on the outcomes (swings, money graphs etc.)

Not a single soul makes a post about how high the variance of the distribution of his opponents range is after a big downswong. That's totally irrelevant.

I hope this clears things up better.

[Sorcerer808]

tyty RC

[Ratamahatta]

[ QUOTE ]

8) if we lose, we win -1 . The deviation from the arithmetic mean is -0.5. The square deviation is -0.5*-0.5=0.25


[/ QUOTE ]

FYP

Edit: and the mean is 0 and I get variance = 1 if X={-1,1}.

[Rainclouds]

[ QUOTE ]
[ QUOTE ]

8) if we lose, we win -1 . The deviation from the arithmetic mean is -0.5. The square deviation is -0.5*-0.5=0.25


[/ QUOTE ]

FYP

Edit: and the mean is 0 and I get variance = 1 if X={-1,1}.

[/ QUOTE ]
Well that's another interpretation which is correct as well.
I expressed the amount in which part of the pot we get, you expressed it in the amount of stacks we win or lose.

We could also say that the stacks are 100 and then the variance is 10000.

The concept remains the same so it doesn't matter anyway.

[Check_The_Nuts]

rata has obviously been corrected on the variances not being the same. They clearly have different variances, different standard deviations. Your really just averaging numbers (even tho their equity).

The distributions of the sample are different. They aren't the same at all.

Two samples:

0 1
.75 .75 0

They have the same mean, 0.50, they have different standard deviations/variances and distributions. If you plotted the 3 on a bar graph the two graphs would not be the same.

If you look at http://en.wikipedia.org/wiki/Variance, scroll down to population variance and sample variance, there is a very clear formula:

variance=1/n*sum(xi-mean)^2

since we know the entire population of results (they are given), you don't need to use (n-1) as is used for samples.

So for the two samples:

variance1=1/2*((0-0.5)^2+(1-0.5)^2)=0.25
variance2=1/* 3 ((0.75-0.5)^2+(0.75-0.5)^2+(0-0.5)^2)=0.125

So the second example has half the variance as the first, which makes sense if you look at the two distributions.

I could look in my stats book and figure out the E(X) notation. I took another stats course which had the crap in it. But I don't see why I should bother.

edit: I made a mistake first time around, the 3 was a 2. It was calculated using 3 tho.

[Rainclouds]

[ QUOTE ]
rata has obviously been corrected on the variances not being the same. They clearly have different variances, different standard deviations. Your really just averaging numbers (even tho their equity).

The distributions of the sample are different. They aren't the same at all.

Two samples:

0 1
.75 .75 0

They have the same mean, 0.50, they have different standard deviations/variances and distributions. If you plotted the 3 on a bar graph the two graphs would not be the same.

If you look at http://en.wikipedia.org/wiki/Variance, scroll down to population variance and sample variance, there is a very clear formula:

variance=1/n*sum(xi-mean)^2

since we know the entire population of results (they are given), you don't need to use (n-1) as is used for samples.

So for the two samples:

variance1=1/2*((0-0.5)^2+(1-0.5)^2)=0.25
variance2=1/* 3 ((0.75-0.5)^2+(0.75-0.5)^2+(0-0.5)^2)=0.125

So the second example has half the variance as the first, which makes sense if you look at the two distributions.

I could look in my stats book and figure out the E(X) notation. I took another stats course which had the crap in it. But I don't see why I should bother.

edit: I made a mistake first time around, the 3 was a 2. It was calculated using 3 tho.

[/ QUOTE ]
[ QUOTE ]
Oh yeah, I'd like to add one thing:

Some posters are trying to calculate the variance of the distribution of villains hand.
The only thing that matters to the poker player though is the variance of the outcomes of the hands.

If we speak about variance in the cheese thread or in bbv, we are always addressing the variance on the outcomes (swings, money graphs etc.)

Not a single soul makes a post about how high the variance of the distribution of his opponents range is after a big downswong. That's totally irrelevant.

I hope this clears things up better.

[/ QUOTE ]

[Check_The_Nuts]

and to answer the question, the best distribution is:

0.75 0.75 0

I like this distribution the most because I'm pretty sure its the most skewed to the right of the mean. I would think you'd want a distribution that looks like "/" as opposed to the other way, "\".

[Oct0puz]

[ QUOTE ]
rata has obviously been corrected on the variances not being the same. They clearly have different variances, different standard deviations. Your really just averaging numbers (even tho their equity).

The distributions of the sample are different. They aren't the same at all.

Two samples:

0 1
.75 .75 0

They have the same mean, 0.50, they have different standard deviations/variances and distributions. If you plotted the 3 on a bar graph the two graphs would not be the same.

If you look at http://en.wikipedia.org/wiki/Variance, scroll down to population variance and sample variance, there is a very clear formula:

variance=1/n*sum(xi-mean)^2

since we know the entire population of results (they are given), you don't need to use (n-1) as is used for samples.

So for the two samples:

variance1=1/2*((0-0.5)^2+(1-0.5)^2)=0.25
variance2=1/* 3 ((0.75-0.5)^2+(0.75-0.5)^2+(0-0.5)^2)=0.125

So the second example has half the variance as the first, which makes sense if you look at the two distributions.

I could look in my stats book and figure out the E(X) notation. I took another stats course which had the crap in it. But I don't see why I should bother.

edit: I made a mistake first time around, the 3 was a 2. It was calculated using 3 tho.

[/ QUOTE ]

Your and many others here are making the misstake of thinking that you can win a fraction of the pot. You always win 100% or 0% so the variance is the same.

And I have no idea why are you are referring to population variance and sample varince when we obviously know everything about the distribution.