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Old 12-04-2005, 01:17 AM
LockLow34 LockLow34 is offline
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Default What are the odds of 100 hands no PP?

Just played in a home tourney. I made it to the 8th level. With each round lasting 20 minutes and 6-10 players at the tables at various times I'm figuring I saw about 100 hands or so. Out of those 100 hands I had 7 hands with 2 broadway cards (AK, AQ, AT, KQ, KQs, KJ, QJ) and 0 pocket pairs.

What are the odds of having this bad run? More specifically, what are the odds of going 100 hands without a single pocket pair?
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  #2  
Old 12-04-2005, 01:21 AM
BruceZ BruceZ is offline
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Default Re: What are the odds of 100 hands no PP?

[ QUOTE ]
Just played in a home tourney. I made it to the 8th level. With each round lasting 20 minutes and 6-10 players at the tables at various times I'm figuring I saw about 100 hands or so. Out of those 100 hands I had 7 hands with 2 broadway cards (AK, AQ, AT, KQ, KQs, KJ, QJ) and 0 pocket pairs.

What are the odds of having this bad run? More specifically, what are the odds of going 100 hands without a single pocket pair?

[/ QUOTE ]

(16/17)^100 =~ 428-to-1.
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  #3  
Old 12-04-2005, 02:17 PM
TheProdigy TheProdigy is offline
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Default Re: What are the odds of 100 hands no PP?

Bruce, I know this isn't my question, but I was wondering if you could provide any information as to what those numbers mean? Like where do you get the 16/17 and such...thanks a bunch

Trying to learn,
TheProdigy
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Old 12-04-2005, 02:44 PM
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Default Re: What are the odds of 100 hands no PP?

The chance of getting a PP is 1/17, so the chance of not getting a PP is 16/17. The chance of something happening n times in a row is x^n, where x is the probability of the event. The chance of not getting a PP 100 hands in a row is then (16/17)^100.
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  #5  
Old 12-04-2005, 02:49 PM
BruceZ BruceZ is offline
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Default Re: What are the odds of 100 hands no PP?

[ QUOTE ]
Bruce, I know this isn't my question, but I was wondering if you could provide any information as to what those numbers mean? Like where do you get the 16/17 and such...thanks a bunch

Trying to learn,
TheProdigy

[/ QUOTE ]

The probability of being dealt a pair is 52/52 * 3/51 = 1/17, so 16/17 is the probability of not getting dealt a pair. The probability of not getting a pair 100 times in a row is (16/17)^100.
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  #6  
Old 12-04-2005, 10:33 PM
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Default Re: What are the odds of 100 hands no PP?

If you have trouble understanding where it comes from, stick with the raw numbers:

(Any first card, or 52 out of 52) and (a non-pairing second card, which is 48 cards out of the remaining 51)
(52/52) * (48/51)
.9411765
94.11765%

100 times in a row:
.9411765 * .9411765 * .9411765 * [. . .] * .9411765
.9411765^100
.002329
.2329%
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