Stupid question from a math illiterate

[Kevin J]

In hold'em when you're dealt two non-paired cards, you have a 32.43% chance of flopping at least a pair. But if you're against 1 opponent (assuming he is not holding one of your ranks), is there now a 64.86% chance that ONE of you will flop a pair? And if so, does this mean that if you do NOT flop a pair, the chances are better than 50/50 that your opponent DID flop a pair? Or is it that if you don't flop a pair, then the chances are still 32.43% that your opponent did?

I know this is easy stuff for you guys, I just really suck at math and probabilities. Is there a FAQ guide anywhere that explains how to calculate probabilities as they relate to hold'em? I'm lazy and don't want to go through the process of learning all about how to calculate combinational probabilities in general. I just want to know how to do it as it pertains to hold'em. Thanks.

[lossage]

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In hold'em when you're dealt two non-paired cards, you have a 32.43% chance of flopping at least a pair.

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Almost. Your chances of pairing at least one of your cards is 32.43%. If your cards are close together, you might also get "a pair or better" by, say, flopping a straight; similarly, if they're soooted.

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But if you're against 1 opponent (assuming he is not holding one of your ranks), is there now a 64.86% chance that ONE of you will flop a pair?

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It's more like 58.7%. You see, your opponent and you have occupied 4 cards together (all different ranks by assumption), so there are C(48,3) = 17296 possible flops. There are 12 cards left in the deck that pair one of you. So, there are C(48-12,3) = 7140 flops that help neither of you. The probability that at least one of you will get paired is, therefore, 1 - 7140/17296 = 0.587.

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And if so, does this mean that if you do NOT flop a pair, the chances are better than 50/50 that your opponent DID flop a pair? Or is it that if you don't flop a pair, then the chances are still 32.43% that your opponent did?

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You can probably guess from the above discussion that it's neither.

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Is there a FAQ guide anywhere that explains how to calculate probabilities as they relate to hold'em? I'm lazy and don't want to go through the process of learning all about how to calculate combinational probabilities in general. I just want to know how to do it as it pertains to hold'em.

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There are online calculators that can answer the typical questions, like, "Given the following players with certain cards, who wins what fractions of the time?" However, they can't answer arbitrary questions like yours. If I were you, I would try to take the opportunity to learn something. There are plenty of people to answer your questions.

[Kevin J]

Thanks! You obviously know what you're talking about. Could you please help me with this..

"so there are C(48,3) = 17296 possible flops."

I don't understand the symbols "C(48,3)". I know that if you hold two cards there are 50*49*48= 117,600 possible flops. So if you're trying to figure out your chances of NOT flopping a set it would be 48/50*47/49*46*48. Is that right?

I see what you're saying that we changed the nature of 117,600 flops for TWO cards, because we are now concerned with two more. But wouldn't that be, 48*47*46 for 103,776 possible flops for 4 cards?

I'm sure you can now see how and why I'm so terrible at math.

Also, do you know where I can learn more about how to calculate this stuff myself? Like how to calculate runner-runner possibilities, etc. It would be helpful so that I can think on my own without having to embarrass myself with stupid questions like these every time I want to figure something out.

Lastly, that was a cool site you game me, but I don't know how to input data into the calculator. Is there another page which explains how to format the data? Thanks.

[lossage]

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Thanks! You obviously know what you're talking about.

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Thank you for the compliment.

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Could you please help me with this... "so there are C(48,3) = 17296 possible flops."

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Sure, I'll try. There's this function C(n,r). It's defined, usually, when n and r are integers, n is positive, and 0 <= r <= n; the formula is C(n,r) = n!/(r!(n-r)!). (However, you can find calculators for that kind of thing on the web.)

One usually uses "C" for the function's name, because it means "combinations;" specifically, C(n,r) is the number of distinct combinations of r things that can be chosen out of a set of n objects.

So, getting 'round to your actual question, if four cards are out, then you have 48 remaining from which the flop will be chosen. Since the flop is a set of 3 cards, the number of possible flops is C(48,3) = 17296.

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I know that if you hold two cards there are 50*49*48= 117,600 possible flops. So if you're trying to figure out your chances of NOT flopping a set it would be 48/50*47/49*46*48. Is that right?

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Not 117600. Your idea of 50*49*48 is right, but that counts all the ways the flop could be dealt onto the board. However, dealing A [img]/images/graemlins/club.gif[/img], A [img]/images/graemlins/spade.gif[/img], A [img]/images/graemlins/diamond.gif[/img] is the same for our purposes as dealing A [img]/images/graemlins/spade.gif[/img], A [img]/images/graemlins/club.gif[/img], A [img]/images/graemlins/diamond.gif[/img], and so on. For any three cards on the board, there are 3! = 6 ways of arranging them, so the number of distinct flops is 117600/6 = 19600. By no coincidence, C(50,3) = 19600 as well!

Back to your question, if you hold, say, pocket 3s in hold'em, then there are 48 cards left that don't give you a set (or quads), everything but the two remaining 3s, so there are C(48,3) flops that don't contain a 3. There are C(50,3) flops total. So, the probability of not flopping a set or quads is C(48,3)/C(50,3) = 0.882. The probability of flopping a set or quads is 1 - 0.882 = 0.118.

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I'm sure you can now see how and why I'm so terrible at math.

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Naw, just need the proper instruction.

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Also, do you know where I can learn more about how to calculate this stuff myself...

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Other than taking my discrete math class? [img]/images/graemlins/cool.gif[/img]

Seriously, I'm afraid I don't.

[Rasputin]

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Also, do you know where I can learn more about how to calculate this stuff myself? Like how to calculate runner-runner possibilities, etc. It would be helpful so that I can think on my own without having to embarrass myself with stupid questions like these every time I want to figure something out.

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There must be a local college or community college near where you live. Contact them for a course catalog and look into an introductory statistics class.

[Sporky]

a combinatorics book would also be of great help. i took a class in combinatorial enumerations in college, and the whole class was calculating things like this. it has a silly name, but the class was really helpful and was based on counting out different combination possibilities. i'm sure if you read through the first few chapters of a related book it would help a great deal. good luck.