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  #1  
Old 08-09-2005, 09:51 AM
brianmarc brianmarc is offline
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Default Major Problem with Bill Chin\'s Article on Variance

I believe there is a major, possibly even fatal flaw with all the work currently done on gambling variance - the use of standard deviation as the measure of the variance being measured. SD assumes the distribution of observations is symmetrical around the mean; i.e., that the normal distribution is the correct statistical model to use. However, this is often not the case: a successful player will have a significantly positively skewed distribution of returns, and vice versa for the consistent loser. This means that even if the better player has a high SD, much of it will be "good" variance. Again, vice versa for the loser. The conclusion is that a better measure of this variance is the 2nd degree lower-partial moment (LPM2). LPM2 essentially captures just the variances below the mean and, since it is distribution-free it avoids the fatal flaw of SD in an asymmetric world. In the financial investment world where this measure has replaced SD, LPM2 is known as "Downside Risk", and has now become the risk measure of choice for the non-normal investment distributions that are typical for derivative and hedge fund strategies. For addition material on this concept see my website www.investmenttechnologies.com.
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  #2  
Old 08-09-2005, 01:06 PM
RedManPlus RedManPlus is offline
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Default Finally - A Real World Perspective

I agree with this.

Finally a lone voice of reason...
In repsonse to the chorus that...
"You gotta play 5,000,000 SNGs before your results mean anything".

Your point in analogous to using the Sharpe Ratio...
To quantify risk/reward for a hedge fund...
Which is ridiculous...
Because the Sharpe Ratio penalizes upward/downward variance equally.

I manage a small hedge fund (about $1 million)...
And if I'm averaging a 25% return...
But have a big year and do 50%...
The Sharpe Ratio penalizes me for "variance".

Simply analyzing your drawdowns...
To quantify the most important thing - "downside risk"...
And is far more meaningful then misapplied stats.

You are very perceptive to have a gut feeling...
That the way the "geniuses" around here...
Apply off-the-shelf freshman stats...
To very exotic games such as SNGs...
Is equally ridiculous.

The dogma they push...
Would result in the conclusion...
That winning the WSOP is not "statistically significant"...
Becuase you only play about (a guess) 2000-3000 hands to win.

I suspect that smart, good players...
Have far less variance...
Than what is "conventional wisdom" around here.

rm+

[img]/images/graemlins/cool.gif[/img] [img]/images/graemlins/cool.gif[/img] [img]/images/graemlins/cool.gif[/img]
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  #3  
Old 08-09-2005, 05:26 PM
fyodor fyodor is offline
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Default Re: Major Problem with Bill Chin\'s Article on Variance

For a statistical ingnoramus how would one go about calculating downside risk (LPM2) with PokerTracker numbers?
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  #4  
Old 08-09-2005, 06:36 PM
AaronBrown AaronBrown is offline
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Default Re: Major Problem with Bill Chin\'s Article on Variance

Without disputing your argument, the rationale for standard deviation is generally based on the Central Limit Theorem rather than the symmetry of the underlying distribution. If you add up enough independent observations, whether Poker hand outcomes or daily portfolio returns, only the mean and variance matter. The shape of the distribution washes out.

Say your successful player with the positively skewed distribution loses $10 on nine hands in ten and wins $200 on the tenth hand. The unsuccessful one has the opposite pattern, making $10 on nine hands in ten and losing $200 on the tenth hand. All hands are independent. The successful one makes $10 per hand, the unsuccessful loses $10, but both have the same $63 standard deviation.

After 1,000 hands, the distribution of total profit and loss is almost indistinguishable from Normal in both cases. Average profit per hand will be plus or minus $11 with a standard deviation of $2. The probability that the successful player is between $9 and $11 is 34.7%, versus a Normal approximation of 34.1%. The probability that the unsuccessful player is one standard deviation below his mean (-$13 to -$11) is 33.6%. The probability that the successful player is between $7 and $9 is 14.1%, versus 13.6% in a Normal, and also 13.6% for the unsuccessful player being two standard deviations below his mean ($15 to -$13). So knowing the standard deviation per hand, and the mean, tells you almost all you need to know about the long term distribution of outcome.

I think the more important issue is independence. Poker hands are not independent, neither are hedge fund returns. Even small deviations can skew these calculations.
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  #5  
Old 08-09-2005, 06:49 PM
AliasMrJones AliasMrJones is offline
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Default Re: Finally - A Real World Perspective

[ QUOTE ]
I suspect that smart, good players...
Have far less variance...
Than what is "conventional wisdom" around here.

[/ QUOTE ]

And...
I suppose...
The fact that smart, good players...
with significant gains...
over a large sample of hands...
have had the significant downswings...
predicted by the "freshman statistics"...
and, in fact, are usually the ones warning about variance...
means nothing.
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  #6  
Old 08-09-2005, 07:21 PM
StellarWind StellarWind is offline
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Default Re: Major Problem with Bill Chin\'s Article on Variance

Well I see that your post is still here.

Maybe I was mistaken in thinking this was just a clever off-topic plug for your investment business website [img]/images/graemlins/confused.gif[/img]?

[ QUOTE ]
I believe there is a major, possibly even fatal flaw with all the work currently done on gambling variance - the use of standard deviation as the measure of the variance being measured. SD assumes the distribution of observations is symmetrical around the mean; i.e., that the normal distribution is the correct statistical model to use. However, this is often not the case:

[/ QUOTE ]
I'm very surprised to hear that the normal distribution is not a good approximation for ring-game poker, blackjack, and similar gambling activities. Do you have any evidence or explanation for this?

