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Wahrscheinlichkeit Set over Set / Flush over Flush

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Wahrscheinlichkeit Set over Set / Flush over Flush - 27-05-2013, 17:33
(#1)
Benutzerbild von HuiRan
Since: Sep 2009
Posts: 4
Hi Leute,

hatten in unserer cg-runde die diskussion über wahrscheinlichkeiten. Die frage dabei, was ist in einem fullring cg wahrscheinlicher, das es eine flush over flush oder set over set situation am flop gibt?

Habe im twoplustwo mal gesucht und folgende werte gefunden:

Zitat:
The probability of 2 or more flushes is exactly:

0.212546574414913%
I get 0.20995%.
select count(handsHaving(exactHandType, flop, flush) >= 2) from game='holdem', p1='**', p2='**', p3='**', p4='**', p5='**', p6='**', p7='**', p8='**', p9='**', p10='**'

If I include straight flushes, I get 0.21296%, again quite close to your number.
select count(handsHaving(exactHandType, flop, flush) >= 2 or (handshaving(exactHandType, flop, flush) >= 1 and handshaving(exactHandType,flop, straightflush) >= 1) or (handshaving(exactHandType, flop, straightflush) >= 2)) from game='holdem', p1='**', p2='**', p3='**', p4='**', p5='**', p6='**', p7='**', p8='**', p9='**', p10='**'

=~ 0.2125%

//
P(2 or more players flopping flush) =

P(1 suit flop) *
[C(10,2)*P(2 particular players have flush | 1 suit flop) -
2*C(10,3)*P(3 particular players have flush | 1 suit flop) +
3*C(10,4)*P(4 particular players have flush | 1 suit flop) -
4*C(10,5)*P(5 particular players have flush | 1 suit flop)]

4*C(13,3)/C(52,3) *
[C(10,2)*C(10,4)/C(49,4) -
2*C(10,3)*C(10,6)/C(49,6) +
3*C(10,4)*C(10,8)/C(49,8) -
4*C(10,5)*C(10,10)/C(49,10)]

=~ 0.2125%
und eine andere erklärung für die set situation:
Zitat:
P(underset on flop and lose) =

P(XYZ flop) *
P(dealt XX, YY, or ZZ) *
[ P(exactly 1 other set out) * P(lose to higher set) *
(1-P(quads and no quads for opponent)) +
P(2 other sets out) * P(lose to higher set) *
(1-P(quads and no quads for opponents) ]

Note that the probability on each line is conditional on the previous lines.

52/52 *48/51 * 44/50 *
9/C(49,2) *
[ (9*6/C(47,2) - 2*C(9,2)*6*3/C(47,2)/C(45,2)) * 1/2 *
(1-1*43/C(45,2)) +
C(9,2)*6*3/C(47,2)/C(45,2) * 2/3 *
(1-1*41/C(43,2)) ]

=~ 0.00015 or 1 in 6657
Werd da nicht richtig schlau, ist es also 0,00015 gegen 0.2125 - somit wahrscheinlicher flush over flush als set over set am flop??

set over set twoplustwo

flush over flush twoplustwo

vielen dank
gruß

Geändert von HuiRan (27-05-2013 um 17:51 Uhr).
 
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27-05-2013, 18:07
(#2)
Benutzerbild von Alanthera
Since: May 2010
Posts: 4.298
Ich rechne das sicherlich nicht nach, aber der Wert fürn einfachen gefloppten Flush is schon unter 1% und dasses nen Set vs Set-Flop gibt is glaub ich ziemlich genau bei 1%.

Bei den Berechnungen muss man natürlich dann auch die Fragestellung korrekt auffassen:
1. P(2 or more players flopping flush)
2. P(underset on flop and lose)

Die zweite Frage geht halt viel weiter als nur nach Set vs Set am Flop.
 
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27-05-2013, 23:06
(#3)
Benutzerbild von HuiRan
Since: Sep 2009
Posts: 4
Hab ich vollkommen überlesen.
Hier nochmal eine Ausführung:

1) Chances to get PP:
there are (52*51)/2=1326 theoretically possible holdings, 13 PPs and 6 ways to be dealt each which leads to 13*6/1326=0.058823 or 5.88% or (1-to-17) 1/17.

2) Chances that at least one of your opponents has different PP himself:
first we must take 2 (your hole) cards out of the deck so there are (50*49)/2=1225 possible holdings and remaining 12 PPs, 6 ways to be dealt each -> 12*6/1225=0.058775 or 1/17.01 AGAINST 1 OPPONENT (HU).
With 9 OPPs (FR): (1 - (16/17)^9) = 1/2.38
With 5 OPPs (6m): (1 - (16/17)^5) = 1/3.82

By now we have two different PPs preflop. Since we want to know how often will our OPP have the better one (OBV half of the time), we need to divide these with 2 ->

HU: (1/2)*(1/17.01)= 1/34.02
6m: (1/2)*(1/3.82)= 1/7.64
FR: (1/2)*(1/2.38)= 1/4.76

3) Chances of you flopping a set:
first we must take 4 known (hole) cards out of the deck -> 1-((46/48)*(45/47)*(44/46))=0.12234 or 1/8.17

4) Chances of your OPP flopping a set:
now we must take 5 known cards out of the deck: 4 hole cards and one flop card that brought us a set which also means there are 2 remaining flop cards that OPP might hit set with -> 1-((45/47)*(44/46))=0.084181 or 1/11.88

5) Putting it all together:
now we just have to multiply all of the above:
(HU) (1/17)*(1/34.02)*(1/8.17)*(1/11.88)= 1/56133
(6m) (1/17)*(1/7.64)*(1/8.17)*(1/11.88)= 1/12606
(FR) (1/17)*(1/4.76)*(1/8.17)*(1/11.88)= 1/7854

Steht dann 0,2125% zu ~1,27%. Set over Set also wahrscheinlicher als Flush over Flush.. Problem gelöst, danke!

Geändert von HuiRan (27-05-2013 um 23:09 Uhr).