Probability of starting hands

JustBeginning · #1 06-13-2007, 09:07 AM

How do I work out the chances of their being a pocket pair(for example) at the table at a table of 9(again for example). Or for their to be two pocket pairs. I'm not really interested in the actual answer, more in the workings out of it.

kinghippo423 · #2 06-15-2007, 10:16 PM

The chances to get a pocket pair is (52/52 * 3/51). Why?

On the first card, you have a 52/52 chance to have a part of a pocket pair

On the second card, assuming to got a 7, you have 3/51 to get another 7.

P(get a pocket pair) = 52/52 * 3/51 = 5.88%

For the other question, I really don't know sorry...

tarheeljks · #3 06-16-2007, 12:05 AM

exactly 2 pp's or at least 2?

filsteal · #4 06-17-2007, 03:10 AM

If you want an EXACT answer for the probability of two pocket pairs, it's a tremendous mess.

I'll try to work it out if there's any interest.

JustBeginning · #5 06-19-2007, 08:30 AM

I know it's 16/1 on you getting a PP, so is it therefore 16/9 on there being a PP in any given hand if there are 9 people at the table?

I'm not really that interested in the actual odds of there being 2 or 3 PP's at a table to be honest. I just want to know how to go about working it out. If it's too much hassle it's fine.

Thanks.

filsteal · #6 06-19-2007, 12:40 PM

[ QUOTE ]
I know it's 16/1 on you getting a PP, so is it therefore 16/9 on there being a PP in any given hand if there are 9 people at the table?

[/ QUOTE ]

No. I think you're asking what the probability is that at least one player will be dealt a PP in a nine-handed game.

To figure that out, first assume that each player getting a PP is independent of each other player getting a PP. (This isn't quite true, but it's close enough.) Then the probability of NO PPs being dealt is (16/17)^9, so the probability of at least one PP being dealt is [1-(16/17)^9] ~= 42%.