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This session explores the application of probability in poker, focusing on high-stakes scenarios like the match between Farha and Greenstein, with analyses of specific hand probabilities (e.g., AA vs. KK) and the implications of continuous random variables. Delve into concepts such as density functions, the uniform and normal distributions, and the law of large numbers (LLN) alongside the central limit theorem (CLT). Understand how to strategically approach betting in poker by considering the likelihood of opponents' hands and the overall statistical dynamics at play.
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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
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%.
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.
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%.
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. •
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.
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.
Truth: -49 to 51, exp. value = 1.0 Estimated as X +/- 1.96 s/√n = .95 +/- 0.28
* 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.
5) High Stakes Poker, Gold vs. Helmuth … to be continued
