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June 16th, 2017 at 10:16:20 AM
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I am new to this forum and apologize in advance if this question is too simple or dumb.

You are seated at a 9-handed poker table, playing Texas Hold'em.

The probability that a player will be dealt a pocket pair is about 1 out of 17.

Therefore, when all 9 players are dealt a hand, and you haven't seen any of them yet, the odds that at least one player at the table of nine players is dealt a pocket pair are 9 out of 17.

The hands are dealt. Seven players confirm that were not dealt a pocket pair.

Are the odds that one of the remaining two players has a pocket pair still 2 out of 17, the same as before the hand was dealt, or does the fact that you now know that the 7 players were not dealt a pocket pair make it more likely that one of the remaining two players has a pocket pair, since the odds were 9 out of 17 that at least one player would be dealt a pocket pair?

You are seated at a 9-handed poker table, playing Texas Hold'em.

The probability that a player will be dealt a pocket pair is about 1 out of 17.

Therefore, when all 9 players are dealt a hand, and you haven't seen any of them yet, the odds that at least one player at the table of nine players is dealt a pocket pair are 9 out of 17.

The hands are dealt. Seven players confirm that were not dealt a pocket pair.

Are the odds that one of the remaining two players has a pocket pair still 2 out of 17, the same as before the hand was dealt, or does the fact that you now know that the 7 players were not dealt a pocket pair make it more likely that one of the remaining two players has a pocket pair, since the odds were 9 out of 17 that at least one player would be dealt a pocket pair?

June 16th, 2017 at 10:40:05 AM
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a few corrections

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

78/1326=

1/17

probability NONE have a pair = (16/17)^9 = about 0.579481 = A

so at-least-one = 1-A

about 0.4205

9/17 = about 0.52941

way too high

see what others have to say too

that should help with the last question

fun question

Sally

actually it is exactly 1 in 17Quote:BDSattvaI am new to this forum and apologize in advance if this question is too simple or dumb.

You are seated at a 9-handed poker table, playing Texas Hold'em.

The probability that a player will be dealt a pocket pair is about 1 out of 17.

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

78/1326=

1/17

not right there.Quote:BDSattvaTherefore, when all 9 players are dealt a hand, and you haven't seen any of them yet, the odds that at least one player at the table of nine players is dealt a pocket pair are 9 out of 17.

probability NONE have a pair = (16/17)^9 = about 0.579481 = A

so at-least-one = 1-A

about 0.4205

9/17 = about 0.52941

way too high

see what others have to say too

that should help with the last question

fun question

Sally

I Heart Vi Hart

June 16th, 2017 at 10:43:49 AM
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Quote:BDSattvaI am new to this forum and apologize in advance if this question is too simple or dumb.

You are seated at a 9-handed poker table, playing Texas Hold'em.

The probability that a player will be dealt a pocket pair is about 1 out of 17.

Therefore, when all 9 players are dealt a hand, and you haven't seen any of them yet, the odds that at least one player at the table of nine players is dealt a pocket pair are 9 out of 17.

That's not a very good estimation. It is true the probability any one player has a pair is 3/51 = 1/17. The probability at least one player out of 9 has a pair could be closely approximated as 1-(1-(1/17))^9 = 42.05%. Before a perfectionist jumps down my throat, let me emphasize that this is just an approximation.

Quote:Are the odds that one of the remaining two players has a pocket pair still 2 out of 17, the same as before the hand was dealt, or does the fact that you now know that the 7 players were not dealt a pocket pair make it more likely that one of the remaining two players has a pocket pair, since the odds were 9 out of 17 that at least one player would be dealt a pocket pair?

As just an educated guess, if the first seven confirm no pair, I think the odds don't change much for players 8 and 9. If anything, the odds of a pair should go down slightly, compared to a full deck.

It's not whether you win or lose; it's whether or not you had a good bet.

June 16th, 2017 at 11:09:30 AM
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Yeah there is the effect of card removal and all that fun too so the odds are different.

-Tim

-Tim

-----
You want the truth! You can't handle the truth!

