Odds of same pocket pair three times in a row.
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
November 10th, 2018 at 6:55:13 PM
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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.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
November 10th, 2018 at 7:14:02 PM
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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
Simplicity is the ultimate sophistication - Leonardo da Vinci
November 11th, 2018 at 5:06:55 AM
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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.
All animals are equal, but some are more equal than others
November 11th, 2018 at 5:41:34 AM
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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.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
November 11th, 2018 at 6:32:50 AM
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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.
All animals are equal, but some are more equal than others
November 11th, 2018 at 9:22:53 AM
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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.
It’s all about making that GTA
November 11th, 2018 at 9:53:30 AM
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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
I Heart Vi Hart
November 11th, 2018 at 10:40:04 AM
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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
I Heart Vi Hart
November 13th, 2018 at 3:36:59 PM
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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.
December 28th, 2018 at 5:59:16 AM
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I just posted this question and my solution in Ask the Wizard column #311.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
December 28th, 2018 at 6:04:08 AM
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Quote: mustangsally
see
Sally, I was just making an Ask the Wizard question out of this and was thinking that with 120 total hands, what we should solve for is T^119. That will show us the probability of being in each state 119 hands after the first hand.
That said, I'm changing my answer to 0.0001402719. Do you agree?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
December 29th, 2018 at 12:31:27 PM
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I disagree.Quote: WizardSally, I was just making an Ask the Wizard question out of this and was thinking that with 120 total hands, what we should solve for is T^119. That will show us the probability of being in each state 119 hands after the first hand.
That said, I'm changing my answer to 0.0001402719. Do you agree?
because your transition matrix does actually start with a 0 state.
(I think I have been corrected by those that know on this enough to know)
instead of starting with a "start" state (they are the same as per my understanding)
even though P and P^1 are the same. that is why I think this works at 120.
Let us see with my transition matrix
matrix = P
some photos (orange cell is what power the matrix is raised to)
P^1 is after the first trial
P^2 is after the 2nd trial
P^3 is after the 3rd trial
shows we can get 3 in a row after 3 trials
17*221*221
from that transition matrix starting at state 0
we get the correct answer for 3 in a row over 3 trials
now, WITH A START STATE
P^120 is after the 120th trial
looks like the same result be starting at
state 0 or
state start
Happy New Year!
Sally
I Heart Vi Hart
December 29th, 2018 at 5:13:20 PM
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I also think it should be 120 iterations (for a probability of 1 in ~7,069) and I don’t understand the reasoning behind using 119.
On the very first hand you have a 1/17 chance of going to the state of 1 pair or 16/17 of state of no pair. And the 120th hand also counts in this case (it wouldn’t if the problem involved achieving success on that hand).
Just can’t see why you’d deduct 1 from 120.
On the very first hand you have a 1/17 chance of going to the state of 1 pair or 16/17 of state of no pair. And the 120th hand also counts in this case (it wouldn’t if the problem involved achieving success on that hand).
Just can’t see why you’d deduct 1 from 120.
It’s all about making that GTA
December 29th, 2018 at 6:01:23 PM
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I think it comes from this thread where the Wizard made a transition matrix that already started after trial 1Quote: Ace2Just can’t see why you’d deduct 1 from 120.
https://wizardofvegas.com/forum/questions-and-answers/math/32095-last-open-number-roulette/2/#post689723
then the explanation in the post below that by my Uncle
I agree on both those posts
Sally
Happy NEw Year!
(Mom wants to be in Connecticut instead of Miami for her Christmas present)
I Heart Vi Hart
December 29th, 2018 at 6:36:27 PM
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Thank you Sally and Ace for your corrections. I guess I was confusing the state at the beginning of the hand and the end of the hand.
This problem appears in Ask the Wizard column #311.
This problem appears in Ask the Wizard column #311.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
