Odds of someone having a....
 

I just played this hand...
Party Poker 2/4 Hold'em (8 handed) converter

Preflop: Hero is BB with J[img]/images/graemlins/club.gif[/img], T[img]/images/graemlins/club.gif[/img].
UTG calls, UTG+1 folds, MP1 calls, MP2 calls, CO calls, Button folds, SB completes, Hero checks.

Flop: (6 SB) 9[img]/images/graemlins/club.gif[/img], 2[img]/images/graemlins/diamond.gif[/img], J[img]/images/graemlins/diamond.gif[/img] <font color="#0000FF">(6 players)</font>
<font color="#CC3333">SB bets</font>, <font color="#CC3333">Hero raises</font>, UTG calls, MP1 calls, MP2 folds, CO folds, SB calls.

Turn: (7 BB) 8[img]/images/graemlins/heart.gif[/img] <font color="#0000FF">(4 players)</font>
SB checks, <font color="#CC3333">Hero bets</font>, UTG calls, MP1 calls, SB calls.

River: (11 BB) 5[img]/images/graemlins/heart.gif[/img] <font color="#0000FF">(4 players)</font>
SB checks, <font color="#CC3333">Hero bets</font>, UTG folds, MP1 folds, SB folds.

Final Pot: 12 BB

I'm curious what are the odds that nobody has a J??? How do I calculate that?

Re: Odds of someone having a....
 

Schizo,

There is no exact way to solve this problem without assuming that your oppos were dealt random cards. This assumption is false, though, because it ignores starting hand requirements as well as information implied by the way they played their hands.

Nonetheless, assuming random cards, after the flop there are 2 J left and 47 cards left. The odds that none of 6 oppos has a J are:

(45 choose 2)*(43 choose 2)*...*(35 choose 2)
--------------------------------------------
(47 choose 2)* (45 choose 2)*...*(37 choose 2)

which reduces to:

(35 choose 2)/(47 choose 2)=.55

So the odds that at least 1 person has a J are 45%.

Again, don't put too much stock in the calculation.

gm