#1
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open message to gaming_mouse
gm-
Have you considered writing an article for the new 2+2 magazine? I have found your posts very informative in the past and you really have a great knowlege of probability and stats. I think your input would be great. Anyways, thanks for your posts in the past and going forward. |
#2
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Re: open message to gaming_mouse
J LU,
Thanks for the kind words. I actually have thought about it, but I don't yet have a specific topic. I need to find something that is: 1. Interesting to me 2. Interesting to others 3. Useful, in the sense that it clears up a common misconception or provides an insight most decent players don't already have. If you have any ideas, let me know. I'll keep thinking too. gm |
#3
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Re: open message to gaming_mouse
[ QUOTE ]
If you have any ideas, let me know. I'll keep thinking too. [/ QUOTE ] Clearly, it would have to be about some obscure math. [img]/images/graemlins/smile.gif[/img] I'm disappointed nobody has suggested that I should write an article. [img]/images/graemlins/grin.gif[/img] |
#4
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Re: open message to gaming_mouse
[ QUOTE ]
I'm disappointed nobody has suggested that I should write an article. [img]/images/graemlins/grin.gif[/img] [/ QUOTE ] If you hand't replaced that sweet yin/yang avatar, I would have. [img]/images/graemlins/shocked.gif[/img] gm |
#5
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Re: open message to gaming_mouse
I agree. Gaming mouse, you have really informative posts. Not only are your explanations clear and concise, but you frequently relate them to thought processes/opponent play. Have you ever thought of writting a book?
Maybe you and BruceZ can tackle it together. |
#6
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Re: open message to gaming_mouse
I'd like to see an article on how to calculate the math behind the probabilities. A primer on basic thought process with examples.
Cover combinations, and some examples. Say, lay out the math on why a set is 1 in 8 to hit on flop, why flushdraw gets there 35% of time. How and why to account for ordering. Maybe show some common mistakes people make in doing calcs and the correct thinking. You could toss in some Omaha calcs too maybe. |
#7
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Re: open message to gaming_mouse
Explain in stupid poker player (me) terms what (24 choose 3) means [img]/images/graemlins/laugh.gif[/img]
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#8
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Re: open message to gaming_mouse
[ QUOTE ]
Explain in stupid poker player (me) terms what (24 choose 3) means [img]/images/graemlins/laugh.gif[/img] [/ QUOTE ] I'd love to take a crack at that - I figure as a novice player myself if I can explain it, then I really understand it. Question is, can you bear to read it? [img]/images/graemlins/laugh.gif[/img] At a high level (24 choose 3) means how many ways can you pick three items out of a group of 24 items, if the resulting order doesn't matter and if we don't replace the items as we pick them. In 2+2 context it's the total number of _different_ hands in a deal of 3 cards from a deck of only 24. How do we calculate (24 choose 3)? Well, that requires a brief digression into the world of permutations & combinations. In permutations, order is important, but in combinations like (24 choose 3), order is not important. The permutation of 5[img]/images/graemlins/heart.gif[/img] then 2[img]/images/graemlins/club.gif[/img] then K[img]/images/graemlins/spade.gif[/img] would be considered a different result than if you got the same cards in a different order. Clearly 5[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/club.gif[/img] K[img]/images/graemlins/spade.gif[/img] is not a different hand than 2[img]/images/graemlins/club.gif[/img] K[img]/images/graemlins/spade.gif[/img] 5[img]/images/graemlins/heart.gif[/img] in poker, so we're usually not all that interested in permutations, except where they help us understand combinations The "choose" in (24 choose 3) is shortand for a combination. Intuitively you can imagine that there are fewer combinations (order unimportant) than permutations (order matters), but the key is how to calculate how many fewer combinations there are than permutations. Calculating permutations is useful as a starting point to understand combinations. How many ways can you choose the first card from a deck of 24? Of course, 24 different ways. Now 23 cards remain, so there are 23 ways to choose the 2nd card, for a total of 24 * 23 = 552 different permutations. Now what if we said the order doesn't matter? Well, every permutation like 7[img]/images/graemlins/club.gif[/img] 3[img]/images/graemlins/heart.gif[/img] can be ignored if we already listed 3[img]/images/graemlins/heart.gif[/img] 7[img]/images/graemlins/club.gif[/img]. In other words, the list of permutations can be cut in half for the list of combinations in this case. So if (24 permute 2) is 552, (24 choose 2) is [(24 permute 2) / 2 ] or 276. What happens with three cards? You already probably realize that (24 permute 3) = 24 * 23 * 22, but how many "similar permutations" do we get to eliminate when we try to calculate the combinations of (24 choose 3)? Well, consider one 3-card example: 2[img]/images/graemlins/club.gif[/img] 5[img]/images/graemlins/heart.gif[/img] K[img]/images/graemlins/spade.gif[/img]. How many ways can you arrange (permute) these three cards into your three-card hand? Well, from this three-card group, you can place any of the 3 cards in position 1, 2 cards in position 2, and one remaining card in position 3. So there are 3 * 2 * 1 = 6 "similar permutations". So (24 choose 3) must be equal to [ (24 permute 3) / 6 ] = 2024. If you want to further generalize, it's convenient to use the shorthand notation of factorials. For big N, N! means (N) * (N-1) * (N-2) * ... (3) * (2) * (1). 1! is just one, and 0! is also one. We saw earlier that (24 permute 3) = 24 * 23 * 22, which looks something like the first three terms of 24 factorial ( 24! ). In fact, the generalization is that [N permute R] is [ N! / (N-R)! ]. If you write out some examples, you'll see that all the denominator terms from N-R down to one cancel the same term in the numerator. Trying (N,R) = (9,2) for example, shows that (9 permute 2) is (9! / 7!) which is just 9*8. Notice when we calculated (24 choose 3) we divided the permutations by 3! = 6. The general rule is that (N choose R) is [ (N permute R) / R! ], and given what we know about permutations now, it's also [ N! / ( (N-R)! R! ) ]. Uh-oh, my second post here is a lengthy hijack of this thread. Apologies, but I hope this info helps someone. |
#9
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Re: open message to gaming_mouse
Encore! Encore! This is exactly the kind of stuff that I, and I assume numerous others, want/need. As I have read along through various threads about odds and such, I sort of follow the math, but often don't have an understanding of some of the "shorthand", the formulas and math that are second nature to many of you. I also assume that this is the kind of thing being sought by those asking for good math books on probability.
But whatever, this was a superlative post, I think. THANKS! |
#10
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Re: open message to gaming_mouse
Zapped,
Nice post. Starting with permutations, and then counting repeats, is the best way to explain combinations, and you did a very clear job of it. Cheers, gm |
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