Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Deal-making and expected value 2.Odds ratios, revisited 3.Variance and SD 4.Booth vs. Ivey Hand in hw2!!!   u 

1. Deal-making. (Expected value, game theory) Game-theory: For a symmetric-game tournament, the probability of winning is approx. optimized by the myopic rule (in each hand, maximize your expected number of chips), and P(you win) = your proportion of chips. For a fair deal, the amount you win = the expected value of the amount you will win.

For instance, suppose a tournament is winner-take-all, for $8600. With 6 players left, you have 1/4 of the chips left. An EVEN SPLIT would give you $8600 ÷ 6 = $1433. A PROPORTIONAL SPLIT would give you $8600 x (your fraction of chips) = $8600 x (1/4) = $2150. A FAIR DEAL would give you the expected value of the amount you will win = $8600 x P(you get 1st place) = $2150. But suppose the tournament is not winner-take-all, but pays $3800 for 1st, $2000 for 2nd, $1200 for 3rd, $700 for 4th, $500 for 5th, $400 for 6th. Then a FAIR DEAL would give you $3800 x P(1st place) + $2000 x P(2nd) +$1200 x P(3rd)+$700 x P(4th) +$500 x P(5th) +$400 x P(6th). Hard to determine these probabilities. But, P(1st) = 25%, and you might roughly estimate the others as P(2nd) ~ 20%, P(3rd) ~ 20%, P(4th) ~ 15%, P(5th) ~10%, P(6th) ~ 10%, and get $3800 x 25% + $2000 x 25% +$1200 x 20% + $700 x 15% + $500 x 10% +$400 x 5% = $1865. If you have 40% of the chips in play, then: EVEN SPLIT = $1433. PROPORTIONAL SPLIT = $3440. FAIR DEAL ~ $2500!

Another example. Before the Wasicka/Binger/Gold hand, Gold had 60M, Wasicka 18M, Binger 11M. Payouts: 1st place $12M, 2nd place $6.1M, 3rd place $4.1M. Proportional split: of the total prize pool left, you get your proportion of chips in play. e.g. $22.2M left, so Gold gets 60M/(60M+18M+11M) x $22.2M ~ $15.0M. A fair deal would give you P(you get 1st place) x $12M + P(you get 2nd place) x $6.1M + P(3rd pl.) x $4.1M. *Even split: Gold $7.4M, Wasicka $7.4M, Binger $7.4M. *Proportional split: Gold $15.0M, Wasicka $4.5M, Binger $2.7M. *Fair split: Gold $10M, Wasicka $6.5M, Binger $5.7M. *End result: Gold $12M, Wasicka $6.1M, Binger $4.1M.

2. Odds ratios, revisited: Odds ratio of A = P(A)/P(A c ) Odds against A = Odds ratio of A c = P(A c )/P(A). An advantage of probability over odds ratios is the mult. rule: P(A & B) = P(A) x P(B|A), but can’t multiply odds ratios. Example: High Stakes Poker, 1/25/07, on GSN: Blinds $300/$600 + $1200 “straddle” + eight $100 antes = $2900. Gabe Kaplan: K  K , raises to $4200. Chen folds, Doyle folds, Matusow calls with A  9 . Gold folds, Alaie folds, Sheikhan folds, Ramdin folds. Flop: T  8  J . Kaplan checks, Matusow bets $8000, Kaplan calls. Turn: A . (both players check) River: 6 . Kaplan bets $16000, Matusow folds! (pot $43,300)

Gabe Kaplan: K  K . Mike Matusow: A  9 . Flop: T  8  J . To win, Matusow needs a 7, Q, or A. (But not Q9 or QA, and they tie with 79…. And not two more  s. ) And, Matusow can also win with two more  s. So, part of figuring out P(Matusow winning) would be to get: Given their cards and the flop, what’s P(Matusow makes a flush)? P(  on turn AND  on river) = P(  turn) x P(  river |  turn) = 9/45 x 8/44 = 3.6%. Odds against it? 96.3% ÷ 3.6% = : 1. Odds against  on turn = 4 : 1. Odds against  on river |  on turn = 4.5 : 1. Multiplying, you’d get (4 x 4.5) = 18. Moral: you can’t multiply odds ratios!

3. Variance and SD For a discrete random variable X, E(X) = ∑ f(y) x y = µ. For instance, if you roll a die, and you lose $100 (i.e. win $-100) if you roll a 1 or 2, you win $50 if you roll 3-5, and you win $20 if you roll a 6. If X = your winnings, then E(X) = $-100 x 2/6 + $50 x 3/6 + $20 x 1/6 = $-5. If you roll three dice, and you pay $10 to play, but you get back $10 for each 6 you roll, then if X = your profit after playing this game once, then E(X) = $-10 x P(zero 6s) + $0 x P(one 6) + $10 x P(two 6s) + $20 x P(three 6s) = $-10 x 5/6 x 5/6 x 5/ $10 x 3 x 1/6 x 1/6 x 5/6 + $20 x 1/6 x 1/6 x 1/6 = $-5. But the two games seem different: the first has higher variance,  2 :  2 =V(X) = E(X-µ) 2 = ∑ f(y) x (y-µ) 2 = E(X 2 ) - µ 2.  = standard deviation = √  2. Indicates how far a typical observation is from µ. For instance, in game 1, E(X 2 ) = ($-100) 2 x 2/6 + ($50) 2 x 3/6 + ($20) 2 x 1/6 = $2150. µ 2 = (-$5) 2 = $25, so  2 = $ $25 = $2125, and  = √$2125 ~ $46. Game 2: E(X 2 ) = ($-10) 2 x (5/6) ($10) 2 x 3 x 1/6 x 1/6 x 5/6 + ($20) 2 x (1/6) 3 ~ $66.7. µ 2 = (-$5) 2 = $25, so  2 = $ $25 = $41.7, and  = √$41.7 ~ $6.50.