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  #21  
Old 11-01-2007, 11:32 PM
TNixon TNixon is offline
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Posts: 616
Default Re: A tough variance problem (at least for me)

[ QUOTE ]
I read your example. It was a confused mess, and the method does not work.

[/ QUOTE ]
Wow. On the drive home, I realized what a true maroon I was being here.

std. deviation is a measure of the variation of individual results around the mean, but for my comparison to work, I would have had to be able to compare against the variation of *accumulated* results around the "accumulated" mean, or average expectation.

There's probably a way to do that, but my brain hurts, and it's definitely not as simple as I was trying to make it.

Which is exactly what pzhon was saying here:

[ QUOTE ]
Having a downswing of a particular size is very different from losing that amount in a fixed number of hands.

[/ QUOTE ]

He just said it far too simply for me to actually understand.

You can all stop laughing now. [img]/images/graemlins/smile.gif[/img]

Ah well. I told you all I was over my head here, and the easiest way to gain understanding is through massive bouts of stupidity, right?

...

RIGHT?

[img]/images/graemlins/smile.gif[/img]
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  #22  
Old 11-03-2007, 12:33 AM
rufus rufus is offline
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Join Date: Aug 2005
Posts: 425
Default Re: A tough variance problem (at least for me)

[ QUOTE ]
Ah well. I told you all I was over my head here, and the easiest way to gain understanding is through massive bouts of stupidity, right?

[/ QUOTE ]

I'm not sure if it's the easiest, but it's certainly a popular route. Struggling puts you in good company.

'Problems worthy of attack, prove their worth by hitting back' - Piet Hein
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  #23  
Old 11-04-2007, 01:57 PM
pococurante pococurante is offline
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Join Date: Jul 2007
Location: VA
Posts: 138
Default Re: A tough variance problem (at least for me)

I think this can be simplified to some extent, and still get numbers that are semi accurate.

In your average pot, you either gain $1 or lose $1. Over 100 hands, you average a $16 profit. Therefore, you win 58% of the hands and lose 42% of them.

So to find the odds for 1000 hands...

sqrt{(1000*.58*(.42)}
sqrt(243.6)
standard deviation = 15.61
expected wins = 580
actual wins (for a $500 loss) = 250
# of deviations = 14.7

Something that's 7 deviations away from the EV has odds of about 1 in 385,000,000,000 of happening... so you can imagine what 14.7 deviations means.

So anyway, according to these numbers:

68% chance to be within $15.60 of your expected $160 profit
95% chance to be within $31.20
99% chance to be within $46.80

Now that I see the results... those are too consistent for how poker really works. I realize now it's because I didn't count the variance on your 58% winrate, and the variance on the $1 pot size. If I had to guess... just double the above numbers (65% to be within $31, for instance) and that sounds reasonable.

Bottom line: no way in hell are you going to lose $500 in 1000 hands. Unless you're the type to tilt and then move way up in stakes so you can win it back faster... then it's all but guaranteed to happen.

It's pretty obvious when you look at the basic facts, too. With those kinds of pots, what are you playing? Probably 10nl. Who the hell has a 50 buyin downswing in 1000 hands? Unless you're tuff_fish, losing a buyin every 20 hands is pretty hard to do.
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  #24  
Old 11-04-2007, 02:05 PM
pococurante pococurante is offline
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Default Re: A tough variance problem (at least for me)

BTW given your example of WLLWLWLLW in the original post, I was assuming this is a heads up game.
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