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#1
11-12-2007, 10:44 PM
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Razz - paired cards
Lets assume that the villain has precisely a 3-card bike hand (three unduplicated cards 5 or lower) on third street. On 4th street the villain catches a 5. What is the probability that the 5 paired the villain assuming that there are 12 bike cards unaccounted for remaining in the deck including the three remaining 5's? Please include the formula, that would help me bunches because I think i calculated this incorrectly. Thanks in advance!
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#2
11-13-2007, 12:33 AM
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Re: Razz - paired cards
Villain's possible hands, + whether the 5 paired him:
234 - no 235 - yes 245 - yes 345 - yes There's a 75% chance the 5 paired him. (I don't know just what a bike card is, but since you said there are 12 left after 4 were dealt, I assume it means 2, 3, 4, 5)
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#3
11-13-2007, 01:15 AM
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Re: Razz - paired cards
a bike is A2345. Theer are 12 bike cards unaccounted for in the deck, 3 of them are 5's. I assumed it would be just 3/12=25% but I dont know if we have to factor in both cards. This comes from a thread in the Stud forum if you want to read it, Betgo's "Big Pot" right at the top of the forum.
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#4
11-13-2007, 01:33 AM
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Re: Razz - paired cards
Hmm.... possibly the best way is just to count all the possible combinations including suits.
We know he holds the 2 of diamonds. The other potential cards he could hold: three 5s, four 4s, three 3s, two As. A3 x6 - no A4 x8 - no A5 x6 - yes 34 x12 - no 35 x9 - yes 45 x12 - yes That's 27 hands where the 5 pairs him, and 26 hands where it doesn't. 27/53 = there's a 50.94% chance he got paired on 4th street.
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#5
11-13-2007, 03:02 AM
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Re: Razz - paired cards
[ QUOTE ]
Hmm.... possibly the best way is just to count all the possible combinations including suits. We know he holds the 2 of diamonds. The other potential cards he could hold: three 5s, four 4s, three 3s, two As. A3 x6 - no A4 x8 - no A5 x6 - yes 34 x12 - no 35 x9 - yes 45 x12 - yes That's 27 hands where the 5 pairs him, and 26 hands where it doesn't. 27/53 = there's a 50.94% chance he got paired on 4th street. [/ QUOTE ] Right except that there is only 1 ace unaccounted for. The possibilities then become: A3 x3 - no A4 x4 - no A5 x3 - yes 34 x12 - no 35 x9 - yes 45 x12 - yes This gives a probability of pairing the five of 24/43 =~ 55.8%.
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#6
11-13-2007, 03:53 AM
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Re: Razz - paired cards
Bruce - pococurante got it right, there are two A's unaccounted for. By using combinations the probability of the villain pairing his 5 is 50.9%.
But now comes the second part of my question, is there a shortcut for calculating this probability on the fly rather than typing out the combinations by hand?
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#7
11-13-2007, 06:18 PM
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Re: Razz - paired cards
[ QUOTE ]
Lets assume that the villain has precisely a 3-card bike hand (three unduplicated cards 5 or lower) on third street. On 4th street the villain catches a 5. What is the probability that the 5 paired the villain assuming that there are 12 bike cards unaccounted for remaining in the deck including the three remaining 5's? Please include the formula, that would help me bunches because I think i calculated this incorrectly. Thanks in advance! [/ QUOTE ] Simplify. Villain shows an ace on third street. A. The remaining 12 cards are 3,3,3, and 3 of each. Then it's 50/50 the five paired him. B. The remaining 12 cards are 3,4,3, and 2. Then it's 27 ways to pair and 26 ways to not pair. 51% the five paired him.
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