Odds for this?

Piper Tim · #1 08-18-2005, 10:17 PM

What are the odds of being dealt pocket aces (with no diamond) and then all 5 cards being diamonds.

If possible, show your math so I can see how it is calculated.

Party Poker Pot-Limit Hold'em, $ BB (10 handed) converter
Preflop: Hero is UTG+2 with A[img]/images/graemlins/club.gif[/img], A[img]/images/graemlins/spade.gif[/img].
Flop: 8[img]/images/graemlins/diamond.gif[/img], 5[img]/images/graemlins/diamond.gif[/img], 3[img]/images/graemlins/diamond.gif[/img] Turn: 2[img]/images/graemlins/diamond.gif[/img] River: 4[img]/images/graemlins/diamond.gif[/img]

KJL · #2 08-18-2005, 11:25 PM

C(3,2)/C(52,2) * C(13,5)/C(50,5) is the formula, I think.

LetYouDown · #3 08-19-2005, 12:20 AM

Yup...looks right to me about 727654 to 1. Assuming you already have the A-A, the first part drops out and you just have C(13,5)/C(50,5) which is a more reasonable 1645.3 to 1.

hukilai · #4 08-19-2005, 12:33 AM

He asked for odds for 1st AND 2nd things happen, so the first part should stay there...

LetYouDown · #5 08-19-2005, 01:09 AM

I'm well aware. He also asked for an explanation. I took it a step further and explained that the odds increase dramatically due to the A-A being a factor. My apologies for being thorough...and thanks for your analysis.