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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: Farha vs. Greenstein, AA/KK WPT Example Continuous Random Variables, Density, Uniform, Normal LLN & CLT. Gold vs. Helmuth NOTE: No class Tuesday Nov 6!!!.   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: Farha vs. Greenstein, AA/KK WPT Example Continuous Random Variables, Density, Uniform, Normal LLN & CLT. Gold vs. Helmuth NOTE: No class Tuesday Nov 6!!!  u   u 

  2. Farha/Greenstein: P(SOMEONE has AA, given you have KK)? Out of your 8 opponents? Note that given that you have KK, P(player 2 has AA & player 3 has AA) = P(player 2 has AA) x P(player 3 has AA | player 2 has AA) = choose(4,2) / choose(50,2) x 1/choose(48,2) = 0.0000043, or 1 in 230,000. So, very little overlap! Given you have KK, P(someone has AA) = P(player2 has AA or player3 has AA or … or pl.9 has AA) ~ P(player2 has AA) + P(player3 has AA) + … + P(player9 has AA) = 8 x choose(4,2) / choose(50,2) = 3.9%, or 1 in 26. ----------- What is exactly P(SOMEONE has an Ace | you have KK)? (8 opponents) (or more than one ace) P(SOMEONE has an Ace) = 100% - P(nobody has an Ace). P(nobody has an Ace) = P(pl2 doesn’t have one & pl.3 doesn’t & … & pl.9 doesn’t) = P(pl.2 doesn’t) x P(pl.3 doesn’t | pl.2 doesn’t) x … x P(pl.9 doesn’t | 2-8 don’t) = choose(46,2)/choose(50,2) x choose(44,2)/choose(48,2) x … x ch(32,2)/ch(36,2) [ = choose(46,16)/choose(50,16) ] = 20.1%. So P(SOMEONE has an Ace) = 79.9%.

  3. 2) 11/4/05, Travel Channel, World Poker Tour, $1 million Bay 101 Shooting Star. 4 players left, blinds $20,000 / $40,000, with $5,000 antes. Avg stack = $1.1 mil. 1st to act: Danny Nguyen, A 7. All in for $545,000. Next to act: Shandor Szentkuti, A K. Call. Others (Gus Hansen & Jay Martens) fold. (66% - 29%). Flop: 5 K 5 . (tv 99.5%; cardplayer.com: 99.4% - 0.6%). P(tie) = P(55 or A5 or 5A) = (2/45 x 1/44) + (2/45 x 2/44) + (2/45 x 2/44) = 0.505%. 1 in 198. P(Nguyen wins) = P(77) = 3/45 x 2/44 = 0.30%. 1 in 330. [Note: tv said “odds of running 7’s on the turn and river are 274:1.” Given Hansen/Martens’ cards, 3/41 x 2/40 = 1 in 273.3). ] Turn: 7. River: 7! * Szentkuti was eliminated next hand, in 4th place. Nguyen went on to win it all.

  4. 11/4/05, Travel Channel, World Poker Tour, $1 million Bay 101 Shooting Star. 3 players left, blinds $20,000 / $40,000, with $5,000 antes. Avg stack = $1.4 mil. (pot = $75,000) 1st to act: Gus Hansen, K 9. Raises to $110,000. (pot = $185,000) Small blind: Dr. Jay Martens, A Q. Re-raises to $310,000. (pot = $475,000) Big blind: Danny Nguyen, 7 3. Folds. Hansen calls. (tv: 63%-36%.) (pot = $675,000) Flop: 4 9 6. (tv: 77%-23%; cardplayer.com: 77.9%-22.1%) P(no A nor Q on next 2 cards) = 37/43 x 36/42 = 73.8% P(AK or A9 or QK or Q9) = (9+6+9+6) ÷ (43 choose 2) = 3.3% So P(Hansen wins) = 73.8% + 3.3% = 77.1%. P(Martens wins) = 22.9%.

  5. 1st to act: Gus Hansen, K 9. Raises to $110,000. (pot = $185,000) Small blind: Dr. Jay Martens, A Q. Re-raises to $310,000. (pot = $475,000) Hansen calls. (pot = $675,000) Flop: 4 9 6. P(Hansen wins) = 77.1%. P(Martens wins) = 22.9%. Martens checks. Hansen all-in for $800,000 more. (pot = $1,475,000) Martens calls. (pot = $2,275,000) Vince Van Patten: “The doctor making the wrong move at this point. He still can get lucky of course.” Was it the wrong move? His prob. of winning should be ≥ $800,000 ÷ $2,275,000 = 35.2%. Here it was 22.9%. So, if Martens knew what cards Hansen had, he’d be making the wrong move. But given all the possibilities, it seems very reasonable to assume he had a 35.2% chance to win. (Harrington: 10%!) River: 2. * Turn: A! • * Hansen was eliminated 2 hands later, in 3rd place. Martens then lost to Nguyen. •   

  6. 3) Continuous Random Variables, Density, Uniform, Normal Density (or pdf = Probability Density Function) f(y): ∫B f(y) dy = P(X in B). Expected value (µ) = ∫ y f(y) dy. (= ∑ y P(y) for discrete X.) Example 1: Uniform (0,1). f(y) = 1, for y in (0,1). µ = 0.5. s = 0.29. P(X is between 0.4 and 0.6) = ∫.4 .6 f(y) dy = ∫.4 .6 1 dy = 0.2. Example 2: Normal. mean = µ, SD = s, 68% of the values are within 1 SD of µ 95% are within 2 SDs of µ Example 3: Standard Normal. Normal with µ = 0, s = 1.

  7. 95% between -1.96 and 1.96

  8. 4) Law of Large Numbers, CLT Sample mean (X) = ∑Xi / n iid: independent and identically distributed. Suppose X1, X2 , etc. are iid with expected value µ and sd s , LAW OF LARGE NUMBERS (LLN): X ---> µ . CENTRAL LIMIT THEOREM (CLT): (X - µ) ÷ (s/√n) ---> Standard Normal. Useful for tracking results. Note: LLN does not mean that short-term luck will change. Rather, that short-term results will eventually become negligible.

  9. 95% between -1.96 and 1.96

  10. Truth: -49 to 51, exp. value = 1.0

  11. Truth: -49 to 51, exp. value = 1.0 Estimated as X +/- 1.96 s/√n = .95 +/- 0.28

  12. * Poker has high standard deviation. Important to keep track of results. * Don’t just track ∑Xi. Track X +/- 1.96 s/√n . Make sure it’s converging to something positive.

  13. 5) High Stakes Poker, Gold vs. Helmuth … to be continued

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