1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Review of random walks. 2.Ballot theorem. 3.P( avoiding zero ) for a random walk. 4.Cada vs. Ivey. 5.Martingale games. 6. Chip proportions and induction. * Computer Project B will be due on Monday, Nov 30, at 8:00pm. * No class Thursday, Dec 3. Last class is Dec 1. * Final is Thursday, Dec 10, 3-6pm, in class, open note. * Bring a calculator and a pen or pencil. * about 20-25 multiple choice + 2-3 open-answers + an extra-credit with 2 open-answer parts.   u 

2. TEAMS FOR PROJECT B: team a CoreyC AraD team b WooO JiweiS team c DaiN SethB team d JasonG BrianY team e SarahW MichaelD team f ZiqiM BenjaminN team g LongL AmandaD team h AvinashC ChrisB team i DavidC RafalJ team j HoL ErikM team k JamesL RyanK team l JesseC JeffC SamanthaS

3. 1. Random walks. Suppose that X 1, X 2, …, are iid and that Y k = X 1 + … + X k for k = 1, 2, …. Y 0 = 0. Then the totals {Y 1, Y 2, …} is a random walk. The classical example is when each X i is 1 or -1 with probability ½ each. For the classical random walk,.P(Y k = j) = Choose{k, (k+j)/2} (1/2) k..If j and k are > 0, then the number of paths from (0,j) to (n,k) that touch the x-axis = the number of paths from (0,-j) to (n,k). [reflection principle].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). [ballot theorem].P(Y 1 ≠ 0, Y 2 ≠ 0, …, Y 2n ≠ 0) = P(Y 2n = 0).

4. 2. Ballot theorem. Suppose that in an election, candidate X gets x votes, and candidate Y gets y votes, where x > y. The probability that X always leads Y throughout the counting is (x-y) / (x+y). Proof. We know that, after counting n = x+y votes, the total difference in votes is x-y = a. We want to count the number of paths from (1,1) to (n,a) that do not touch the x-axis. By the reflection principle, the number of paths from (1,1) to (n,a) that do touch the x-axis equals the total number of paths from (1,-1) to (n,a). So the number of paths from (1,1) to (n,a) that do not touch the x-axis equals the number of paths from (1,1) to (n,a) minus the number of paths from (1,-1) to (n,a) = Choose(n-1,x-1) – Choose(n-1,x) = (n-1)! / [(x-1)! (n-x)!] – (n-1)! / [x! (n-x-1)!] = (n-1)! x / [x! (n-x)!] – (n-1)! (n-x) / [x! (n-x)!] = {n! / [x! (n-x)!]} (x/n) – {n! / [x! (n-x)!]} (n-x)/n = (x – y) / (x+y) {n! / [x! (n-x)!]} = (x – y) / (x+y) {Number of paths from (0,0) to (n,a)}. And each path is equally likely.

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

6. 4. Cada vs. Ivey. (Bluffing with medium-strength hands? Questionable.) Bluffing, and catching bluffs, can have huge impacts on equity and expected profit. Equity = pot * P(you win the pot), assuming no folding. Your expected profit = your equity minus how much you put in. (assuming no folding.) If Ivey folds, his profit is surely -1.25 million. If Ivey calls, he will have put in 4.05 million, and his equity would be 80%*9.2 million = 7.36 million. So, his expected profit if he calls = 7.36 million - 4.05 million = 3.31 million. Difference is 3.31 million minus -1.25 million = 4.56 million!

7. 5. Martingale games. If a game is fair, then E(X) = 0, regardless of your wager. {E(aX) = aE(X) = a*0 = 0.} Suppose E(X i ) = 0, for i = 1, 2, …, and suppose you wager $1 on game 1. If you win, you stop. If you lose, you wager $2 on game 2. If you win, you stop. If you lose, you wager $4 on game 3, etc. What is your expected profit? After n steps, you will either have lost $1+$2+$4 = $2 n -1 or profited $1. Expected profit = [$-(2 n -1)][1/2] n + [$1][1 - (1/2) n ] = $-2 n /2 n + 1/2 n + 1 - 1/2 n = $0. Alternate way: Let Y i = your profit on game i. E(Y i ) = 0 no matter how much you wager. So E(Y 1 + Y 2 + …) = E(Y 1 ) + E(Y 2 ) + … = 0 + 0 + … = 0.

8. 6. 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, for j = 0, 1, and 2, p j = j p 1. (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.

9. 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.) Exp. return = 0.54 10 ($102,400) = $215.89. So exp. profit = $115.89.