#11
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Re: Razz - paired cards
Make that, THREE aces are known, with one folded and two raising.
What bike hands do we fold here, having completed in early position on 3rd? |
#12
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Re: Razz - paired cards
[ QUOTE ]
Does it make sense to call with bike cards WIHTOUT the Ace, if we see 2 [sic - actually 3] Aces? [/ QUOTE ] No. But assuming that most razz players will act sensibly is a bad assumption, and will usually get you into trouble. Given this particular villain, any 3-card bike or 6 (no matter how rough) would be worth an UTG completion and then a cap. |
#13
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Re: Razz - paired cards
Note that 23, 24 and 25 are not listed above. There may be others that are not listed also. I think I got it right in my reply to that thread, and I think I also showed an quick method that gets you pretty close. So the answer above is actually not quite right. I think it was really more like 46%
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#14
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Re: Razz - paired cards
Oh, I see, we left the 2 out on purpose. Sorry.
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#15
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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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