Wizard
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Wizard
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November 10th, 2018 at 6:55:13 PM permalink
Quote: AZDuffman

FWIW 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.
It's not whether you win or lose; it's whether or not you had a good bet.
Ayecarumba
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November 10th, 2018 at 7:14:02 PM permalink
Were they the same suits each time? What are the odds of a pair with the same rank regardless of suit appearing three times in a row in a span of 120 hands?

If same suits three times in a row is ~1:7000, then two other available cards make it twice as common? 1:3500
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AZDuffman
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November 11th, 2018 at 5:06:55 AM permalink
Quote: Wizard

If 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.
All animals are equal, but some are more equal than others
Wizard
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Wizard
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November 11th, 2018 at 5:41:34 AM permalink
Quote: AZDuffman

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.



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.
It's not whether you win or lose; it's whether or not you had a good bet.
AZDuffman
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November 11th, 2018 at 6:32:50 AM permalink
Quote: Wizard

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.



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.
All animals are equal, but some are more equal than others
Ace2
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November 11th, 2018 at 9:22:53 AM permalink
Quote: AZDuffman

Quote: Wizard

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.



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 donít 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.
Itís all about making that GTA
mustangsally
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November 11th, 2018 at 9:53:30 AM permalink
Quote: Ace2

Hope you donít mind if I chime in. 1/221 is for a specific pair.

yes, the probability the event happens on one trial.
Quote: Ace2

For instance, the odds of getting a pair of Jacks on the next hand are 4/52 * 3/51 = 1/221.

the ODDS?
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
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mustangsally
mustangsally
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November 11th, 2018 at 10:40:04 AM permalink
Quote: AZDuffman

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.

makes it easier to see the event at least 1 time
or less probability one will NOT see the event.
in 3 rounds is simple 1 in 17*221*221
rounds1 in
3830,297.00
10104,199.91
3029,783.83
909,478.06
1207,068.61
2004,212.87
5001,675.32
1000836.23


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
I Heart Vi Hart
ThatDonGuy
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November 13th, 2018 at 3:36:59 PM permalink
Quote: Wizard

If 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.
Wizard
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December 28th, 2018 at 5:59:16 AM permalink
I just posted this question and my solution in Ask the Wizard column #311.
It's not whether you win or lose; it's whether or not you had a good bet.

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