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#1
05-11-2003, 10:27 AM
 Mike Haven Senior Member Join Date: Sep 2002 Location: Northern Ireland Posts: 2,288
Odds question.

What are the odds of finishing with quads in hold'em where you also have to use both your pocket cards to meet the conditions of a site's bonus, please?

#2
05-11-2003, 03:36 PM
 Carl_William Member Join Date: Dec 2002 Location: CA & Ohio USA Posts: 70
Re: Odds question.

What are the odds of finishing with quads in hold'em where you also have to use both your pocket cards to meet the conditions of a site's bonus, please?

There is a very fine line between "hobby" and "mental illness."

Assuming you start with a pair: (using hypergeometric distribution)
50 cards you have not seen before flop; two cards required on board make you quads; &amp;
of course 5 cards on board after river….

Chances of making quads = Combin(2,2)*Combin(48,3)/Combin(50,5) = 1/122.5

------------------------------------
Testing Wilson Holdem Software:
Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=45): Probability = 8509/1000000 = 117.5

Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=57): Probability = 8476/1000000=117.98

Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=649): Probability = 8597/1000000 =1/116.3

Monte Carlo (10^6) trials Wilson Holdem software (pair Q’s seed=349): Probability = 8434/1000000=1/118.57

Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=911): Probability = 8471/1000000=118.05

For the Wilson Monte Carlo trials I listed: you can see that the trials predict Quads a little more frequently than the Hypergeometric distribution probability.

Assuming you get a pair in holdem about 1 in 17 hands. Then the probability of quads (before hand) on the river is (1/17)*(1/122.5) = 1/ 2082.5 = 2 in 4165.

Like Buzz says: just my opinion &amp; subject to correction if in my haste I goofed….
#3
05-11-2003, 03:43 PM
 switters Junior Member Join Date: Apr 2003 Location: Austin, TX Posts: 22
Re: Odds question.

if you start with a pocket pair, then for both cards to play you must make quads in the rank you hold. this will happen C[48,3]/C[50,5] = 17296/2118760 = 0.816% of the time you start with a pocket pair.

If you start with any other hand, you will make quads where both cards play precisely when you get three of one of your ranks on the board and the other two cards below (or equal to, in most jackpots) the other rank in your hand.

for example, if you hold KJ, the odds of making quad J's with your K playing are C[43,2]/C[50,5] = 0.043%

this is derived by:
a) from the 3 remaining J's, choose all 3 (C[3,3] = 1)
b) from the 47 remaining cards, subtract out the ones that invalidate your kicker (the 4 A's) and choose 2 (C[43,2] = 903)
c) divide by number of possible boards (C[50,5]=2118760)

these are the chances of each rank playing as a kicker when the other card (x) makes quads, and x is the smaller rank:

Ax 0.05102%
Kx 0.04262%
Qx 0.03497%
Jx 0.02808%
Tx 0.02195%
9x 0.01657%
8x 0.01194%
7x 0.00807%
6x 0.00496%
5x 0.00260%
4x 0.00099%
3x 0.00014%

these are the chances of each rank playing as a kicker when the other card (x) makes quads, and x is the larger rank (if x is the larger rank, then four of the cards that would counterfeit your kicker are gone anyway!):

Kx 0.05102%
Qx 0.04262%
Jx 0.03497%
Tx 0.02808%
9x 0.02195%
8x 0.01657%
7x 0.01194%
6x 0.00807%
5x 0.00496%
4x 0.00260%
3x 0.00099%
2x 0.00014%

you can just sum up the numbers from these two columns to figure out any particular hand, for instance -- with AK you will make quads with a playing kicker 0.10204% of the time, with 27 you will make quads with a playing kicker 0.00821% of the time.

SO, if what you wanted to know was, if I play any two random cards, and see it to the river no matter what, without even looking, what percentage of the time will i flip it over at the end to find quads with a kicker that plays?

well, using the procedure above, excel tells me the answer is: 0.0942% of the time, or just shy of one in a thousand.

-switters
#4
05-11-2003, 04:56 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Exactly correct *nm*

#5
05-11-2003, 06:48 PM
 Mike Haven Senior Member Join Date: Sep 2002 Location: Northern Ireland Posts: 2,288
Re: Odds question.

Assuming you get a pair in holdem about 1 in 17 hands. Then the probability of quads (before hand) on the river is (1/17)*(1/122.5) = 1/ 2082.5 = 2 in 4165.

thank you very much - this is the one i wanted

williamhill (a uk internet poker site) is offering a "prize" of \$50 for quads using a pair in your hand, on monday only

i didn't think this was generous by any stretch of the imagination, even though i suppose i can't really complain about any sort of freebie

but in fact, i've just seen it is worse than i thought - it has to be quads A K Q or J! what are the odds on you getting one of those (before hand)?
#6
05-11-2003, 06:56 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Odds question.

but in fact, i've just seen it is worse than i thought - it has to be quads A K Q or J! what are the odds on you getting one of those (before hand)?

Well obviously 4/13 of what we calculated for any quads.
#7
05-11-2003, 08:09 PM
 Mike Haven Senior Member Join Date: Sep 2002 Location: Northern Ireland Posts: 2,288
Re: Odds question.

= 8/54,145 = 1/6,768?
#8
05-11-2003, 09:25 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
yes *nm*

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