Poker Math Question
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August 9th, 2011 at 8:23:16 PM
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Hello,
I am brand new here so (mods) please forgive me if I have placed this question in the wrong section.
For any of you mathematically minded folks out there here is a Texas hold Em math question.
In a heads up game what are the odds of one particular player (not combining both players...one player only) getting nothing less than a Queen (that is A, K, Q only) as one of their pocket cards in 20 out of 21 hands?
Would love to know the mathematical odds of this happening.
Thanks in advance.
Southern Belle
I am brand new here so (mods) please forgive me if I have placed this question in the wrong section.
For any of you mathematically minded folks out there here is a Texas hold Em math question.
In a heads up game what are the odds of one particular player (not combining both players...one player only) getting nothing less than a Queen (that is A, K, Q only) as one of their pocket cards in 20 out of 21 hands?
Would love to know the mathematical odds of this happening.
Thanks in advance.
Southern Belle
August 10th, 2011 at 4:40:14 AM
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3*4 cards of 52 (12/52) ,
then 11 of 51 (12/52*11/51)
happening 20 times (12/52*11/51)^20
one of 20 ways 20*(12/52*11/51)^20
About 1.8*10^-25. A very small number.
then 11 of 51 (12/52*11/51)
happening 20 times (12/52*11/51)^20
one of 20 ways 20*(12/52*11/51)^20
About 1.8*10^-25. A very small number.
"When two people always agree one of them is unnecessary"
August 10th, 2011 at 6:47:52 AM
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One, not both.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
August 10th, 2011 at 6:54:30 AM
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I think A,Q would qualify as valid. I get 1 in 85,685,896,462
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
August 10th, 2011 at 7:09:53 AM
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Maybe OP means Q-high or better? As in, Q-2 would also count. Then the chance is more reasonable
Wisdom is the quality that keeps you out of situations where you would otherwise need it
August 10th, 2011 at 7:57:59 AM
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One in 85 billion is not reasonable!
I had to recalculate but it's still one in 71.6 billion.
I had to recalculate but it's still one in 71.6 billion.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
August 10th, 2011 at 7:59:41 AM
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The odds of getting at least one Queen or higher are 12/52 plus 11/51 = 0.446 455 505 279
The odds of that happening 20 times in a row are 0.000 000 098 984 or 1 in 10,102,674.
For what it's worth, the odds of getting pocket deuces thru pocket jacks are 40/52 * 3/51 = 0.045 248 868 778
The totoal odds of getting at least one queen, or pocket pairs, is almost 50%: 0.491 704 374 057
The odds of it happening 20 times in a row are 0.000 000 682 467 or 1 in 1,465,273.
For the record, I'm not sure if any of these are exactly what is being asked in the original post.
The odds of that happening 20 times in a row are 0.000 000 098 984 or 1 in 10,102,674.
For what it's worth, the odds of getting pocket deuces thru pocket jacks are 40/52 * 3/51 = 0.045 248 868 778
The totoal odds of getting at least one queen, or pocket pairs, is almost 50%: 0.491 704 374 057
The odds of it happening 20 times in a row are 0.000 000 682 467 or 1 in 1,465,273.
For the record, I'm not sure if any of these are exactly what is being asked in the original post.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
August 10th, 2011 at 8:14:02 AM
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Oops. My thinking was wrong. 12/52 + 11/51 is only part of a complex formula.
It's so complex, that it's actually easier to work it backwards. I.E. Take the odds of NOT getting any card higher than a Jack, and subtract that from 1. Therefore the odds of getting at least one Queen is 1 - ( 40/52 * 39/51 ) = 0.411 764 705 882
The odds of that happening 20 times is 0.000 000 019 633 or 1 in 50,935,154.
That makes the odds of getting a queen or better or a pocket pair 0.457 013 574 661
The odds of that happening 20 times is 0.000 000 157 973 or 1 in 6,330,208.
By comparison, the odds of getting Red on a double zero Roulette table is 0.473 684 210 526
The odds of getting 20 Reds in a row is 0.000 000 323 428 or 1 in 3,091,874.
