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08-01-2002, 06:39 PM
Most card problems can be done by using the intuitively more obvious method of parlays or multiplying fractions, without bringing in permutations and combinations. Take the problem of determining the chance of flopping two pairs with no pair on board. Say you have QJ.

The probability that the first card is a queen or a jack is 6/50. If it is, the probability that the second card is the other card (Q or J) is 3/49. If it is, the probability that the third card is something else is 44/48. Multiplying, we get 792/117600. But that is the chances assuming the off card comes last. Since it could also be first or second, we triple these results to 2376/117600 or about 2.02%. Feel free to ask me how to do other poker problems avoiding combinations.

08-01-2002, 08:04 PM
How do you do other poker problems avoiding combinations? /images/smile.gif No seriously, any useful quick estimation techniques would be helpful. Also helpful would be some ideas on what to think about using probability either at or away from the table. For example, you have figured out what starting hands should be played in the blinds against other starting hands, and when they should continue to be played depending on what flops. Stuff like that.

Here is a very rapid approximation. If you need to get a particular denomination to match one already in your hand, you can estimate that it is 6% per card on the board. So if you have an A in your hand, you would estimate that there is a 30% chance of catching another A by the river. The exact probability is

1-C(47 5)/C(50 5) = 27.6%. Pretty close for something you can do in your head in a second. You can make it even closer if you remember your estimate will be too high by around 3%.

08-01-2002, 08:05 PM
I am doubtless going to display my ignorance, and some conceptual errors, now and in many future posts on this new forum. But better late than never.

You are tripling the results based on the possible positions of the third card, but there are six orders the cards could fall in...why are we multiplying the result by three instead of by six?

08-01-2002, 08:30 PM
He wants either a Q or J in two positions, and any card in the other position. Since the any card can occur in one of 3 positions, there are really only 3 distinct rearangements, since he's treating Qs and Js and as if they are indistinguishable. So the only orders are xQJ, QxJ and QJx because xJQ, JxQ, and JQx are the same thing.

Sometimes probability is tricky, and when you're first getting started it's very easy to justify the wrong method. Especially when it comes to counting things.

But you are correct that we often want to divide by 6 to get the number of flops independent of order. For example, the total number of possible flops is 52*51*51/6. The numerator is the number of ways to get all the flops in any order, and we divide by 6 to get unique flops. In general, to get the number of unique combinations of N cards from the total number of ordered permutations, we divide by N! = N*(N-1)*(N-2)...*1

08-01-2002, 08:35 PM
I should say this better: the possible orders are xYY, YxY, and YYx where Y is a new denomination made up of Qs and Js. It would be the same as if he wanted 2 aces (though the probability would be different because there are fewer As than Ys).

08-01-2002, 08:58 PM
"For example, the total number of possible flops is 52*51*51/6."

Did you mean 52*51*50/6? Typo?

Anyway I think I see what you are saying about the QJ in any order since it is already built into the calculation by specifying 3/49.

Thanks Bruce.

08-01-2002, 09:36 PM
Here's another way to do it. The probability of getting exactly QJx in that order is

(3/50)(3/49)(44/48). Now we do multiply by 6 since Qs and Js are now distinguishable. Perhaps this is less confusing.