simple Q for you math geniuses

[j-cor]

needless to say, i'm not very good at math

what's the probability that in a 9 handed game, 2 players are dealt pocket pairs? 3 players? 4 players?

what is the probability that 2 players hit a set on the same flop? by the turn?

thanks in advance [img]/images/graemlins/smile.gif[/img]

[gaming_mouse]

[ QUOTE ]

what's the probability that in a 9 handed game, 2 players are dealt pocket pairs? 3 players? 4 players?

[/ QUOTE ]

The chance that two players are both dealt pocket pairs can be approximated very accurately with the first term of inclusion-exclusion:

(9 choose 2)*(13*(4 choose 2)/(52 choose 2))*((12*(4 choose 2)+1)/(50 choose 2))= .126

About 12.6%

Similarly, for three players (a small additional approximation is used here by assuming that the second players PP is of a different rank than the first players):

(9 choose 3)*(13*6/(52 choose 2))*((12*6+1)/(50 choose 2))*((11*6+2)/(48 choose 2))= .0177

About 1.8%


[ QUOTE ]
what is the probability that 2 players hit a set on the same flop? by the turn?

[/ QUOTE ]

Given that they already have pocket pairs, or before the hand is dealt?

[j-cor]

gaming_mouse,

thanks for the help, i appreciate it

regarding your question, i would actually like to know the answers to both of them

thanks again [img]/images/graemlins/smile.gif[/img]

[RacersEdge]

[ QUOTE ]
The chance that two players are both dealt pocket pairs can be approximated very accurately with the first term of inclusion-exclusion:

(9 choose 2)*(13*(4 choose 2)/(52 choose 2))*((12*(4 choose 2)+1)/(50 choose 2))= .126

About 12.6%

[/ QUOTE ]

I did it the quick and dirty way - (9 choose 2)*.065^2*.0935^7 = 9.5%

I can see it underestimates the answer slightly - didn't think it would be that much though..

[gaming_mouse]

[ QUOTE ]


I did it the quick and dirty way - (9 choose 2)*.065^2*.0935^7 = 9.5%

I can see it underestimates the answer slightly - didn't think it would be that much though..

[/ QUOTE ]

keep in mind that if you subtract off the second term of inclusion exclusion, the answers may be alot closter.

also, i don't get the .065 and .0935. shouldn't those numbers sum to 1?

[BrassGuru]

Before getting into this, I would like to have somebody critique the logic I'm using here. I feel it's a very good approximation, but I could be missing something.

Here is the way I would calculate the probabilty: The chance of any one person being dealt a pocket pair is 13 * (4C2)/52C2 (mathematically, this means 13 times "4 Choose 2" divided by "52 Choose 2". I get this to be 5.88%.

If you model this as a Bernoulli trial with a binomial distribution (and I think it's a good approximation), this means a "success" {i.e., pocket pair} will occur 5.88% on any one particular hand and a "failure" {i.e. a non-pair} will occur 94.12% of the time on any one particular hand.

Using this assumption, I get the following results at a nine-person table:

Chance of No Pocket Pairs: (0.9412)^9 = 58.0%

Chance of Exactly 1 Pocket Pair:
(9C1) * (.9412)^8 * (.0588) = 32.6%

Chance of Exactly 2 Pocket Pairs:
(9C2) * (.9412)^7 * (.0588)^2 = 8.1%

Chance of Exactly 3 Pocket Pairs:
(9C3) * (.9412)^6 * (.0588)^3 = 1.2%

Chance of Exactly 4 Pocket Pairs:
(9C4) * (.9412)^5 * (.0588)^4 = 0.1%


John <font color="black"> </font> <font color="blue"> </font> <font color="blue"> </font> <font color="black"> </font>

[gaming_mouse]

[ QUOTE ]
Before getting into this, I would like to have somebody critique the logic I'm using here. I feel it's a very good approximation, but I could be missing something.

Here is the way I would calculate the probabilty: The chance of any one person being dealt a pocket pair is 13 * (4C2)/52C2 (mathematically, this means 13 times "4 Choose 2" divided by "52 Choose 2". I get this to be 5.88%.

If you model this as a Bernoulli trial with a binomial distribution (and I think it's a good approximation), this means a "success" {i.e., pocket pair} will occur 5.88% on any one particular hand and a "failure" {i.e. a non-pair} will occur 94.12% of the time on any one particular hand.

Using this assumption, I get the following results at a nine-person table:

Chance of No Pocket Pairs: (0.9412)^9 = 58.0%

Chance of Exactly 1 Pocket Pair:
(9C1) * (.9412)^8 * (.0588) = 32.6%

Chance of Exactly 2 Pocket Pairs:
(9C2) * (.9412)^7 * (.0588)^2 = 8.1%

Chance of Exactly 3 Pocket Pairs:
(9C3) * (.9412)^6 * (.0588)^3 = 1.2%

Chance of Exactly 4 Pocket Pairs:
(9C4) * (.9412)^5 * (.0588)^4 = 0.1%


John <font color="black"> </font> <font color="blue"> </font> <font color="blue"> </font> <font color="black"> </font>

[/ QUOTE ]

The approximation will be fairly accurate. I'll do the second term of inclusion exclusion to compare the approximation for 2 pocket pairs. It has:

((9 choose 2) choose 2) = 630 terms.

Of those, (9 choose 4)*3 = 378 are non-overlapping (ie, they represent 4 people all having pocket pairs)

The chance of 3 particular people all having a pocket pair is (using a very slight and negligible cheat by assuming that all 4 are of different ranks):

(13*6/(52 choose 2))*((12*6+1)/(50 choose 2))*((11*6+2)/(48 choose 2)) = 0.00021131857

And for 4:

0.00021131857*(10*6+3)/(46 choose 2) = .0000128628695

The second term of inclusion exclusion is therefore:

378*.0000128628695 + (630 -378)*0.00021131857= 0.0581144443

Which means that the 1 term approximation is not very accurate after all. The 2nd term approximation is:

12.6 - 5.8 = 6.8

but is only guarenteed to be accurate within 5.8%. I don't feel like going after the 3rd term, as it only get messier. But it looks like the binomial answer will be pretty close after we add back on the 3rd term.

[BrassGuru]

Mouse,

I believe that using a binomial distribution would be 100% accurate if Player 1 was dealt two cards, and then those two cards were put back into the deck, the deck was reshuffled, and then 2 cards were dealt to Player 2. Then those two cards were put back into the deck, the deck was reshuffled and then two cards were dealt to Player 3, etc. The lack of REPLACEMENT prevents the Bernoulli trial (binomial distribution) assumption from being 100% accurate.

You seem to have excellent knowledge of probability theory -- is what I'm saying above accurate? I would think there would be a more straightforward way to calculate the probability of, say, 3 out of 9 people being dealt a pocket pair. But I'm no statistician.

There are a finite combination of hands that can be dealt at a nine person table (although it's a really big number), so it would probably just be easiest to let a computer crunch out the numbers. Do you know of any website that would calculate something like this?

John

[gaming_mouse]

[ QUOTE ]
Mouse,

I believe that using a binomial distribution would be 100% accurate if Player 1 was dealt two cards, and then those two cards were put back into the deck, the deck was reshuffled,

John

[/ QUOTE ]

This is correct.

We say that the without reshuffling the hands are loosely dependent. The looser the dependence, the better the binomial approximation. In this case it should be fairly accurate -- within 1, or at most 2, percent of the true answer.

That method also has the advantage of simplicity, as you can see from my messy answers.

[Cobra]

http://www.math.sfu.ca/~alspach/mag29

This is an interesting web site with many good articles.

Cobra