It's so complex, that it's actually easier to work it backwards. I.E. Take the odds of NOT getting any card higher than a Jack, and subtract that from 1. Therefore the odds of getting at least one Queen is 1 - ( 40/52 * 39/51 ) = 0.411 764 705 882
The odds of that happening 20 times is 0.000 000 019 633 or 1 in 50,935,154.
That makes the odds of getting a queen or better or a pocket pair 0.457 013 574 661
The odds of that happening 20 times is 0.000 000 157 973 or 1 in 6,330,208.
By comparison, the odds of getting Red on a double zero Roulette table is 0.473 684 210 526
The odds of getting 20 Reds in a row is 0.000 000 323 428 or 1 in 3,091,874.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
August 10th, 2011 at 8:18:58 AM
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I wasn't including queens, D'oh!Quote: s2dbakerOne in 85 billion is not reasonable!
I had to recalculate but it's still one in 71.6 billion.
Here's my math:
Chance of not pulling a Q or better with the first cad is .76923....
Chance of not pulling a Q or better on the second card is. 76471....
Combined = .58824...
Inverse to get remaining (desired result) = .41176...
20 times in a row = 1 in 50,935,154
Better than a lottery ticket.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
August 10th, 2011 at 8:32:04 AM
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I calculate that the probability of getting any Queen or higher in two cards is 1-combin(40,2)/combin(52,2). Actually, I'm calculating the probability of getting two cards that are jacks or lower (which is a simpler calculation) and then subtracting that from 1 to get the odds of getting any queen or higher.
p(queens+ in one hand) = 0.411764706
The original question said what is the probability of this happening in 20 out of 21 hands.
p(queens+ in 20 of 21 hands) = 0.411764706^20 * (1-0.411764706)^1 * combin(21,1) = 2.42523E-07
Therefore, the odds of this happening is 1 : 4,123,322.026, which is four times better than hitting the top award on most slot machines.
p(queens+ in one hand) = 0.411764706
The original question said what is the probability of this happening in 20 out of 21 hands.
p(queens+ in 20 of 21 hands) = 0.411764706^20 * (1-0.411764706)^1 * combin(21,1) = 2.42523E-07
Therefore, the odds of this happening is 1 : 4,123,322.026, which is four times better than hitting the top award on most slot machines.
I heart Crystal Math.
August 10th, 2011 at 8:55:02 AM
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I will also assume that the original post meant "at least one of the two cards is a Queen or better".
The probability of it happening once is 1 minus the probability that both cards are 2 through Jack.
There are (52 * 51) / 2 different sets of pocket cards, of which (40 * 39) / 2 are both 2-J, so the probability of getting at least one A/K/Q is 1 - 780/1326 = 0.4117647.
The probability of it happening exactly 20 times out of 21 is (the probability of it happening once)^20 * (the probability of it not happening, once) * 21 (since any of the 21 hands can be the "bad" one) = 1 / 4,123,322.
If one of the two cards is AKQ and the other is 2-J, there are 12 * 40 = 480 sets of these pocket cards, so the probability of getting exactly one A/K/Q is 480/1326 = 0.362, and the probability of this happening exactly 20 times out of 21 = 0.362^20 * 0.638 * 21 = about 1 / 50,000,000.
The probability of it happening once is 1 minus the probability that both cards are 2 through Jack.
There are (52 * 51) / 2 different sets of pocket cards, of which (40 * 39) / 2 are both 2-J, so the probability of getting at least one A/K/Q is 1 - 780/1326 = 0.4117647.
The probability of it happening exactly 20 times out of 21 is (the probability of it happening once)^20 * (the probability of it not happening, once) * 21 (since any of the 21 hands can be the "bad" one) = 1 / 4,123,322.
If one of the two cards is AKQ and the other is 2-J, there are 12 * 40 = 480 sets of these pocket cards, so the probability of getting exactly one A/K/Q is 480/1326 = 0.362, and the probability of this happening exactly 20 times out of 21 = 0.362^20 * 0.638 * 21 = about 1 / 50,000,000.