June 16th, 2017 at 4:52:07 PM
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42% would be a goodQuote:WizardThat's not a very good estimation. It is true the probability any one player has a pair is 3/51 = 1/17. The probability at least one player out of 9 has a pair could be closely approximated as 1-(1-(1/17))^9 = 42.05%. Before a perfectionist jumps down my throat, let me emphasize that this is just an approximation.

approximation to quote William Feller

op made a BASIC probability error that many make (even eye).

He got 9/17 adding the probabilities.

(actually is a binomial probability distribution)

by his math thought, if there were say 18 players, 18/17 would B the probability of at least 1 pair.

maybe not a good example there...

same with rolling a die.

at least 1 six with 3 dice is NOT 1/6 + 1/6 + 1/6 (that would be the average)

8 dice would give 8/6 (no, no, no)

we should grasp basics with truth

and always use them correctly

(no using screwdriver like a hammer, like me)

Sally

Last edited by: mustangsally on Jun 16, 2017

I Heart Vi Hart

June 16th, 2017 at 5:08:39 PM
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Years ago, I used to play Sic Bo under a false impression. I figured that if I choose one number, I have a one in six chance of getting it. With three die, I have a 3/6 chance and since the game pays double for two of the same number, that I had a better than 50% chance of winning.

I played blissful in my ignorance, and before moving up in stakes I wrote to a fairly new message board asking for confirmation of my thoughts. The first two posters actually agreed with my math and I was all happy when The Wizard himself dropped the bomb and showed the error of my ways.

Here's the funny thing though- Until he shot me down, I though I was playing a winning game and the results backed me up. After I found out the house had a fairly big edge, I never did well at the game again.

I played blissful in my ignorance, and before moving up in stakes I wrote to a fairly new message board asking for confirmation of my thoughts. The first two posters actually agreed with my math and I was all happy when The Wizard himself dropped the bomb and showed the error of my ways.

Here's the funny thing though- Until he shot me down, I though I was playing a winning game and the results backed me up. After I found out the house had a fairly big edge, I never did well at the game again.

It's what you do and not what you say
If you're not part of the future then get out of the way

June 16th, 2017 at 6:01:43 PM
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Probability analyses ab initio are no longer accurate following partial disclosure of the results. Wizard, I am confident that you are familiar with the famous "Monte Hall" apparent paradox. One of three door choices is selected, after which one unchosen door is opened to reveal nothing. The probability of correctly choosing the door concealing the big prize waiting behind either of two remaining closed doors is not 50 percent! Rather, the contestant should switch her preference to the previously unchosen door.

This theoretical poker problem regarding unrevealed pocket pairs following partial disclosure of some of the hands is analogous. It must be analyzed using Bayesian techniques.

This theoretical poker problem regarding unrevealed pocket pairs following partial disclosure of some of the hands is analogous. It must be analyzed using Bayesian techniques.

June 16th, 2017 at 7:15:59 PM
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Quote:pwcrabbWizard, I am confident that you are familiar with the famous "Monte Hall" apparent paradox. One of three door choices is selected, after which one unchosen door is opened to reveal nothing. The probability of correctly choosing the door concealing the big prize waiting behind either of two remaining closed doors is not 50 percent! Rather, the contestant should switch her preference to the previously unchosen door.

I can tell you that if I'm last to act at a full holdem table looking at AK suited pre flop and I know no one has a pocket pair, there is no way I'm trading those two cards for another random two out of the deck.

June 16th, 2017 at 9:47:10 PM
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noQuote:BDSattvaAre the odds that one of the remaining two players has a pocket pair still 2 out of 17, the same as before the hand was dealt,

this is also easily simulated and it is very close to the binomial probability value

of about

0.1142

here is a table

should be real close (simulation values very close too)

players remain | prob at least 1 pair |
---|---|

9 | 0.420518532 |

8 | 0.38430094 |

7 | 0.345819749 |

6 | 0.304933484 |

5 | 0.261491826 |

4 | 0.215335065 |

3 | 0.166293507 |

2 | 0.114186851 |

1 | 0.058823529 |

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