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  #1  
Old 08-16-2005, 04:11 AM
ThinkQuick ThinkQuick is offline
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Location: Alberta, Canada
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Default P (being a winning player)

Just wanted to check this:

I have a person's hourly rate (which is negative) and hourly SD.
Is calculating the probability that he is greater than a 1.0BB/hr player as simple as setting up a standard curve with Mean = hourly avg and SD = SD, and finding the area above 1BB?

i.e. -1BB average with SD = 9BB: Area above 1.0 = 0.412..

so there is a 41.2% chance that this player is a >1.0BB / hr player.

Also, how valid (if at all) would stats like this be with such a large SD - only 500 hrs of B&M have been played.
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  #2  
Old 08-16-2005, 04:32 AM
BruceZ BruceZ is offline
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Posts: 1,636
Default Re: P (being a winning player)

[ QUOTE ]
Just wanted to check this:

I have a person's hourly rate (which is negative) and hourly SD.
Is calculating the probability that he is greater than a 1.0BB/hr player as simple as setting up a standard curve with Mean = hourly avg and SD = SD, and finding the area above 1BB?

i.e. -1BB average with SD = 9BB: Area above 1.0 = 0.412..

so there is a 41.2% chance that this player is a >1.0BB / hr player.

Also, how valid (if at all) would stats like this be with such a large SD - only 500 hrs of B&M have been played.

[/ QUOTE ]

If 9 bb is his SD for 1 hr, then you must divide this by sqrt(500) to get the standard deviation of his hourly rate, or standard error SE.

SE = 9/sqrt(500) =~ 0.4 bb.

A win rate of -1 bb/hr is almost 5 standard deviations below the mean for a 1 bb/hr winner, and this would have a probability of about 1 in 3 million. Note that this not the same as saying that there is a 1 in 3 million probability that he is a 1 bb/hr winner. That kind of statement would require knowledge about the win rates of players in general. Still, it is very unlikely that a 1 bb/hr winner would be this far behind after 500 hours. The correct statement is that we have 99.99997% confidence that his true win rate is less than 1 bb/hr. Also, we have 99.4% confidence that he is not playing winning poker.
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  #3  
Old 08-16-2005, 04:50 AM
ThinkQuick ThinkQuick is offline
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Join Date: Jan 2004
Location: Alberta, Canada
Posts: 97
Default Re: P (being a winning player)

[ QUOTE ]
[ QUOTE ]
Just wanted to check this:

I have a person's hourly rate (which is negative) and hourly SD.
Is calculating the probability that he is greater than a 1.0BB/hr player as simple as setting up a standard curve with Mean = hourly avg and SD = SD, and finding the area above 1BB?

i.e. -1BB average with SD = 9BB: Area above 1.0 = 0.412..

so there is a 41.2% chance that this player is a >1.0BB / hr player.

Also, how valid (if at all) would stats like this be with such a large SD - only 500 hrs of B&M have been played.

[/ QUOTE ]

If 9 bb is his SD for 1 hr, then you must divide this by sqrt(500) to get the standard deviation of his hourly rate, or standard error SE.

SE = 9/sqrt(500) =~ 0.4 bb.

A win rate of -1 bb/hr is almost 5 standard deviations below the mean for a 1 bb/hr winner, and this would have a probability of about 1 in 3 million. Note that this not the same as saying that there is a 1 in 3 million probability that he is a 1 bb/hr winner. That kind of statement would require knowledge about the win rates of players in general. Still, it is very unlikely that a 1 bb/hr winner would be this far behind after 500 hours. The correct statement is that we have 99.99997% confidence that his true win rate is less than 1 bb/hr. Also, we have 99.4% confidence that he is not playing winning poker.

[/ QUOTE ]


Thank you very much for your prompt and explanatory reply. It makes perfect sense and makes me feel like editing my original so I can look less silly. Of course I should compare where he is to the standard curve of a winning player.

As you also said, he's even more than 2SDs below the mean for a break even player.

Anyways thanks Bruce
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