The board shows 4 cards of the same suit, whats is the % of

[Vannek]

Lets say you hold 2c 3h, the board has 4 clubs, its HU, what are the chances he has a club, by what ratio does it increase by the number of opponents(Is there any rule of thumb like like +~8% per opponent)

[BruceZ]

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Lets say you hold 2c 3h, the board has 4 clubs, its HU, what are the chances he has a club

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On the river:

1 - C(37,2) / C(45,2) = 32.7%

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by what ratio does it increase by the number of opponents(Is there any rule of thumb like like +~8% per opponent)

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1 - C(37,2n) / C(45,2n)

where n is the number of opponents.

<font class="small">Code:</font><hr /><pre>
# opps. P(club)

1 32.7%
2 55.7%
3 71.5%
4 82.1%
5 89.1%
6 93.6%
7 96.3%
8 98.0%
9 99.0%</pre><hr />

[emp1346]

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1 - C(37,2) / C(45,2) = 32.7%

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could you please explain where you got 37 and 45... i understand the concept but i just don't see where those are coming from entirely...

thanks...

[BruceZ]

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[ QUOTE ]
1 - C(37,2) / C(45,2) = 32.7%

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could you please explain where you got 37 and 45... i understand the concept but i just don't see where those are coming from entirely...

thanks...

[/ QUOTE ]

There are 45 total unseen cards (52 minus 5 board cards minus 2 cards in our hand = 45). 37 of these are non-clubs since there are 4 clubs on the board and 1 in our hand, so that leaves 13 - 4 - 1 = 8 clubs, and 45 - 8 = 37 non-clubs remaining.

C(45,2) is the total number of hands our opponent can hold.

C(37,2) is the number of hands with no clubs that our opponent can hold.

C(37,2) / C(45,2) is the probability that our opponent has no clubs.

1 - C(37,2) / C(45,2) is the probability that our opponent has at least 1 club.

[MortalWombat]

Keep in mind that the probabilities listed below do not take into consideration that a person who is holding at least one club is more likely to stick it out to the river once clubs start showing up on the board. So in real life, these numbers are probably a bit low (at least the probabilities where only one or a few opponents are involved) in the scenario proposed. If there four clubs showing at the river, then this means that there were at least two at the flop and at least three showing at the turn, thus increasing the probability that someone with at least one club would have stayed in after the turn.

[ QUOTE ]
[ QUOTE ]
Lets say you hold 2c 3h, the board has 4 clubs, its HU, what are the chances he has a club

[/ QUOTE ]

On the river:

1 - C(37,2) / C(45,2) = 32.7%

[ QUOTE ]
by what ratio does it increase by the number of opponents(Is there any rule of thumb like like +~8% per opponent)

[/ QUOTE ]

1 - C(37,2n) / C(45,2n)

where n is the number of opponents.

<font class="small">Code:</font><hr /><pre>
# opps. P(club)

1 32.7%
2 55.7%
3 71.5%
4 82.1%
5 89.1%
6 93.6%
7 96.3%
8 98.0%
9 99.0%</pre><hr />

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