60 likes | 267 Views
Stat 35b: Introduction to Probability with Applications to Poker Outline for the day Review list. Gold, Farha, and Bayes’s rule. P(flop 2 pairs). P(suited king). Midterm is this Friday Feb 21. Homework 3 is due Feb 28. Project B is due Mar 8, 8pm, by email to frederic@stat.ucla.edu .
E N D
Stat 35b: Introduction to Probability with Applications to Poker Outline for the day • Review list. • Gold, Farha, and Bayes’s rule. • P(flop 2 pairs). • P(suited king). • Midterm is this Friday Feb 21. • Homework 3 is due Feb 28. • Project B is due Mar 8, 8pm, by email to frederic@stat.ucla.edu. • Same teams as project A.
1. Review List • Basic principles of counting. • Axioms of probability, and addition rule. • Permutations & combinations. • Conditional probability. • Independence. • Multiplication rules. P(AB) = P(A) P(B|A) [= P(A)P(B) if ind.] • Odds ratios. • Random variables (RVs). • Discrete RVs and probability mass function (pmf). • Expected value. • Pot odds calculations. • Luck, skill, and deal-making. • Variance and SD. • Bernoulli RV. [0-1. µ = p, s= √(pq). ] • Binomial RV. [# of successes, out of n tries. µ = np, s= √(npq).] • Geometric RV. [# of tries til 1st success. µ = 1/p, s= (√q) / p. ] • Negative binomial RV. [# of tries til rth success. µ = r/p, s= (√rq) / p. ] • E(X+Y), V(X+Y) (ch. 7.1). • Bayes’s rule (ch. 3.4). • We have basically done all of ch. 1-5 except 4.6, 4.7, and 5.5.
2. Gold vs. Farha. Gold: 10u 7 Farha: Qu Q Flop: 9u 8 7 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%. 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)]
3. P(flop two pairs). If you’re sure to be all-in next hand,what isP(you will flop two pairs)? This is a tricky one. Don’t double-count (4 4 9 9 Q) and (9 9 4 4 Q)! There are choose(13,2) possibilities for the NUMBERS of the two pairs. For each such choice (such as 4 & 9), there are choose(4,2) choices for the suits of the lower pair, and same for the suits of the higher pair. So, choose(13,2) * choose(4,2) * choose(4,2) different possibilities for the two pairs. For each such choice, there are 44 [52 - 8 = 44] different possibilities for your fifth card, so that it’s not a full house but simply two pairs. So, P(flop two pairs) = choose(13,2) * choose(4,2) * choose(4,2) * 44 / choose(52,5) ~ 4.75%, or 1 in 21.
4. What is the probability that you will be dealt a king and another card of the same suit as the king? 4 * 12 / C(52,2) = 3.62%. The typical number of hands until this occurs is ... 1/.0362 ~ 27.6. (√96.38%) / 3.62% ~ 27.1. So the answer is 27.6 +/- 27.1.