Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: E(cards til 2 nd king). Negative

1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: • E(cards til 2nd king). • Negative binomial. • Rainbow flops examples, binomial and negative binomial. • Gold/Farha and Bayes rule. • P(2 pairs). • 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   u 

2. 1. E(cards til 2nd king). Z = the number of cards til the 2nd king. What is E(Z)? Let X1 = number of non-king cards before 1st king. Let X2 = number of non-kings after 1st king til 2nd king. Let X3 = number of non-kings after 2nd king til 3rd king. Let X4 = number of non-kings after 3rd king til 4th king. Let X5 = number of non-kings after 4th king til the end of the deck. Clearly, X1 + X2 + X3 + X4 + X5 + 4 = 52. By symmetry, E(X1) = E(X2) = E(X3) = E(X4) = E(X5). Therefore, E(X1) = E(X2) = 48/5. Z = X1 + X2 + 2, so E(Z) = E(X1) + E(X2) + 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) = p1 qk - 1, µ = 1/p, s= (√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) pr qk - r, for k = r, r+1, …. e.g. say r=3 & k=7: P(X = 7) = choose(6,2) p3 q4. 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) p2 q4. P(pair on 7th) = p. If X is negative binomial (r,p), then µ = r/p, and s= (√rq) ÷ p. e.g. Suppose X = the number of hands til your 12th pocket pair. P(X = 100)? E(X)? s? X = Neg. binomial (12, 5.88%). P(X = 100) = choose(99,11) p12 q88 = choose(99,11) * 0.0588 ^ 12 * 0.9412 ^ 88 = 0.104%. E(X) = r/p = 12/0.0588 ~ 204. s= 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 4th 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) pr qk-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. s= 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.