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celt71

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How do you calculate the odds of being dealt a specific poker hand?

Want to learn how to solve these types of problems for math class.

Playing poker against 1 opponent, if you are holding J(s) & 10(h) and the flop is K(s),10(s),J(d). The turn is Q(s). The next card was already burnt, so the next card is the river. What is the probability of hitting a
1)Straight
2)Flush
3)Full house
4)Striaght flush
5)Royal Flush

not knowing what cards your opponent has or which cards have been burnt thoughout this hand.

Can someone help explain this.?

Posted 787 day ago

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Answers (3)

montie 		Okay, I don't know all the Lingo, so I'll just add a little Probability stuff.

There are 52 cards in the Deck, so 1 in 52 (approx 2%, a little under) of a Particular card coming out.
Except 6 are Accounted for, so a Particular card is 1 in 46 (a little over 2%)
1) looks like 10-K is drawn, so A or 9 any suite will fill, or 8 of the 46....1 in 6
2) 13 cards are (s), but 4 are out, 9 in 46, 1 in 5
3)10 or J any suit left, and there are 2 of each-1 of those get the 3, 1 in 11..(.but there are other combos, 1 10 or J plus 1 K or Q, or 2 K or 2 Q...if that meant 1 card left to draw-only the First is do-able.)
4)A(s)or 9(s) 1 in 23 (note 1 in 4 of the straight, 2 in 9 of the flush)
5)A(s) 1 in 46

The cards in the Hand, or Burnt, do not Change the Probability, as they are Unknown, thay are Counted in the 46.
2 OpHand+1 Brnt+ 43 deck
In Reality, any Card could be in the Opponant hand or Burnt and Out, making the Probability of it being drawn 0%, BUT you can Only work with the Cards you Know. That is why you count the 46, each card, Including the Out Cards, has the same Chance 1 in 46, of being a Particular card. It's like tossing a Coin 50 times, and Guessing what the 31st will be. The other 49 tosses or the Results are Immaterial-it's a 50% chance, even if you throw 30 Heads in a row before it.

I hope:
1) I understood this Right
2)Did my Math Right
3)Explained Right
4) Helped
5) That you have a Great Day!

<Hugs>

Posted 787 day ago

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smithy 		There is a quick method of approximating the probability of hitting a winning hand known as "counting outs".

You outs is the number of unseen cards that will give you the winning hand if you hit any one of them on the next card.

In your example (1), the number of outs to hit a straight is 8 since either an Ace or a Nine will give you a straight. (There are 4 Aces and 4 nines so you have a total of 8 outs.)

Your chance of hitting a straight is given by this formula:

Approximated Probability = (2 x Number of Outs) + 1 = (2 x 8) + 1 = 17%

See http://www.suntzupoker.com/counting-outs.aspx for more examples of this quick hand computation

Posted 787 day ago

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ger2000 		head over to pokerisrigged.com and ask one of our helpful memebers any of your poker questions they will answer asap as for now ill post this question up on our forums and you can check it out for the best responses from pros.

Posted 787 day ago

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