August 10th, 2011 at 9:19:47 AM
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Quote: CrystalMathI calculate that the probability of getting any Queen or higher in two cards is 1-combin(40,2)/combin(52,2). Actually, I'm calculating the probability of getting two cards that are jacks or lower (which is a simpler calculation) and then subtracting that from 1 to get the odds of getting any queen or higher.
p(queens+ in one hand) = 0.411764706
The original question said what is the probability of this happening in 20 out of 21 hands.
p(queens+ in 20 of 21 hands) = 0.411764706^20 * (1-0.411764706)^1 * combin(21,1) = 2.42523E-0.
I agree with this calculation.. Once the p = .411764706 is established, the next calculation is a simple binomial probability of 20 successes in 21 trials, except the binomial uses combin(21,20) instead of combin(20,1) and both give the same result.
August 17th, 2011 at 3:11:55 PM
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Quote: dwheatleyMaybe OP means Q-high or better? As in, Q-2 would also count. Then the chance is more reasonable
Yes...this is what I meant.
In 20 out of 21 hands straight ONE of the players cards was never lower than a queen. However remember that I am speaking of the same player as well. The other was any card from 2 thru Ace.
EXAMPLE: Heads up poker. 2 players only.
Pocket Cards:
Player 1 A-6 Hand 1
Player 1 A-9 Hand 2
Player 1 Q-7 Hand 3
Player 1 K-10 Hand 4
Player 1 Q-J Hand 5
Player 1 A-J Hand 6
Player 1 K-8 Hand 7
Player 1 K-Q Hand 8
Player 1 Q-6 Hand 9
Player 1 A-10 Hand 10
Player 1 K-7 Hand 11
Player 1 A-J Hand 12
Player 1 A-10 Hand 13
Player 1 K-4 Hand 14
Player 1 Q-5 Hand 15
Player 1 2-3 Hand 16 (The only hand that does NOt have an A-Q in the pocket)
Player 1 K-J Hand 17
Player 1 A-9 Hand 18
Player 1 K-4 Hand 19
Player 1 A-Q Hand 20
Player 1 Q-3 Hand 21
Recalculations anyone?
Thanks in advance.
August 17th, 2011 at 6:10:43 PM
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Technically, the Non-Queen-or-better hand could not be the first or last hand since that would make it 20 in a row.Quote: ThatDonGuyThe probability of it happening exactly 20 times out of 21 is (the probability of it happening once)^20 * (the probability of it not happening, once) * 21 (since any of the 21 hands can be the "bad" one) = 1 / 4,123,322.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
August 17th, 2011 at 6:24:46 PM
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Granted, the chances are remote (Mike - please help this thread...)
Since we're talking about poker - did we consider the flop here, and how many hands were won/lost?
You hold 9-8 of diamonds, and the flop is 9h,8s, and 3c, you might want to stay in...
You play your hand, not just your hole cards...
Since we're talking about poker - did we consider the flop here, and how many hands were won/lost?
You hold 9-8 of diamonds, and the flop is 9h,8s, and 3c, you might want to stay in...
You play your hand, not just your hole cards...
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
January 17th, 2012 at 11:48:35 PM
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Quote: s2dbakerTechnically, the Non-Queen-or-better hand could not be the first or last hand since that would make it 20 in a row.
20 in a row would still be 20 out of 21, whatever the denominator (number of trials) is specified.
January 17th, 2012 at 11:53:01 PM
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Quote: s2dbakerwith the first cad....
Boston pokah playah: These caads ah maaked! LOL.
January 18th, 2012 at 4:17:45 AM
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OP was using a real world example. If it was 20 in a row, he would have said it.Quote: NowTheSerpent20 in a row would still be 20 out of 21, whatever the denominator (number of trials) is specified.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
January 18th, 2012 at 4:20:24 AM
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Many of my missives are written on a phone on a bouncy Long Island Rail Road train. I think I do pretty well considering the circumstances.Quote: NowTheSerpentQuote: s2dbakerwith the first cad....
Boston pokah playah: These caads ah maaked! LOL.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
