1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.E(cards til 2 nd king). 2.Negative binomial. 3.Rainbow flops examples, binomial and negative binomial. 4.Gold/Farha and Bayes rule. 5.P(2 pairs). 6.Project B example Zelda. Midterm is Feb 21, in class. 50 min. Open book plus one page of notes, double sided. Bring a calculator!   u 

2. 1. E(cards til 2 nd king). Z = the number of cards til the 2nd king. What is E(Z)? Let X 1 = number of non-king cards before 1 st king. Let X 2 = number of non-kings after 1 st king til 2 nd king. Let X 3 = number of non-kings after 2 nd king til 3 rd king. Let X 4 = number of non-kings after 3 rd king til 4 th king. Let X 5 = number of non-kings after 4 th king til the end of the deck. Clearly, X 1 + X 2 + X 3 + X 4 + X 5 + 4 = 52. By symmetry, E(X 1 ) = E(X 2 ) = E(X 3 ) = E(X 4 ) = E(X 5 ). Therefore, E(X 1 ) = E(X 2 ) = 48/5. Z = X 1 + X 2 + 2, so E(Z) = E(X 1 ) + E(X 2 ) + 2 = 48/5 + 48/5 + 2 = 21.2.

3. 2. Negative Binomial Random Variables, ch 5.4. Recall: if each trial is independent, and each time the probability of an occurrence is p, and X = # of trials until the first occurrence, then: X is Geometric (p), P(X = k) = p 1 q k - 1, µ = 1/p,  = (√q) ÷ p. Suppose now X = # of trials until the rth occurrence. Then X = negative binomial (r,p). e.g. the number of hands you have to play til you’ve gotten r=3 pocket pairs. Now X could be 3, 4, 5, …, up to ∞. pmf: P(X = k) = choose(k-1, r-1) p r q k - r, for k = r, r+1, …. e.g. say r=3 & k=7: P(X = 7) = choose(6,2) p 3 q 4. Why? Out of the first 6 hands, there must be exactly r-1 = 2 pairs. Then pair on 7th. P(exactly 2 pairs on first 6 hands) = choose(6,2) p 2 q 4. P(pair on 7th) = p. If X is negative binomial (r,p), then µ = r/p, and  = (√rq) ÷ p. e.g. Suppose X = the number of hands til your 12th pocket pair. P(X = 100)? E(X)?  ? X = Neg. binomial (12, 5.88%). P(X = 100) = choose(99,11) p 12 q 88 = choose(99,11) * 0.0588 ^ 12 * 0.9412 ^ 88 = 0.104%. E(X) = r/p = 12/0.0588 ~ 204.  = sqrt(12*0.9412) / 0.0588 = 57.2. So, you’d typically expect it to take 204 hands til your 12th pair, +/- around 57.2 hands.

4. 3. Rainbow flops. P(Rainbow flop) = choose(4,3) * 13 * 13 * 13 ÷ choose(52,3) choices for the 3 suits numbers on the 3 cards possible flops ~ 39.76%. Q: Out of 100 hands, what is the expected number of rainbow flops? +/- what? X = Binomial (n,p), with n = 100, p = 39.76%, q = 60.24%. E(X) = np = 100 * 0.3976 = 39.76 SD(X) = √( npq) = sqrt(23.95) = 4.89. So, expect around 39.76 +/- 4.89 rainbow flops, out of 100 hands.

5. Rainbow flops, continued. P(Rainbow flop) ~ 39.76%. Q: Let X = the number of hands til your 4 th rainbow flop. What is P(X = 10)? What is E(X)? What is SD(X)? X = negative binomial (r,p), with r = 4, p = 39.76%, q = 60.24%. P(X = k) = choose(k-1, r-1) p r q k-r. Here k = 10. P(X = 10) = choose(9,3) 39.76% 4 60.24% 6 = 10.03%. µ = E(X) = r/p = 4 ÷ 0.3976 = 10.06 hands.  = SD(X) = (√rq) / p = sqrt(4*0.6024) / 0.3976 = 3.90 hands. So, you expect it typically to take around 10.06 +/- 3.90 hands til your 4th rainbow flop. Gold/Farha.