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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day:

Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: Straight draws. HW2 clarification. Greenstein vs. Farha AA vs. KK. Odds ratios. Random variables. Read through chapter 2! No class Mon Jan 20, MLK day.   u    u .

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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day:

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  1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: Straight draws. HW2 clarification. Greenstein vs. Farha AA vs. KK. Odds ratios. Random variables. Read through chapter 2! No class Mon Jan 20, MLK day.  u   u 

  2. 1. More counting problems: straight draw example World Series of Poker Main Event 2005, Day 1, from cardplayer.com: 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? 

  3. 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 about 1 in 15.

  4. A: For J-T, you need: KQx, Q9x, 98x, AQ8, K97, KKQ, KQQ, QQ9, Q99, 998, or 988. So the probability is: 3 x [4 x 4 x 34] + 2 x [4 x 4 x 4] + 6 x [C(4,2) x 4] C(50,3) = 9.71%, or about 1 in 10.

  5. 2. Hw2: 2.6, 2.9, 2.10, 2.18, 3.2, 3.6. Read up through ch.3. On 2.18, let me explain a bit what I mean in this problem. Suppose the board is K Q 10u 6 3u and you have A J. Then you have the nuts, and your 5-card hand is an ace-high straight (AKQJT).. Suppose instead that the board is K Q 7u 6 3u. Then if you have 5 4u you have the nuts, and your 5-card hand is a 7-high straight (76543). A 7-high straight is worse than an ace-high straight, in the sense that it is outranked by an ace-high straight in the hand rankings. Now, if the board is K 10 7u 3 2u, and you have Ku K, then you have the nuts, and your 5-card hand is three kings: KKKT7, which is a worse hand than a 7-high straight. This question is asking: how far can you go with this? Can you think of a different board, so that it is possible that the nuts is a 5-card hand even worse than KKKT7? What is the worst possible?

  6. 3. Greenstein and Farha. 4. Odds ratios: Odds ratio of A = P(A)/P(Ac) Odds against A = Odds ratio of Ac = P(Ac)/P(A). Ex: (from Phil Gordon’s Little Blue Book, p189) Day 3 of the 2001 WSOP, $10,000 No-limit holdem championship. 613 players entered. Now 13 players left, at 2 tables. Phil Gordon’s table has 5 other players. Blinds are 3,000/6,000 + 1,000 antes. Matusow has 400,000; Helmuth has 600,000; Gordon 620,000. (the 3 other players have 100,000; 305,000; 193,000). Matusow raises to 20,000. Next player folds. Gordon’s next, in the cutoff seat with K K and re-raises to 100,000. Next player folds. Helmuth goes all-in. Big blind folds. Matusow folds. Gordon’s decision…. Fold! Odds against Gordon winning, if he called and Helmuth had AA?

  7. What were the odds against Gordon winning, if he called and Helmuth had AA? P(exactly one K, and no aces) = 2 x C(44,4) / C(48,5) ~ 15.9%. P(two Kings on the board) = C(46,3) / C(48,5) ~ 0.9%. [also some chance of a straight, or a flush…] Using www.cardplayer.com/poker_odds/texas_holdem, P(Gordon wins) is about 18%, so the odds against this are: P(Ac)/P(A) = 82% / 18% = 4.6 (or “4.6 to 1” or “4.6:1”)

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