1. (Day 14 was review. Day 15 was the midterm.) Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Return and review midterm 2.Bayes’ rule 3.Gold vs. Farha 4.Doyle vs. Elezra 5.Variance, CLT, and prop bets Computer Project B will be due on Wednesday, Nov 28, at 5:00pm.   u 

2. 2. Bayes’ Rule Say B 1, B 2, B n are disjoint events and that exactly one of them must occur. Suppose you want P(B 1 | A), but you only know P(A | B 1 ), P(A | B 2 ), etc. Suppose you also know P(B 1 ), P(B 2 ), …, P(B n ). Bayes’ Rule: If B 1, …, B n are disjoint events with P(B 1 or … or B n ) = 1, then P(B i | A) = P(A | B i ) * P(B i ) ÷ [ ∑P(A | B j )P(B j )]. Why? Recall: P(X | Y) = P(X & Y) ÷ P(Y). So P(X & Y) = P(X | Y) * P(Y). P(B 1 | A) = P(A & B 1 ) ÷ P(A) = P(A & B 1 ) ÷ [ P(A & B 1 ) + P(A & B 2 ) + … + P(A & B n ) ] = P(A | B 1 ) * P(B 1 ) ÷ [ P(A | B 1 )P(B 1 ) + P(A | B 2 )P(B 2 ) + … + P(A & B n )P(B n ) ].

3. 3. Gold vs. Farha Gold: 10 u 7  Farha: Q u Q  Flop: 9 u 8  7  a)Who really is the favorite (ignoring all other players’ cards)? Gold’s outs: J, 6, 10, 7. (4 + 4 + 3 + 2 = 13 outs, 32 non-outs) P(Gold wins) = P(Out Out or Jx [x ≠ 10] or 6x or 10y [y≠Q,9,8] or 7z [z≠Q]) = [choose(13,2) + 4*28 + 4*32 + 3*24 + 2*30] ÷ choose(45,2) = 450 ÷ 990 = 45.45%. b)What would you guess Gold had? Say he’d do that 50% of the time with a draw, 100% of the time with an overpair, and 90% of the time with two pairs. (and that’s it) Using Bayes’ rule, P(Gold has a DRAW | Gold raises ALL-IN) = P(all-in | draw) * P(draw) ----------------------------------------------------------------------------------------------- [P(all-in | draw) P(draw)] + [P(all-in | overpair) P(overpair)] + [P(all-in | 2pairs) P(2 pairs)] = 50% * P(draw) ------------------------------------------------------------------- [50% * P(draw)] + [100% * P(overpair)] + [90% * P(2 pairs)]

4. 4) Doyle vs. Elezra: (simplified) Elezra re-raises you all-in before the flop. Suppose you think he would certainly do that whenever she had AA, KK, or QQ, and would do that 50% of the time with AK or AQ. Nothing else. Given only this (not Doyle’s cards), what’s P(he has AK)? Given nothing, P(AK) = 16/C(52,2) = 16/1326. P(KK) = C(4,2)/C(52,2) = 6/1326. Using Bayes’ rule, P(AK | all-in) = P(all-in | AK) * P(AK) ------------------------------------------------------------------------------- P(all-in|AK)P(AK) + P(all-in|AA)P(AA) + P(all-in|AA)P(AA) + … = 50% * 16/1326 ---------------------------------------------------------------------------------------------- [50% * 16/1326] + [100% * 6/1326] + [1*6/1326] + [1*6/1326] + [50% * 16/1326] (AK)(AA) (KK) (QQ) (AQ) = 23.5%. Compare with 16/1326 ~ 1.21%.

5. 5. Variance, CLT, and prop bets. Central Limit Theorem (CLT): if X 1, X 2 …, X n are iid with mean µ& SD  then (X - µ) ÷ (  /√n) ---> Standard Normal. (mean 0, SD 1). In other words, X has mean µ and SD of  ÷√n. As n increases, (  ÷ √n) decreases. So, the more independent trials, the smaller the SD (and variance) of X. i.e. additional bets decrease the variance of your average. If X and Y are independent, then E(X+Y) = E(X) + E(Y), and V(X+Y) = V(X) + V(Y). Let X = your profit on wager #1, Y = profit on wager #2. If the two wagers are independent, then V(total profit) = V(X) + V(Y) > V(X). So, additional bets increase the variance of your total!