[ QUOTE ]
a successful player will have a significantly positively skewed distribution of returns, and vice versa for the consistent loser. This means that even if the better player has a high SD, much of it will be "good" variance. Again, vice versa for the loser.

[/ QUOTE ]
This is just a restatement of the obvious fact that the distribution of a winning player's results will be centered on a positive mean (his win rate). It in no way shows that the distribution is not normal.

[ QUOTE ]
This means that even if the better player has a high SD, much of it will be "good" variance.

[/ QUOTE ]
Again, why would I expect favorable and unfavorable variance events to be anything other than equal (on average)? Surely you are not going to define "bad" variance to mean losing? A month where you only win half as much as your average rate is certainly an example of unfavorable variance.

All the research I've read on bankroll requirements fully integrates a player's win rate as an important parameter. Everyone knows that a winning player's results are "positively skewed". Unless you can back up your assertion that results are not approximately normally distributed, you have added nothing.
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  #7  
Old 08-09-2005, 09:00 PM
BillC BillC is offline
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Posts: 43
Default Re: Major Problem with Bill Chin\'s Article on Variance

The use of the normal distribution arises from the belief that bets (or an hour of play, or sets of 100 hands) are approximately independent. The net result of independent bets converges (with increasing number of trials) to a normal random variable. If you think the independence of bets or sets of bets does not approximately hold, it might be interesting to hear why.

The beauty of simple games of chance such as poker and blackjack, as opposed to say financial instruments, is that our assertion of approximate normality is much easier to justify (because of the ramdomness of the shuffle. please do not take this as bait).

That being said, it would be interesting to see data to suggest that (say for sets of 100 hands) results are or are not approximately normal or symmetric, i.e. that they are skewed. Anyone have data on this? I would be surprised to see any sinifigant skewing.

The normal/Brownian motion model is a continuous approximation of a discrete reality, and the speed of convergence to the idealized model can be hampered by skewing. In the case of skew, larger samples are needed for accurate predictions.

The poster above may be confusing positive EV with skewing in the sense that the distribution of results is asymmetric about the mean.
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  #8  
Old 08-09-2005, 09:50 PM
AaronBrown AaronBrown is offline
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Location: New York
Posts: 505
Default Re: Major Problem with Bill Chin\'s Article on Variance

[ QUOTE ]
If you think the independence of bets or sets of bets does not approximately hold, it might be interesting to hear why.

[/ QUOTE ]
Players will change their play as they win and lose. One common pattern is to loosen up when you win, but fail to tighten up when you lose. The gambler's ruin theorem means you will almost always be a big loser with this rule, even if you have positive expectation per hand. This violates independence and, with a player like this, I would base my long-term prediction of his wealth on his poor strategy rather than the mean and standard deviation of individual hands.

Another dependence is bluffing. If a player is caught in a bluff, he will be played differently on future hands. That's the point of bluffing.
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  #9  
Old 08-09-2005, 10:10 PM
AliasMrJones AliasMrJones is offline
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Location: Denver, CO
Posts: 377
Default Re: Major Problem with Bill Chin\'s Article on Variance

[ QUOTE ]
[ QUOTE ]
If you think the independence of bets or sets of bets does not approximately hold, it might be interesting to hear why.

[/ QUOTE ]
Players will change their play as they win and lose. One common pattern is to loosen up when you win, but fail to tighten up when you lose. The gambler's ruin theorem means you will almost always be a big loser with this rule, even if you have positive expectation per hand. This violates independence and, with a player like this, I would base my long-term prediction of his wealth on his poor strategy rather than the mean and standard deviation of individual hands.

Another dependence is bluffing. If a player is caught in a bluff, he will be played differently on future hands. That's the point of bluffing.

[/ QUOTE ]

The statements above are complete claptrap. A player with a losing strategy CANNOT have long term positive expectation and a player with a winning strategy CANNOT have negative long term expectation. This is by definition. Therefore, someone who employs the strategy you outline above (loosen up when winning, but fail to tighten up when losing) will not have a winning strategy and WILL NOT have positive per hand expectation. Do you see why?

Just because a player is caught in a bluff one hand and players may play him differently on the next hand does not mean his long-term winrate and SD will change. In fact, a good player will mix bluffs into a well-rounded strategy and will be caught sometimes. This will already have been factored into the players long-term winrate and SD, assuming the sample size is large enough.

Your problem is you seem to be focused on small samples (i.e. per hand vs. long term strategy in the first part and how 1 caught bluff will affect a small number of subsequent hands.)

As I said before, the predicted results of winrate and SD with the regularly discussed formulas have been proven out through observed results of real players with sufficiently large sample sizes and posted about on this forum.
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  #10  
Old 08-10-2005, 01:01 AM
uuDevil uuDevil is offline
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Default Re: Major Problem with Bill Chin\'s Article on Variance

[ QUOTE ]
SD assumes the distribution of observations is symmetrical around the mean; i.e., that the normal distribution is the correct statistical model to use.

[/ QUOTE ]
Whether or not the assumption of normality is justified, calculating the standard deviation requires no knowledge of the shape of the distribution. It would be a valid statistic, whatever the distribution. Or am I crazy?
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