Stat 35b: Introduction to Probability with Applications to Poker Outline for the day, Thur 3/8/12: 0.HAND IN HW3 again! 1.E(X+Y) example corrected. 2.Random walks, ch 7.6. Reflection principle, ballot theorem, and avoiding 0. 3.Hellmuth and Farha. 4.Chip proportions and induction. Read ch7. Also, read pages on optimal play, though it’s not on the final. Homework 4, Stat 35, due Thu March 15, 12:30pm: 6.12, 7.2, 7.8, Project B is due this Fri, March 9, 8pm, by . Many questions on the final will be similar to midterm questions, so go over the midterm questions you missed. Ignore the extra-credit on the midterm, though.

0. HAND IN HW3 again. 1. E(X+Y+Z) example again. Deal til the 2nd king. X = # of cards til the 2nd king. E(X)? Consider any ordering of the deck. e.g. 7 ,Q ,J u,2 , 9 ,K ,3 ,9 ,7 ,A , K ,… For any such ordering, there is another with the cards before the first king exchanged with the cards between the 1st king and the 2nd king. Similarly, for the cards between the 2nd king and 3rd king, and for the cards between the 3rd king and 4th king, and between the 4th king and the end of the deck. So, P(# of cards before 1st king = 5) = P(# of cards bet. 1st and 2nd king = 5), etc. Therefore if X 1 = the # of cards before the 1st king, and X 1 = # of cards between 1st and 2nd king, then E(X 1 ) = ∑ k k P(X 1 =k) = ∑ k k P(X 2 =k) = E(X 2 ). And, X 1 + X 2 + X 3 + X 4 + X 5 = 48, so EX 1 + EX 2 + EX 3 + EX 4 + EX 5 = 48, so E(X 1 ) = E(X 2 ) = E(X 3 ) = E(X 4 ) = E(X 5 ) = 48/5 = 9.6 We want E(X 1 + X 2 + 2) = E(X 1 ) + E(X 2 ) + 2 = 21.2.

2. Random walks, ch 7.6. Suppose that X 1, X 2, …, are iid, and S k = X 0 + X 1 + … + X k for k = 0, 1, 2, …. The totals {S 0, S 1, S 2, …} form a random walk. The classical (simple) case is when each X i is 1 or -1 with probability ½ each. Reflection principle: The number of paths from (0,X 0 ) to (n,y) that touch the x-axis = the number of paths from (0,-X 0 ) to (n,y), for any n,y, and X 0 > 0. Ballot theorem: In n = a+b hands, if player A won a hands and B won b hands, where a>b, and if the hands are aired in random order, P(A won more hands than B throughout the telecast) = (a-b)/n. [In an election, if candidate X gets x votes, and candidate Y gets y votes, where x > y, then the probability that X always leads Y throughout the counting is (x-y) / (x+y).] For a simple random walk, P(S 1 ≠ 0, S 2 ≠ 0, …, S n ≠ 0) = P(S n = 0), for any even n.

Reflection principle: The number of paths from (0,X 0 ) to (n,y) that touch the x-axis = the number of paths from (0,-X 0 ) to (n,y), for any n,y, and X 0 > 0. For each path from (0,X 0 ) to (n,y) that touches the x-axis, you can reflect the first part til it touches the x-axis, to find a path from (0,-X 0 ) to (n,y), and vice versa. Total number of paths from (0,-X 0 ) to (n,y) is easy to count: it’s just C(n,a), where you go up a times and down b times [i.e. a-b = y - (-X 0 ) = y + X 0. a+b=n, so b = n-a, 2a-n=y+x, a=(n+y+x)/2].

