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 multiple choice open-answers + an extra-credit with 2 open-answer parts.   u 

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

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).

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.

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).

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 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 million = 3.31 million. Difference is 3.31 million minus million = 4.56 million!

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 /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.

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.

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 = ($102,400) = $ So exp. profit = $