Mean Value Theorem Question
 

This post is pretty much worhtless, but I was wondering about something today.

For those who haven't taken Calc recently, the Mean Value Theorem states (from Google):

for two points (a, f(a) ) and ( b, f(b) ), on a continuous curve, there is a point c in between where the slope f '(c) is the same as the slope, m, of the line joining the two points.

Which basically means if you average a rate, you have to be going that rate at some point.

Applying this concept to poker, we all know that we go through downswings and upswings. From the Mean Value Theorem we can deduce that if our true winrate is 3 BB/100, then for a certain period of time we have to be winning at truly 3 BB/100. That's about as far as my math skills go. So to all those math majors out there, how long does one truly win at that rate? Is it just an instant? Or does it depend?

Re: Mean Value Theorem Question
 

all the mean value theorem says with respect to poker winrates is this:

look at a poker winrate graph that we've all seen before and take the first point, which is (0 hands, 0 big bets), and the last point, let's say (20000 hands, 150 big bets).

find the slope between these two points. in this case, it is (150 - 0 big bets)/(20000 - 0 hands) = 0.0075 big bets per hand.

this means that somewhere along the way, the instantaneous slope of our graph was 0.0075.

Re: Mean Value Theorem Question
 

It's more of a statistics problem than a calc problem. Think about it like this. Plot a histogram with winrate covering the x axis. what averaging 3 BB/100 means, is that if we took an infinite amount of, say, 10000 hand sessions, the bar at x=3 BB/100 would be the highest; the bars leading up to the 3bb would form what looks like a triangle.

We are more likely to average 3bb/100 than anything else, that's how this can be interpreted.

Re: Mean Value Theorem Question
 

buck: they would not form a triangle. they would form a bell curve.

Re: Mean Value Theorem Question
 

and the bell curve would go from 0bb/100 to 6bb/100 and back to 0bb/100?

Re: Mean Value Theorem Question
 

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this means that somewhere along the way, the instantaneous slope of our graph was 0.0075.

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k, that's what i thought it meant. wanted some confirmation, especailly since i could write sonnets all day about calc, but not solve d(x^2). [img]/images/graemlins/wink.gif[/img]

Re: Mean Value Theorem Question
 

Holy cow I have an analysis test tomorrow whose main star will be the proof of the MVT and this thread is creeping me out.

Re: Mean Value Theorem Question
 

[ QUOTE ]
It's more of a statistics problem than a calc problem. Think about it like this. Plot a histogram with winrate covering the x axis. what averaging 3 BB/100 means, is that if we took an infinite amount of, say, 10000 hand sessions, the bar at x=3 BB/100 would be the highest; the bars leading up to the 3bb would form what looks like a triangle.

We are more likely to average 3bb/100 than anything else, that's how this can be interpreted.

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ya, i think that i was talking about something other than x bar. maybe not...math is gay.

Re: Mean Value Theorem Question
 

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and the bell curve would go from 0bb/100 to 6bb/100 and back to 0bb/100?

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no, call x your true winrate and std your standard deviation

it would go from x - ~4std to x + ~4std

the y-axis is the # of occurrences

Re: Mean Value Theorem Question
 

all it takes is 10 years of non-use and calculus is long forgotten.

Re: Mean Value Theorem Question
 

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buck: they would not form a triangle. they would form a bell curve.



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the bell looks like a triangle though.

Re: Mean Value Theorem Question
 

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It's more of a statistics problem than a calc problem.

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Why would the mean value theorem not be applicable just because the data points come from the "real world?"

Re: Mean Value Theorem Question
 

ok i guess. i think of the bell curve more as an upside down saggy breast.

Re: Mean Value Theorem Question
 

I think it's better to look at with bell curves because it's easier to visualize, that's all.

Re: Mean Value Theorem Question
 

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ok i guess. i think of the bell curve more as an upside down saggy breast.

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Please, don't give me a rubbery one of the bell curve.

Re: Mean Value Theorem Question
 

If I win 6BB/100, lose 3BB/100, win 6BB, lose 3BB and do this for ever.
My winrate is 3BB/100.

At what point along my graph would I ever show +3BB/100?

Re: Mean Value Theorem Question
 

Without bothering to really go at it, I think it won't. It'll converge to 3.0 but not hit it.

Win rate isn't a continuous function so technically the MVT doesn't apply.

Re: Mean Value Theorem Question
 

because poker winrates are stepwise functions, yes, the theorem does not technically apply. the theorem applies to continuous functions only. however if we were to apply a best-fit curve to the winrate, you would at some point show a +3 bb/100 winrate.

Re: Mean Value Theorem Question
 

Hell it'd show a +3 winrate between every hand [img]/images/graemlins/smile.gif[/img]

Re: Mean Value Theorem Question
 

okay let's nitpick. it would show +3 bb/100 between every other hand.

Re: Mean Value Theorem Question
 

Thanks. Without the bestfit curve, I was struggling.

Re: Mean Value Theorem Question
 

So anyway, let L be the line between (a,f(a)) and (b,f(b)). Note that L(a)=a and L(b)=b. But this means L(a)-f(a)=L(b)-f(b)=0, and so we can apply Rolle's Theorem: there exists c in [a,b] such that [L(c)-f(c)]'=0. With some manipulation then we have

L'(c)-f'(c)=0
L'(c)=f'(c) *

But L'(c)=[f(b)-f(a)]/(b-a), and so we're done. Note that * is what you actually posted, which is directly equivalent.

I only needed help for oneline of the proof this time I'm so proud ^_________^