November 9th, 2018 at 7:38:57 PM
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Just got back from Poker Night at the VFW. Got 6-6 in the hole three times in a row! Never had that happen before. While the board never gave me a Nick Perry (older folks from PA know what that means) and thus had to fold all three, it made me wonder the odds. And the right way to figure them.

Odds of any particular pocket pair as we know are about 221/1. But to calculate for three times in a row. is it 221*221*221, or just 221*221 in the same way as the next roulette number being the same as the last is always 50/50 since the first does not matter, just the second matching it?

I am figuring that it is the former, but am I missing that you only count the odds on the second two?

Odds of any particular pocket pair as we know are about 221/1. But to calculate for three times in a row. is it 221*221*221, or just 221*221 in the same way as the next roulette number being the same as the last is always 50/50 since the first does not matter, just the second matching it?

I am figuring that it is the former, but am I missing that you only count the odds on the second two?

All animals are equal, but some are more equal than others

November 9th, 2018 at 7:50:08 PM
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It’s more the latter. The former would be, before a hand starts, the probability of getting a 66 on the next three hands.

You question is more, at some point during the night I will get a pocket pair, and what are the chances the next two hands are also that same pocket pair.

You question is more, at some point during the night I will get a pocket pair, and what are the chances the next two hands are also that same pocket pair.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

November 10th, 2018 at 6:47:28 AM
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I think the way the question should be phrased is, "What is the probability I'll see the same pocket pair three times in a row in one evening?"

Let's say the evening is four hours and 30 rounds per hour, so 120 total hands. There are 118 spans of three consecutive hands in that 120 hands. The probability any given span of three hands has three of the same pair in a row is (3/51)*(4/52)*(3/51)*(4/52)*(3/51) = 1.20439E-06. The probability this not happening is 1 = 1.20439E-06

= 0.999998796.

Granted this next step is an approximation, because the series of three hands overlap each other, but the probability of not getting the same pair three times in a row 118 times is 0.999998796^=118 = 0.999857892. So the probability of it happening at least once is 1-0.999857892= 0.000142108 = 1 in 7037.

A good exercise for the advanced reader is to find the exact probability.

Let's say the evening is four hours and 30 rounds per hour, so 120 total hands. There are 118 spans of three consecutive hands in that 120 hands. The probability any given span of three hands has three of the same pair in a row is (3/51)*(4/52)*(3/51)*(4/52)*(3/51) = 1.20439E-06. The probability this not happening is 1 = 1.20439E-06

= 0.999998796.

Granted this next step is an approximation, because the series of three hands overlap each other, but the probability of not getting the same pair three times in a row 118 times is 0.999998796^=118 = 0.999857892. So the probability of it happening at least once is 1-0.999857892= 0.000142108 = 1 in 7037.

A good exercise for the advanced reader is to find the exact probability.

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 10th, 2018 at 6:48:53 AM
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I did that with 2s the one day and the next couple of hands had a two in it as well

this was holdem

im not patient enough for holdem though, although im good at the psycological perspective, i cant sit still unless playing and blackjack is just continuous when an ASM is involved.

this was holdem

im not patient enough for holdem though, although im good at the psycological perspective, i cant sit still unless playing and blackjack is just continuous when an ASM is involved.

November 10th, 2018 at 10:27:31 AM
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3 / (51 x 221 x 221) = 0.0000012 or 1 in 833,333 of it happening within three hands.

(there is a 100% chance you get your first card. Then a 3/51 chance your second card makes it a pair. Then a 1/221 chance your next hand makes the same pair. Then a 1/221 chance your third hand makes the same pair).

For it to be three straight pairs of 6s, then it would 1 / (221 x 221 x 221)

For the odds of it happening anytime during Poker Night, would have to know how many hands you see.

(there is a 100% chance you get your first card. Then a 3/51 chance your second card makes it a pair. Then a 1/221 chance your next hand makes the same pair. Then a 1/221 chance your third hand makes the same pair).

For it to be three straight pairs of 6s, then it would 1 / (221 x 221 x 221)

For the odds of it happening anytime during Poker Night, would have to know how many hands you see.

November 10th, 2018 at 1:55:03 PM
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Quote:WizardA good exercise for the advanced reader is to find the exact probability.

0.00014027196, or 1 in 7129.013941.

My estimate at 1 in 7037 was not too far off.

Last edited by: Wizard on Dec 28, 2018

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 10th, 2018 at 3:48:06 PM
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Hints:

Here are links to similar problems I've solved before:

Ask the Wizard 253

Ask the Wizard 277

p.s. No wonder women don't talk to me.

Here are links to similar problems I've solved before:

Ask the Wizard 253

Ask the Wizard 277

0.941176 | 0.058824 | 0.000000 | 0.000000 |

0.941176 | 0.054299 | 0.004525 | 0.000000 |

0.941176 | 0.054299 | 0.000000 | 0.004525 |

0.000000 | 0.000000 | 0.000000 | 1.000000 |

p.s. No wonder women don't talk to me.

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 10th, 2018 at 4:30:30 PM
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I am sure they will talk to you. You just need to be at a right places.Quote:Wizardp.s. No wonder women don't talk to me.

You may not have the looks that Elvis had, wow!!!

but you have looks.

I agree on the method and result

see

extending that method will show the mean (average) number of rounds to play to see this event is

834,071

the probability it happens at least one time over that many rounds is only about 0.632120779

Sally

I Heart Vi Hart

November 10th, 2018 at 4:47:38 PM
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Quote:mustangsallybut you have looks.

Thanks Sally!

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

November 10th, 2018 at 6:42:33 PM
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Quote:TomG

For the odds of it happening anytime during Poker Night, would have to know how many hands you see.

FWIW 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.

All animals are equal, but some are more equal than others