Quote:AZDuffmanFWIW I never thought about time played. Just curious about three of the same pocket pair in a row.

If you want to know about the probability, we have to assume something about number of hands played.

Quote:WizardIf you want to know about the probability, we have to assume something about number of hands played.

I think I confused the issue with wording. I was more just wondering the odds of *any* pocket pair three times in a row in an independent trial.

Reason being that I have had 2 Royal Flushes in a lifetime of well over 100,000 hands online and live but this was the first time I had three pockets in a row. Had two too many times to remember, but this was the first time three in a row.

Quote:AZDuffmanI think I confused the issue with wording. I was more just wondering the odds of *any* pocket pair three times in a row in an independent trial.

Well, it's easy to say that the probability of any pair happening three times in a row, including different pairs from each other, is (3/51)^3.

The odds of the same pair three times in a row is (3/51)*[(4/52)*(3/51)]^2.

I needed to brush up on my matrix algebra, which is why I muddied the waters with asking about it happening anywhere in a session.

Quote:WizardWell, it's easy to say that the probability of any pair happening three times in a row, including different pairs from each other, is (3/51)^3.

The odds of the same pair three times in a row is (3/51)*[(4/52)*(3/51)]^2.

I needed to brush up on my matrix algebra, which is why I muddied the waters with asking about it happening anywhere in a session.

Can you better explain where the numbers come from? I always thought the odds of any given pocket pair are about 1/221. Been 25+ years since my last stats class so am rusty.

Hope you don’t mind if I chime in. 1/221 is for a specific pair. For instance, the odds of getting a pair of Jacks on the next hand are 4/52 * 3/51 = 1/221. For any pair the odds are 52/52 * 3/51 = 1/17 since the first card can be any of the 52 cards and the second can be any of the 3 remaining cards of rank established by the first card.Quote:AZDuffmanQuote:WizardWell, it's easy to say that the probability of any pair happening three times in a row, including different pairs from each other, is (3/51)^3.

The odds of the same pair three times in a row is (3/51)*[(4/52)*(3/51)]^2.

I needed to brush up on my matrix algebra, which is why I muddied the waters with asking about it happening anywhere in a session.

Can you better explain where the numbers come from? I always thought the odds of any given pocket pair are about 1/221. Been 25+ years since my last stats class so am rusty.

yes, the probability the event happens on one trial.Quote:Ace2Hope you don’t mind if I chime in. 1/221 is for a specific pair.

the ODDS?Quote:Ace2For instance, the odds of getting a pair of Jacks on the next hand are 4/52 * 3/51 = 1/221.

is that Ur final answer?

then what is: 220 to 1 against?

my Math teacher had a major melt down when I said things like that.

(Because he said "I should have known better" Now I see where he got that

I should have known better with a girl like you. OMG!)

for the probabilities, here is what I get using combinations [C() or COMBIN()]

for a specific pair

C(4,2)/ C(52,2)

C(4,2) choose 2 of the 4 Jacks (in the example) can also be read 4 Jacks choose 2

C(52,2) choose 2 from the 52 total cards

6 / 51*26 = 6/1326 = 1/221

http://www.wolframalpha.com/input/?dataset=&equal=Submit&i=C(4,2)%2F+C(52,2)

For any pair:

C(13,1) * C(4,2)/ C(52,2)

C(13,1) choose 1 of the 13 ranks

C(4,2) choose 2 of the 4 (4 of each rank)

C(52,2) choose 2 from the 52 total cards

13*6 / 51*26 = 78/1326 = 1/17

still not a very common event

even if cards are not shuffled very good (well)

Sally

makes it easier to see the event at least 1 timeQuote:AZDuffmanFWIW I never thought about time played. Just curious about three of the same pocket pair in a row.

Though the time played makes it interesting in a different way.

or less probability one will NOT see the event.

in 3 rounds is simple 1 in 17*221*221

rounds | 1 in |
---|---|

3 | 830,297.00 |

10 | 104,199.91 |

30 | 29,783.83 |

90 | 9,478.06 |

120 | 7,068.61 |

200 | 4,212.87 |

500 | 1,675.32 |

1000 | 836.23 |

I have thrown away 2,7 offsuit many many times just to see 2,2 or 7,7 on the flop

would NOT have won all of those...

even a few times LOL seeing 2,7,7 flopped

Sally

Quote:WizardIf you want to know about the probability, we have to assume something about number of hands played.

I tried to come up with a straight equation based on how many hands, using the Eigenvalue/Eigenvector matrix method, but I end up with a cubic equation for three of the Eigenvalues (the fourth is 1), and I have a feeling that two of them are complex.

It's probably much easier just to crunch the numbers in Excel.