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  #1  
Old 10-04-2005, 06:58 AM
arod4276 arod4276 is offline
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Default probablity question

What are the odds of a winning , or breakeven player losing exactly 50 sessioins in a row? Assume each session had 1k hands and lasted 8 hours. Lastly how do you come to those odds? thanks alot
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  #2  
Old 10-04-2005, 07:11 AM
Little Fishy Little Fishy is offline
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Default Re: probablity question

you're going have to tell us your EV and standard deviation

ie: if you suck it's a hellova lot less likely
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  #3  
Old 10-04-2005, 07:19 AM
arod4276 arod4276 is offline
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Default Re: probablity question

Ok thanks for the reply.For simplicity sake lets take a 100 percent breakeven player over 500k hands
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  #4  
Old 10-04-2005, 08:01 PM
AaronBrown AaronBrown is offline
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Default Re: probablity question

A breakeven player has a 50% chance of losing in any session (ignoring the probability of coming out exactly even). To do that 50 sessions in a row, regardless of session length, is 1 chance in 2^50 or about 1 quintillion. That's the same ballpark as dealing three Poker hands from a well-shuffled deck and getting a royal flush and two other straight flushes.

For a winning player, it's even less likely.

Therefore, based on the evidence, I bet this is a losing player.
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  #5  
Old 10-05-2005, 01:09 AM
arod4276 arod4276 is offline
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Default Re: probablity question

thank you very much for the reply.W e agree that even for the worst player on the planet it would be quite a feat to accomplish?
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  #6  
Old 10-05-2005, 09:56 PM
alThor alThor is offline
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Default Re: probablity question

[ QUOTE ]
A breakeven player has a 50% chance of losing in any session (ignoring the probability of coming out exactly even). To do that 50 sessions in a row, regardless of session length, is 1 chance in 2^50 or about 1 quintillion.


[/ QUOTE ]

There would be a little bit of skew for short sessions, making it slightly more likely. But I'm getting picky. And 1k hands is obviously not a short session, so your numbers make sense for the original question.
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