View Full Version : What are the odds FH vs. FH vs. FH?

04-29-2005, 07:56 PM
PokerStars No-Limit Hold'em Tourney, Big Blind is t30 (8 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)

saw flop|<font color="#C00000">saw showdown</font>

BB (t1575)
UTG (t810)
<font color="#C00000">UTG+1 (t850)</font>
MP1 (t1380)
MP2 (t1550)
<font color="#C00000">CO (t2510)</font>
Button (t3010)
<font color="#C00000">Hero (t1815)</font>

Preflop: Hero is SB with 9/images/graemlins/diamond.gif, 9/images/graemlins/heart.gif.
<font color="#666666">1 fold</font>, <font color="#CC3333">UTG+1 raises to t90</font>, <font color="#666666">2 folds</font>, CO calls t90, <font color="#666666">1 fold</font>, Hero calls t75, <font color="#666666">1 fold</font>.

Flop: (t300) T/images/graemlins/club.gif, K/images/graemlins/club.gif, 7/images/graemlins/heart.gif <font color="#0000FF">(3 players)</font>
Hero checks, UTG+1 checks, CO checks.

Turn: (t300) 9/images/graemlins/spade.gif <font color="#0000FF">(3 players)</font>
<font color="#CC3333">Hero bets t150</font>, UTG+1 calls t150, <font color="#CC3333">CO raises to t1020</font>, Hero calls t870, UTG+1 calls t610 (All-In).

River: (t3100) K/images/graemlins/heart.gif <font color="#0000FF">(3 players, 1 all-in)</font>
<font color="#CC3333">Hero bets t705 (All-In)</font>, CO calls t705.

Final Pot: t4510

Results in white below: <font color="#FFFFFF">
UTG+1 has Ts Th (full house, tens full of kings).
CO has Kd Td (full house, kings full of tens).
Hero has 9d 9h (full house, nines full of kings).
Outcome: CO wins t4510. </font>

05-01-2005, 11:02 PM
Well, the odds depend on the board.

If the board doesn't pair, the odds are 0%, because there will be no FH, period.

If the board has one pair -let's say Kings- then there are two ways to generate a FH: a pocket pair that hits one of the three non-King cards on the board [3 * (3*2)/(47*46) = 0.832%], or a King plus a pairing non-king ( (2*9)/(47*46) = 0.832%]. With one board pair there can be up to five FH hands at showdown: the two outstanding kings and one pocket pair for the three remaining cards on the board (Do you see why?) The exact odds will depend on several factors, like how many players were dealt, the game you were playing (e.g. it appears you are playing Hold-em, where you can get a FH by playing only one of your cards, but in other games, like OMaha, you must play two -- and the odds change). In crude terms, we could estimate the chances in a TXHE game 0.8%/FH, so three FH would be 0.00832^3 = 1:1,732,804

If the board has two hands [let's say Ks and Qs], there is only one non-paired board card. In TXHE, you would get a FH with any K or Q that isn't a pocket pair [ 2* (4*43)/(47*46) = 15.91%] or a pocket pair of the non-paired card [(3*2)/(47*46) = 0.278%] so the chances of three FH would be crudely (0.1591)^3 + 3*(0.00278*0.1591*0.1591) = 0.424% or about 1:235. It's so much easier to get three FH with two board pairs, that we'll probably all see this in our online poker careers. In a B&amp;M or home game, probably not.

Of course, for the a priori probabilities, you'd have to factor in the chances of having one or two board pairs respectively. There are other factors to take into account, too, but I'm not sure *exactly* what chances you are asking for, so I'll leave that for you to fiddle with.