Short probability calculation

[Ghazban]

I was just at a no-limit table where a guy held pocket 10s two hands in a row and flopped a set both times. What are the odds of this happening (ballpark figure would be fine)? Thanks in advance.

[uuDevil]

(P(being dealt specific pr)*P(flopping set))^2= ((4/52)*(3/51)*(1-(48/50)*(47/49)*(46/48))^2=2.83E-7

Or about 3.53 million to 1 against.

I can guarantee that the probability this is correct is >=0.

[img]/images/graemlins/tongue.gif[/img]

[Ghazban]

If you make it more general and don't specify the rank of the pair (that then flops a set), you just multiply the percentage by 13, right? So the odds of getting the same pair twice in a row and flopping a set both times would be (roughly) 270,000:1 correct? Guess I shouldn't bank on it happening to me any time soon.... [img]/images/graemlins/tongue.gif[/img] Thanks for doing the math; I didn't trust myself to formulate it correctly.

[Shawsy]

[ QUOTE ]

(P(being dealt specific pr)*P(flopping set))^2= ((4/52)*(3/51)*(1-(48/50)*(47/49)*(46/48))^2=2.83E-7

Or about 3.53 million to 1 against.

I can guarantee that the probability this is correct is >=0.

[img]/images/graemlins/tongue.gif[/img]

[/ QUOTE ]

I go about it a little differently, but come up with a result in the same ball park.

Probability of being dealt any specific pair and flopping a set (but not quads):
C(4,2)/C(52,2)*C(2,1)*C(48,2)/C(50,3)~.005208
Probability of this happening two hands in a row = Answer^2 ~2.71E-7 or 1 in 3.67 Million.

[uuDevil]

[ QUOTE ]
If you make it more general and don't specify the rank of the pair (that then flops a set), you just multiply the percentage by 13, right? So the odds of getting the same pair twice in a row and flopping a set both times would be (roughly) 270,000:1 correct?

[/ QUOTE ]

I think this is correct. BTW, my number includes the times you flop quads rather than just a set, so Shawsy's number is better.