Resolved: Holding a pocket pair what is the expectation of only a set.

[VivaLaViking]

I posed the question, if holding a pocket pair what is the expectation of flopping only a set without two pair, quads or a full house.

LetYouDown correctly responded:

C(2,1)<-Set * C(12,2)<-2Values * C(4,1)*C(4,1)<-2Suits
---------------------------------------------------------

C(50,3)<-Total # of possible flops


=~ 10.78%

TY

[LetYouDown]

Was another thread entirely necessary, LOL? Glad you're happy to have your answer...sorry for the 9 different explanations that were apparently confusing.

...and do we all get partial credit on whoever you're selling this to? LOL

[BruceZ]

[ QUOTE ]
I posed the question, if holding a pocket pair what is the expectation of flopping only a set without two pair, quads or a full house.

LetYouDown correctly responded:

C(2,1)<-Set * C(12,2)<-2Values * C(4,1)*C(4,1)<-2Suits
---------------------------------------------------------

C(50,3)<-Total # of possible flops


=~ 10.78%

TY

[/ QUOTE ]

It's not an expectation, it's a probability.

Everyone should understand all of these methods, and be able to use them ambidextrously:

2*(48*44/2) / C(50,3) =~ 10.78%

2/50 * 48/49 * 44/48 * 3 =~ 10.78%

2*C(12,2)*4*4 / C(50,3) =~ 10.78%

2*[C(48,2) - 12*C(4,2)] / C(50,3) =~ 10.78%

[VivaLaViking]

It wasn't a attemp to start another thread it was an attempt to save others reading also I'm not selling anything, it's being given away. Thanks for all your help and keep waiting for tomorrows question lol. Thanks.

[VivaLaViking]

Bruce,

Thanks for your help and you may be correct about multiple solutions. I wish my barber could split hairs as well as you like expectations vs probability lol.

TY

TY