Stat 35b: Introduction to Probability with Applications to Poker Outline for the day, Tue 3/13/12: 1.Fred Savage hand. 2.Random walks, continued, for 7.14.

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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day, Tue 3/13/12: 1.Fred Savage hand. 2.Random walks, continued, for Pdfs and cdfs, for Optimal play, ch 6.3, pp Project B tournament. Homework 4, Stat 35, due Thu March 15, 12:30pm: 6.12, 7.2, 7.8, I will discuss the answers in class Thursday. Hw4 probably won’t get graded til Tuesday, so I will just give them back in the final exam. Final Exam: Tuesday, March 20, 11:30 AM - 2:30 PM, here in Royce 156. Thur 3/15 will be review and examples. Be sure to submit your reviews of the course online via my.ucla.edu.

1. Fred Savage hand. 2. Random walks, continued, for Theorem 7.6.7, p152, describes another simplified scenario. Suppose you either double each hand you play, or go to zero, each with probability 1/2. Again, P(win a tournament) is prop. to your number of chips. Proof. p 0 = 0, and p 1 = 1/2 p 2 = 1/2 p 2, so again, p 2 = 2 p 1. We have shown that, for j = 0, 1, and 2, p j = j p 1. (induction:) Suppose that, for j ≤ m, p j = j p 1. We will show that p 2m = (2m) p 1. Therefore, p j = j p 1 for all j = 2 k. That is, P(win the tournament) is prop. to number of chips. This time, p m = 1/2 p 0 + 1/2 p 2m. If p j = j p 1 for j ≤ m, then we have mp 1 = 0 + 1/2 p 2m, so p 2m = 2mp 1. Done refers to Theorem 7.6.8, p152. You have k of the n chips in play. Each hand, you gain 1 with prob. p, or lose 1 with prob. q=1-p. Suppose 0<p <1 and p≠0.5. Let r = q/p. Then P(you win the tournament) = (1-r k )/(1-r n ). The proof is again by induction, and is similar to the proof we did last class of Theorem Notice that if k = 0, then (1-r k )/(1-r n ) = 0. If k = n, then (1-r k )/(1-r n ) = 1.

3. Pdfs and cdfs, for Suppose X has the cdf F(c) = P(X ≤ c) = 1 - exp(-4c), for c ≥ 0, and F(c) = 0 for c < 0. Then to find the pdf, f(c), take the derivative of F(c). f(c) = F’(c) = 4exp(-4c), for c ≥ 0, and f(c) = 0 for c<0. Thus, X is exponential with = 4. Now, what is E(X)? E(X) = ∫ -∞ ∞ c f(c) dc = ∫ 0 ∞ c {4exp(-4c)} dc = ¼, after integrating by parts, or just by remembering that if X is exponential with = 4, then E(X) = ¼. For 6.12, the key things to remember are 1. f(c) = F'(c). 2. For an exponential random variable with mean, F(c) = 1 – exp(-1/ c). 3. For any Z, E(Z) = ∫ c f(c) dc. And, V(Z) = E(Z 2 ) – [E(Z)] 2, where E(Z 2 ) = ∫ c 2 f(c) dc. 4. E(X) for exponential = 1/. 5. E(X 2 ) for exponential = 2/ 2. For 7.14, the key things to remember are 1.If p≠0.5, then by Theorem 7.6.8, P(win tournament) = (1-r k )/(1-r n ), where r = q/p. 2.Let x = r 2. If –x 3 + 2x -1 = 0, that means (x-1)(-x 2 - x + 1) = 0. There are 3 solutions to this. One is x = 1. The others occur when x 2 + x - 1 = 0, so x = -1 +/- sqrt(1+4)/2 = or So, x = , 0.118, or 1. Two of these possibilities can be ruled out. Remember that p≠0.5.

4. Optimal play, ch 6.3, pp Project B tournament.

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