Ballot theorem: In n = a+b hands, if player A won a hands and B won b hands, where a>b, and if the hands are aired in random order, P(A won more hands than B throughout the telecast) = (a-b)/n. Proof: We know that, after n = a+b hands, the total difference in hands won is a-b. Let x = a-b. We want to count the number of paths from (1,1) to (n,x) that do not touch the x-axis. By the reflection principle, the number of paths from (1,1) to (n,x) that do touch the x-axis equals the total number of paths from (1,-1) to (n,x). So the number of paths from (1,1) to (n,x) that do not touch the x-axis equals the number of paths from (1,1) to (n,x) minus the number of paths from (1,-1) to (n,x) = C(n-1,a-1) – C(n-1,a) = (n-1)! / [(a-1)! (n-a)!] – (n-1)! / [a! (n-a-1)!] = {n! / [a! (n-a)!]} [(a/n) - (n-a)/n] = C(n,a) (a-b)/n. And each path is equally likely, and has probability 1/C(n,a). So, P(going from (0,0) to (n,a) without touching the x-axis = (a-b)/n.

Avoiding zero. For a simple random walk, for any even # n, P(S 1 ≠ 0, S 2 ≠ 0, …, S n ≠ 0) = P(S n = 0). Proof. The number of paths from (0,0) to (n, j) that don’t touch the x-axis at positive times = the number of paths from (1,1) to (n,j) that don’t touch the x-axis at positive times = paths from (1,1) to (n,j) - paths from (1,-1) to (n,j) by the reflection principle = N n-1,j-1 – N n-1,j+1 Let Q n,j = P(S n = j). P(S 1 > 0, S 2 > 0, …, S n-1 > 0, S n = j) = ½[Q n-1,j-1 – Q n-1,j+1 ]. Summing from j = 2 to ∞, P(S 1 > 0, S 2 > 0, …, S n-1 > 0, S n > 0) = ½[Q n-1,1 – Q n-1,3 ] + ½[Q n-1,3 – Q n-1,5 ] + ½[Q n-1,5 – Q n-1,7 ] + … = (1/2) Q n-1,1 = (1/2) P(S n = 0), because to end up at (n, 0), you have to be at (n-1,+/-1), so P(S n = 0) = (1/2) Q n-1,1 + (1/2) Q n-1,-1 = Q n-1,1. By the same argument, P(S 1 < 0, S 2 < 0, …, S n-1 < 0, S n < 0) = (1/2) P(S n = 0). So, P(S 1 ≠ 0, S 2 ≠ 0, …, S n ≠ 0) = P(S n = 0).

3. Hellmuth and Farha. P(flush | suited)? P(flush | unsuited)? Consider only flushes of your suit. [C(11,3)*C(39,2)+C(11,4)*39+C(11,5)]/C(50,5) = 6.40%. [C(12,4)*38+C(12,4)*38+C(12,5)+C(12,5)]/C(50,5) = 1.85%. 4. Chip proportions and induction. P(win a tournament) is proportional to your number of chips. Simplified scenario. Suppose you either go up or down 1 each hand, with probability 1/2. Suppose there are n chips, and you have k of them. P(win the tournament given k chips) = P(random walk goes from k to n before hitting 0) = p k. Now, clearly p 0 = 0. Consider p 1. From 1, you will either go to 0 or 2. So, p 1 = 1/2 p 0 + 1/2 p 2 = 1/2 p 2. That is, p 2 = 2 p 1. We have shown that p j = j p 1, for j = 0, 1, and 2. (induction:) Suppose that, for j = 0, 1, 2, …, m, p j = j p 1. We will show that p m+1 = (m+1) p 1. Therefore, p j = j p 1 for all j. That is, P(win the tournament) is prop. to your number of chips. p m = 1/2 p m-1 + 1/2 p m+1. If p j = j p 1 for j ≤ m, then we have mp 1 = 1/2 (m-1)p 1 + 1/2 p m+1, so p m+1 = 2mp 1 - (m-1) p 1 = (m+1)p 1.

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. Again, 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. Example (Chen and Ankenman, 2006). Suppose that a $100 winner-take-all tournament has 1024 = 2 10 players. So, you need to double up 10 times to win. Winner gets $102,400. Suppose you have probability p = 0.54 to double up, instead of 0.5. What is your expected profit in the tournament? (Assume only doubling up.) P(winning) = , so exp. return = ($102,400) = $ So exp. profit = $