1 / 3

World Series of Poker Main Event 2005, Day 1, from cardplayer:

World Series of Poker Main Event 2005, Day 1, from cardplayer.com: *Date / Time:* 2005-07-08 01:23:00 With the board showing 10  9  5  Q  , Chris "Jesus" Ferguson moves all in. Kalee Tan calls. Ferguson shows Q-Q for a set of queens,

maik
Download Presentation

World Series of Poker Main Event 2005, Day 1, from cardplayer:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. World Series of Poker Main Event 2005, Day 1, from cardplayer.com: *Date / Time:* 2005-07-08 01:23:00 With the board showing 10 9 5 Q, Chris "Jesus" Ferguson moves all in. Kalee Tan calls. Ferguson shows Q-Q for a set of queens, and Tan flips up J-8 for a queen high straight. Ferguson needs the board to pair in order to stay alive. The river is the 8, no help to Ferguson, and he is eliminated on Day 1. Kalee Tan drags the pot with uncontrollably shaky hands as Ferguson heads to the rail. Q: What is the probability of flopping an open-end straight draw, given you have J-8? What about J-9 or J-T? 

  2. Q: What is the probability of flopping an open-end straight draw, given you have J-8? What about J-9 or J-T? A: For J-8, you need the flop to be KQT or T9x or 976 or 765. Consider the case where x is T or 9 separately (x ≠ Q or 7!). So the probability is: P( KQT or TT9 or T99 or T9x or 976 or 765 ) = 4 x 4 x 4 + C(4,2) x 4 + C(4,2) x 4 + 4 x 4 x 34 + 4x4x4 + 4x4x4 C(50,3) = 4.0%, or 1 in 25. A: For J-9, you need KT7 or T8x or QTx or 876, so it’s P( KT7 or TT8 or T88 or QQT or QTT or T8x or QTx or 876) = 64 + [C(4,2) x 4] x 4 + [4 x 4 x 34] x 2 + 64 C(50,3) = 6.7%, or 1 in 15.

  3. A: For J-T, you need: KQx, Q9x, 98x, AQ8, K97, Q87, KKQ, KQQ, QQ9, Q99, 998, or 988. So the probability is: 3 x [4 x 4 x 34] + 3 x [4 x 4 x 4] + 6 x [C(4,2) x 4] C(50,3) = 10.0%, or 1 in 10.

More Related