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Probability

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Probability - Tue Dec 24, 2002, 01:26 AM
(#1)
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New to the game, trying to calculate slightly more complex probablities.

In Holdem, delt 2 diamonds. ?? Prob. of completing the flush ??
i.e. chances of 3 diamonds being in the 5 community cards.

I know....
Possible combinations = 50! / (5! * 45!) = 2118760
(50 cards I haven't seen, 5 to be delt)

How do I.....
I get turned around on possible successful combination.
How do you calculate that.
(3 diamonds of a possible 11, in 5 chances)

HELP!!!!
 
Old
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Tue Dec 24, 2002, 05:36 AM
(#2)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Caro's Hold'Em Statistics

Your answer, along with a wealth of other information, can be found at the link above.

Have fun!
 
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Tue Dec 24, 2002, 09:43 AM
(#3)
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Thanks for the link, but I'm trying to understand how the number is calculated. I have a basic understanding of stats and am looking for the binomial coefficient and union space that I keep overlooking. Thanks again.

jkr
 
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Tue Dec 24, 2002, 02:49 PM
(#4)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
I haven't done serious statistics of this nature in over 20 years. Hopefully someone more up to date can be of assistance.
 
Old
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Wed Dec 25, 2002, 11:57 AM
(#5)
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Quote:
Originally Posted by riddlejk
Thanks for the link, but I'm trying to understand how the number is calculated. I have a basic understanding of stats and am looking for the binomial coefficient and union space that I keep overlooking. Thanks again.

jkr
I don't know what you said, but pick up Mike Petriv's Hold em's Odds Book, or http://www.math.sfu.ca/~alspach/. Both show how to do the math.

Hope this helps, Den
 
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Sat Dec 28, 2002, 04:53 PM
(#6)
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WARNING: Heavy math content. Not for the squeamish.

Here's how to do it:

The union space is all combinations of 5 cards where at least three (that is, exactly 3, exactly 4 or exactly 5) are diamonds. There's no way to calculate this except to take each case (x=3, x=4 and x=5) separately and add the results.

The type of function this problem represents is called a hypergeometric distribution. It is, by far, the most common type of probability distribution used for cards. It calculates the number of ways of taking a number of items from one pool and the remaining items from a second pool, and divides by the number of ways of taking the total number of items from the two pools combined.

Its basic form is as follows:

f(x) = C(n, x)*C(N-n, R-x)/C(N, R)

where

N is the total number of objects in the pool waiting to be chosen
R is the total number of these objects that will be chosen
n is the total number of objects with the desired characteristic
x is the total number of objects with the desired characteristic that we want to be chosen

and

C(x, y) = x! / y! (x-y)! (as you wrote in your first post).

Now, for each calculation that we do, N, n and R will all be constant:

N = 50 (the total number of cards remaining in the deck)
n = 11 (the total number of diamonds remaining in the deck)
R = 5 (the total number of cards which will be chosen to make up the board).

So, we have f(x) = C(11, x)*C(39, 5-x)/C(50, 5).

(Notice here that the first numbers in the first two combination functions, 11 and 39, add to give the first number in the last combination function, 50. This is also true of the last numbers in the three functions. This is a good way to check that your math is correct.)

Now, since our union space is x >= 3, we need to calculate f(x) for x=3, x=4 and x=5, the add them together:

f(3) = C(11, 3)*C(39, 2)/C(50, 5) = 0.0577
f(4) = C(11, 4)*C(39, 1)/C(50, 5) = 0.0061
f(5) = C(11, 5)*C(39, 0)/C(50, 5) = 0.0002

0.0577 + 0.0061 + 0.0002 = 0.064, or about 15-1.

Chris
 
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Sat Dec 28, 2002, 05:16 PM
(#7)
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Deleted
 
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Sun Dec 29, 2002, 02:14 AM
(#8)
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Uhm, yeah. What Chris said.
 
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Sun Dec 29, 2002, 02:30 AM
(#9)
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My head just exploded.
 
Old
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Sun Dec 29, 2002, 12:43 PM
(#10)
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Darn Chris !!

