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 12-23-2011, 12:45 PM #1 Cangurino Carpal \'Tunnel     Join Date: Apr 2008 Location: The Circus Posts: 13,478 F. Jerome's December Article In his article in the December 2011 issue, Calculating the Probability of Being Dealt Ace-Deuce in Omaha 8 or Better, Frank Jerome does just that. He lists 15 classes of Omaha hands, determines the size of each class, and adds up the sizes of the classes of interest. Nothing wrong with that. Except for the following: Using one of the most fundamental results of combinatorics, the Inclusion-Exclusion Principle, it is absolutely trivial to compute the number of A2 combinations as Code: `binom(52,4) -2 binom(48,4) + binom(44,4)` where "binom(n,k)" is the binomial coefficient n!/(k!(n-k)!). This avoids considering all subcases, and reduces the possibility of making mistakes. It also allows for generalisation to count other classes of hands. I had considered submitting an article as a reply, but since you are "less interested in a new mental trick to calculate probabilities" and no original mathematical work was done I didn't feel it would be appropriate. Of course this raises the question how those criteria were applied to the original article.
12-23-2011, 01:15 PM   #2
Dynasty
Carpal \'Tunnel

Join Date: Sep 2002
Location: Las Vegas
Posts: 25,182
Re: F. Jerome's December Article

Quote:
 Originally Posted by Cangurino I had considered submitting an article as a reply, but since you are "less interested in a new mental trick to calculate probabilities" and no original mathematical work was done I didn't feel it would be appropriate.
Where is that quote coming from?

The only person who would decide if the Magazine is interested in a new article is me. And, I didn't say it.

12-23-2011, 10:30 PM   #3
Buzz
gramps

Join Date: Sep 2002
Location: Los Angeles, California
Posts: 18,256
Re: F. Jerome's December Article

Quote:
 Originally Posted by Cangurino In his article in the December 2011 issue, Calculating the Probability of Being Dealt Ace-Deuce in Omaha 8 or Better, Frank Jerome does just that. He lists 15 classes of Omaha hands, determines the size of each class, and adds up the sizes of the classes of interest.
Yes. That is what I did. I'm glad you followed my reasoning.

Quote:
 Nothing wrong with that.

My object was mainly to explain how to create a table to be used in combinatorics leading to computing probability. The original title was,
"How to Make a Chart to find the Probability of being dealt two particular ranks as part of an Omaha-8 hand."

I used as an example calculating the probability of getting dealt a particular two cards of different ranks. And as a sub-example, the two ranks I chose were aces plus deuces.

But I didn't like that original title, tried some changes, and finally settled on a title I thought would be less confusing and would have more widespread appeal, "Calculating the Probability of Being Dealt Ace-Deuce in Omaha 8 or Better." I think it's that change of title that has led to confusion by you over my purpose in writing the article. My fault. Sorry.

Quote:
 Except for the following: Using one of the most fundamental results of combinatorics, the Inclusion-Exclusion Principle, it is absolutely trivial to compute the number of A2 combinations as Code: `binom(52,4) -2 binom(48,4) + binom(44,4)` where "binom(n,k)" is the binomial coefficient n!/(k!(n-k)!).
That's beautiful. Took me a while to comprehend it, and I don't know how you came upon it, but that's beautiful.

Quote:
 This avoids considering all subcases, and reduces the possibility of making mistakes. It also allows for generalisation to count other classes of hands.
"2 binom(48,4)" evidently means "2*binom(48,4)."

Thus binom(52,4) -2 binom(48,4) + binom(44,4) =
270,725-2*194,580+135,751=17,316.

Very elegant!! Thank you.

And then 17,316/270,725=0.06396

Nifty! Very elegant method! Again, thank you.

From my own perspective, there are often different methods to go about finding probabilities, some more elegant than others.

In the article to which you're referring, I specifically wanted to illustrate the general method of constructing a chart to use in calculation of probability. The chart method is a general method I often use and I thought the method would be useful to others.

Quote:
 I had considered submitting an article as a reply, but since you are "less interested in a new mental trick to calculate probabilities" and no original mathematical work was done I didn't feel it would be appropriate.
I don't know what you're quoting. Doesn't sound like something I'd write. And the magazine editor asserts he didn't write it either. (I believe him).

Quote:
 Of course this raises the question how those criteria were applied to the original article.
I don't understand. I wrote an article illustrating how to construct a chart to find the probability of being dealt a particular two rank combination in a four card Omaha-8 hand. It's something that, in my opinion, is very practical and that might be of interest to serious Omaha-8 players and students of the game. I submitted the article for possible publication, and it was accepted.

I plan to submit another along the same lines for next month. The goal, really, is to illustrate, by example, a method to construct a chart to calculate probabilities. I'll approach the problem from a different perspective. I'm not sure what the title will end up being.

I try to write the Omaha-8 articles I submit on a level sharp, interested readers will understand and will be able to apply. The Omaha-8 articles I write are written for the on-line 2+2 poker magazine rather than for a mathematics journal. (No offense to mathematicians is intended).

