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BDSattva
BDSattva
Joined: Jun 16, 2017
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June 16th, 2017 at 10:16:20 AM permalink
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?
mustangsally
mustangsally
Joined: Mar 29, 2011
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June 16th, 2017 at 10:40:05 AM permalink
a few corrections
Quote: BDSattva

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.

actually it is exactly 1 in 17
C(4,2)*13 / C(52,2) =
78/1326=
1/17
Quote: BDSattva

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.

not right there.
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
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Wizard
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Wizard 
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June 16th, 2017 at 10:43:49 AM permalink
Quote: BDSattva

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.



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

-Tim
----- You want the truth! You can't handle the truth!
mustangsally
mustangsally
Joined: Mar 29, 2011
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June 16th, 2017 at 4:52:07 PM permalink
Quote: Wizard

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.

42% would be a good
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
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billryan
billryan
Joined: Nov 2, 2009
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June 16th, 2017 at 5:08:39 PM permalink
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.
It's what you do and not what you say If you're not part of the future then get out of the way
pwcrabb
pwcrabb
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June 16th, 2017 at 6:01:43 PM permalink
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.
prozema
prozema
Joined: Oct 24, 2016
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June 16th, 2017 at 7:15:59 PM permalink
Quote: pwcrabb

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.



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.
mustangsally
mustangsally
Joined: Mar 29, 2011
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June 16th, 2017 at 9:47:10 PM permalink
Quote: BDSattva

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,

no
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 remainprob at least 1 pair
90.420518532
80.38430094
70.345819749
60.304933484
50.261491826
40.215335065
30.166293507
20.114186851
10.058823529
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