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smoore
12-11-2004, 12:55 AM
1) Ten handed hold'em game. I hold no club. There is a club four flush on the board at the river. There are three people to the river. How do I figure out the probability of someone holding a club. Math only, don't want to take into account the play of the hand in this situation.

2) Four handed hold'em game. I hold KTo. The flop is 9TT. What is the probability of someone holding the other Ten?

I'm sure this is simple probabability, but I don't know how to figure it out.

edit: looking at 1), it appears that it doesn't matter how many people are at the river for this math question. Am I correct?

Mike Haven
12-11-2004, 01:16 AM
1) There are 9 clubs in the 45 unknown cards. There are 6 cards out against you.

There's a 9/45 x 100% = 20% chance of the "first" of the 6 cards being a club.

There's a 9/44 x 80% = 16% chance of the "second" of the 6 cards being a club if the first one wasn't.

There's a 9/43 x 64% = 13% chance of the "third" of the 6 cards being a club if the first two weren't.

And so on.

Add the percentages for the total chance of one of the six being a club.

blank frank
12-11-2004, 01:33 AM
Note that there are 36 cards that aren't clubs among the unknowns. There are 1,947,792 ways to choose 6 cards out of 36, and thus not get a club. There are 8,145,060 ways to choose 6 cards out of all 45 unknowns. Dividing the two gives you a 23.91% chance that none of them has a club, or a 76.09% chance that there is at least one club among the six cards.

In the second case, there are 47 unknowns, one of which is of interest. There are 9,366,819 ways to draw six cards out of 46, and 10,737,573 ways to draw six cards out of 47. That gives a 87.23% chance that no one has the ten, or a 12.77% chance that someone does.

BruceZ
12-11-2004, 01:55 AM
[ QUOTE ]
1) Ten handed hold'em game. I hold no club. There is a club four flush on the board at the river. There are three people to the river. How do I figure out the probability of someone holding a club. Math only, don't want to take into account the play of the hand in this situation.

[/ QUOTE ]

Assuming you are one of the 3 players (2 opponents):

1 - C(36,4)/C(45,4) = 60.5%.

That is 1 minus the probability that your opponents hold no club. C(36,4) is the number of ways to choose 4 opponent's cards out of 36 non-clubs, and C(45,4) is the total number of ways to choose 4 opponent's cards.

If you meant that you have 3 opponents, then:

1 - C(36,6)/C(45,6) = 76.1%.

[ QUOTE ]
2) Four handed hold'em game. I hold KTo. The flop is 9TT. What is the probability of someone holding the other Ten?

[/ QUOTE ]

1 - C(46,6)/C(47,6) = 12.8%.

That is 1 minus the probability your opponents hold no T. There are C(46,6) ways to choose the opponent's 6 cards out of the 46 non-tens, and C(47,6) total ways to choose these 6 cards.

[ QUOTE ]
edit: looking at 1), it appears that it doesn't matter how many people are at the river for this math question. Am I correct?

[/ QUOTE ]

No, the more people at the river, the more likely that a club is out.

Mike Haven
12-11-2004, 07:35 AM
1 - C(36,6)/C(45,6) = 76.1%.

Thanks, Bruce.

Whilst I'm sure this is the correct formula for a mathematician's use, I would doubt if anyone asking the question being answered would be any better off after seeing the formula. I would bet that "C(36/4)" is simply gobbledegook to many more than just me.

blank frank
12-11-2004, 11:38 AM
C(36,6) is just n choose r notation. C(n,r)=n!/(r!(n-r)!), where x!=1*2*3*...*x. C(n,r) is the number of ways you can choose r objects from a set of n objects, if the order you choose them in is not relevant. Bruce calculated his answers the same way I calculated mine, I just read the first question wrong.

Mike Haven
12-11-2004, 01:24 PM
C(36,6) is just n choose r notation. C(n,r)=n!/(r!(n-r)!), where x!=1*2*3*...*x. C(n,r) is the number of ways you can choose r objects from a set of n objects

That's what I meant - this is just sheer gobbledegook to a non-mathematician.

Imo, when someone asks a simple odds question here they need help, not bombardment with formulae that if they knew or understood they wouldn't ask the question in the first place.

KISS.

smoore
12-11-2004, 01:38 PM
yeah, that's greek to me... however I'm also looking for that. Can someone link me to a website that will explain that formula to me? I want to learn how to do this for myself (obviously), but I also want to at least touch on the underlying math.

Thanks for all the helpful responses.

jason1990
12-11-2004, 01:44 PM
[ QUOTE ]
C(36,6) is just n choose r notation. C(n,r)=n!/(r!(n-r)!), where x!=1*2*3*...*x. C(n,r) is the number of ways you can choose r objects from a set of n objects

That's what I meant - this is just sheer gobbledegook to a non-mathematician.

[/ QUOTE ]
Well, I think that's a bit of a stretch. C(n,r) is one of the first things we teach in an introductory course on probability. Very few of the students in a course like that are even math majors and probably less than 0.1% of them go on to become mathematicians.

Most computations in probability become horrendously complicated when you try to avoid using C(n,r). So C(n,r) is a tool that keeps things simple. It is such a fundamental concept, that anyone interested in learning how to compute probabilities on their own would be well-advised to learn it. So perhaps the best way to help someone who posts here is to help them understand something which, previously, they perceived as just "gobbledegook".

jason1990
12-11-2004, 01:51 PM
Try starting with this link:

http://mathforum.org/dr.math/faq/faq.comb.perm.html