You made me take out my programmable calculator out of the closet. Inputed the formula and my variables are ready to be punched in. :roll:

Are there rules on poker tournaments to whipping out a calc and crunching numbers while requesting time? :P

CH
 
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Sun Dec 29, 2002, 12:46 PM
(#11)
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'Ham said:
Quote:
Are there rules on poker tournaments to whipping out a calc and crunching numbers while requesting time?
I don't know but, If you do it at Tunica I'll buy your first beer! :lol:
 
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Mon Dec 30, 2002, 11:47 AM
(#12)
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Free Beer?? You would be buying it if could have made it to Tunica. I would have dug up a sliding calculator and used it as a card marker.

Life is just not fair when work gets on the way of fun.

CH
 
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Quick Thanks - Fri Jan 03, 2003, 11:47 AM
(#13)
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Chris,

Thanks for the detailed answer. Exactly what I was looking for.

jkr
 
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Re: Probability - Mon Jan 06, 2003, 08:50 PM
(#14)
Deleted user
Quote:
Originally Posted by riddlejk
New to the game, trying to calculate slightly more complex probablities.

In Holdem, delt 2 diamonds. ?? Prob. of completing the flush ??
i.e. chances of 3 diamonds being in the 5 community cards.
My second book, "MORE Hold'em Excellence: A Winner For Life" -- available online at www.conjelco.com, and yes, the question was such a perfect set-up I had to get a plug in -- contains a chapter called "Simple Arithmetic" that deals with the question posed above as well as many other common situations one encounters at the poker table.

Since I am neither a statestician nor a math wonk, I wrote this chapter so that anyone with basic understanding of high school aithmetic should be able to follow the logic and run the nubers needed to perform these common calculations.

The following is an excerpt from that chapter:

Let’s say you’re holding AhJh and the flop is 8h7c3h. Since there are thirteen cards of each suit in a deck, and you’ve accounted for four of them, only nine of the remaining 47 cards can be hearts. (Yes, there are three aces and three jacks that may also give you the best hand, but for this exercise we’re only interested in learning how to calculate the chances of making a flush.)

Like the problem about flopping a set when you hold a pair in your hand, it is also simpler to calculate the number of ways to miss your flush, subtract the misses from the universe of possibilities, and the result will be our answer.

On the flop there are 47 unknown cards. Since nine of them are hearts, the remaining 38 will not help you. If you miss your flush on the turn, there are now only 46 unknown cards. Since nine of them are hearts, 37 others won’t help you. Let’s multiply fractions. We’ve done this before, so it should be easy. Multiply the numerator of the first fraction by the numerator of the second, and perform the same calculation for the denominators. The result: 38/47 x 37/46 equals 1406/2162.

If you subtract the number of misses (1406) from the total number of possible events (2162), you are left with 756 combinations that result in a flush. Now divide 756 by 2162. The answer is 0.35 (or 35 percent). If you flop a four-flush, you’ll make your flush 35 percent of the time.

Do you want to convert that percentage into odds? Here’s how to go about it. Subtract 35 percent from 100 percent, and divide that by 35 percent (100 - 35 = 65; 65 / 35 = 1.86). The odds against completing your flush are 1.86:1. If the pot figures to pay 2:1 or more on your investment, it is a profitable draw in the long run — regardless of whether you make the flush now. Even if you were to miss the next twenty-seven times you’re in this situation, drawing to your flush still has a positive expectation under these circumstances.

Since you have no control over the cards that will be dealt, you can only focus on making plays with positive expectations. That’s what having the best of it is all about. It’s also why players can never win over any extended period at craps, roulette, baccarat, and any of the other pit games, with the exception of twenty-one.

Every professional poker player, and every skilled amateur who wins steadily at the game, takes the best of it most of the time. What separates winning players from the rest of the pack is this simple fact: Winning players take the gamble out of poker.
 
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Re: Quick Thanks - Mon Jan 06, 2003, 10:34 PM
(#15)
Deleted user
Quote:
Originally Posted by riddlejk
Thanks for the detailed answer. Exactly what I was looking for.
No problem...I live for this sort of thing.

Kinda sad, actually. 8O

Chris
 
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Tue Jan 07, 2003, 11:45 AM
(#16)
Deleted user
Now I know what BlackAces is doing when time is called when straight and flush draws are on the board! :lol:

Unless Mark adds a "1 year" time button, I will keep guesstimating!

Tim
 

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