Board paired/odds of any player having third card chart?

taylor · #1 04-22-2005, 03:15 PM

is there such a chart from a full table down to heads up anywhere? I'd sure apperciate any info, thanks [img]/images/graemlins/smile.gif[/img]

noob

taylor · #2 04-22-2005, 04:03 PM

and pardon for not introducing myself properly. I'm taylor and I'm a noob here. Hello everyone, Philly born n bred, SoCal baby boomer here.. please initiate me

taylor · #3 04-23-2005, 12:40 PM

Did I ask an improper question?
Was I rude in some way?
Did I post my question in the wrong place?
Am I welcome here at this forum?

spentrent · #4 04-23-2005, 12:45 PM

[ QUOTE ]
Did I ask an improper question?
Was I rude in some way?
Did I post my question in the wrong place?
Am I welcome here at this forum?

[/ QUOTE ]

Howdy.

I didn't post because I don't have an answer to your question. There is no conspiracy against you. But here's a bump of good will.

BradleyT · #5 04-23-2005, 12:45 PM

Welcome to the forums. I think you should ask this in the probabilities forum. They'll be able to show you how to do the calculations and then you can put them into Excel and make a chart. It's probably an easy COMBIN calculation.

Oh and let us know when they answer you so we can get the chart too [img]/images/graemlins/laugh.gif[/img]

microbet · #6 04-23-2005, 12:58 PM

Let me fix this post so that you might get more responses.

[ QUOTE ]

Improper question?
Has rudeness brought me disgrace?
Am I not welcome?

[/ QUOTE ]

taylor · #7 04-23-2005, 02:11 PM

I'm confused. The chances of a random hand holding the Jack if you don't are 3/47 or roughly 1/15 correct? 3/47 + 3/46 = 1/8 are the chances of another J on the board by the river?

I get the going with feel and the assumption noone has the J or Q and the reverse with the 44A situation.

pokerlaw · #8 04-23-2005, 02:32 PM

[ QUOTE ]
I'm confused. The chances of a random hand holding the Jack if you don't are 3/47 or roughly 1/15 correct?

[/ QUOTE ]

No, that is assuming each person gets one hole card. 3/47 + 3/46 = about 1/8 is the relevant number for the J on the QQJ flop.

For the Q, use the same denominators, but decrease the numerator to 2 in each - since only two queens are available - so rougly 1/12 is the odds of a random having the Q.

[ QUOTE ]

3/47 + 3/46 = 1/8 are the chances of another J on the board by the river?

[/ QUOTE ]

yeah.

taylor · #9 04-23-2005, 02:45 PM

Got it completely. I should have forced myself/been able to figure it out. I love this stuff and my minds so full of it lately I was really having a problem with this one. Thanks for taking the time [img]/images/graemlins/smile.gif[/img]

pokerlaw · #10 04-23-2005, 12:55 PM

Hey,
I don't have a chart, but I'll walk you through the math. Check out super system maybe? i recall a decent probabilities chart in the back.

In situations you are talking about (i take it to mean a QQJ flop or the like), there are 3 remaining Jacks in the deck. Assuming you dont have one, we know 5 cards in the deck and one is a jack. Therefore, out of the remaining 47, there are 3 jacks. As such, the chance of a random person having a Jack are = 3/47 + 3/46, or roughly 1/8.

Therefore, multiply the 1/8 * n, where n = number of players at the table other than yourself. Therefore, if you are 5 handed, the odds that someone STARTED with a jack are 50%; if you don't have one on a JQQ flop.

However, it can likely be assumed that even if the pot isn't raised, J2-J7 are folded by the non-blind players (prob J8 also). So the J may no longer be there. Hope this helps....