Buzz
Frank Jerome

12-24-2011, 05:05 AM   #4
Cangurino
Carpal \'Tunnel

Join Date: Apr 2008
Location: The Circus
Posts: 13,478
Re: F. Jerome's December Article

Quote:
 Originally Posted by Dynasty Where is that quote coming from? The only person who would decide if the Magazine is interested in a new article is me. And, I didn't say it.
The quote comes directly from the Contribute to the Magazine page.

12-25-2011, 03:47 AM   #5
Buzz
gramps

Join Date: Sep 2002
Location: Los Angeles, California
Posts: 18,256
Re: F. Jerome's December Article

Quote:
 Originally Posted by Cangurino Using one of the most fundamental results of combinatorics, the Inclusion-Exclusion Principle, it is absolutely trivial to compute the number of A2 combinations as Code: `binom(52,4) -2 binom(48,4) + binom(44,4)` where "binom(n,k)" is the binomial coefficient n!/(k!(n-k)!).
Pretty slick.

I’m interpreting your “binom(52,4)” to mean what I might type as “C(52,4).” Please correct me if I’m wrong.

C(52,4)-2C(48,4)+C(44,4)= 17,316. That's the number of four card hands containing at least one ace plus at least one deuce when Hero's own cards are not considered.

I read what you have written to imply you can get to
“binom(52,4) -2 binom(48,4) + binom(44,4)”
by an “absoluely trivial” application of the inclusion-exclusion principle.

I thought I understood the (seemingly relatively simple) inclusion-exclusion principle, but perhaps I don’t after all. At any rate, it seems to me there must be more to your elegant solution than an "absolutely trivial" application of the inclusion-exclusion principle.

Perhaps you will show us the step(s) in logic involved. (I'd appreciate that).

I hope you don’t regard what I have written as disrespectful. No disrespect is intended. On the contrary, I’m impressed with your expertise.

Thank you.

Buzz
(Frank Jerome)

 12-25-2011, 06:50 AM #6 Cangurino Carpal \'Tunnel     Join Date: Apr 2008 Location: The Circus Posts: 13,478 Re: F. Jerome's December Article Let X be the set of ace-free hands. Since we get them by taking the aces out of the deck and then dealing a four-card hand we have that |X| = C(48,4). Similarly, let Y be the set of deuce-free hands. Again, |Y| = C(48,4). The intersection of X and Y is the set of hands containing neither an ace nor a deuce. So |X ∩ Y| = C(44,4). Now the inclusion-exclusion principle states that |X ∪ Y| = |X| + |Y| - |X ∩ Y| = 2*C(48,4) - C(44,4). However, X ∪ Y is the set of all hands which do not contain both an ace and a deuce. Since we are interested in the complement of this set we subtract its size from the number of all hands, and get C(52,4) - 2*C(48,4) + C(44,4). ------ I like to think of this principle in a more general form where we have a given set of objects - here, Omaha hands - and a set P of properties that these objects can have - here we have two of them, "ace-free" and "deuce-free". For each set S of properties we can determine the number N(S) of objects having all of the properties in the given set, and we are interested in the number N' of objects having none of the given properties. We then get N' = ∑ (-1)^|S| N(S) where the sum is taken over all subsets S of P. The term (-1)^|S| is simply 1 or -1 depending on where the number of elements in S is even or odd. Applying this formula to the original problem we get the result in one step. Last edited by Cangurino; 12-25-2011 at 06:56 AM.
12-25-2011, 06:55 PM   #7
Buzz
gramps

Join Date: Sep 2002
Location: Los Angeles, California
Posts: 18,256
Re: F. Jerome's December Article

Quote:
 Originally Posted by Cangurino Let X be the set of ace-free hands. Since we get them by taking the aces out of the deck and then dealing a four-card hand we have that |X| = C(48,4). Similarly, let Y be the set of deuce-free hands. Again, |Y| = C(48,4). The intersection of X and Y is the set of hands containing neither an ace nor a deuce. So |X ∩ Y| = C(44,4). Now the inclusion-exclusion principle states that |X ∪ Y| = |X| + |Y| - |X ∩ Y| = 2*C(48,4) - C(44,4). However, X ∪ Y is the set of all hands which do not contain both an ace and a deuce. Since we are interested in the complement of this set we subtract its size from the number of all hands, and get C(52,4) - 2*C(48,4) + C(44,4).
Thank you.

I see it now. Very clear. Very impressive.

Quote:
 ------ I like to think of this principle in a more general form where we have a given set of objects - here, Omaha hands - and a set P of properties that these objects can have - here we have two of them, "ace-free" and "deuce-free". For each set S of properties we can determine the number N(S) of objects having all of the properties in the given set, and we are interested in the number N' of objects having none of the given properties. We then get N' = ∑ (-1)^|S| N(S) where the sum is taken over all subsets S of P. The term (-1)^|S| is simply 1 or -1 depending on where the number of elements in S is even or odd. Applying this formula to the original problem we get the result in one step.
Thanks for sharing. Much appreciated.

Buzz

 02-04-2012, 10:10 PM #8 zzbox stranger   Join Date: Feb 2012 Posts: 1 Re: F. Jerome's December Article It is so complicated

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