Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: Turn in Hw2. Hw3 assigned. Expected value and pot odds, continued Violette/Elezra example Yang / Kravchenko Deal-making and expected value Reminder: project A is due by email by Wed, Feb 6, 8pm. You may email me for your teammate’s email address.   u    u 

Teams: team a Sohil Yara team b Chunlin Matthew team c Kyle Vielka team d Bhavik Shannon team e Andrew Megan team f Jasmine Stefan team g Christopher Jon team h Richard Michael team i Wing Shukki team j Alexia Siena team k Casey Binh team l Uddhav Thanh team m Kai Li team n Kaavya Yongxin team o Arash Rocio team p Denise Wenhao

Turn in Hw2. Hw3 is due Thur Feb 28, 11am. 4.7, 4.8, 4.12, 4.16, 5.6, 6.2. Also, read chapter 5. Reminder: project A is due by email by Wed, Feb 6, 8pm. You may email me for your teammate’s email address. Just submit one email per team.   u    u 

2) Pot odds and expected value, continued. From a previous lecture: to call an all-in, need P(win) > B ÷ (B+pot). Expressed as an odds ratio, this is sometimes referred to as pot odds or express odds. If the bet is not all-in & another betting round is still to come, need P(win) > wager ÷ (wager + winnings), where winnings = pot + amount you’ll win on later betting rounds, wager = total amount you will wager including the current round & later rounds, assuming no folding. The terms Implied-odds / Reverse-implied-odds describe the cases where winnings > pot or where wager > B, respectively. See p66.

3) Example: Poker Superstars Invitational Tournament, FSN, October 2005. Ted Forrest: 1 million chips Freddy Deeb: 825,000 Blinds: 15,000 / 30,000 Cindy Violette: 650,000 Eli Elezra: 575,000 * Elezra raises to 100,000  * Forrest folds. * Deeb, the small blind, folds. * Violette, the big blind with Ku Ju, calls. * The flop is: 2u 7 Au * Violette bets 100,000. * Elezra raises all-in to 475,000. (pot = 790,000) So, it's 375,000 more to Violette. She folds. Q: Based on expected value, should she have called? Her chances must be at least 375,000 / (790,000 + 375,000) = 32%.3

Violette has Ku Ju. The flop is: 2u 7 Au. Q: Based on expected value, should she have called? Her chances must be at least 375,000 / (790,000 + 375,000) = 32%. vs. AQ: 38%. AK: 37% AA: 26% 77: 26% A7: 31% A2: 34% 72: 34% TT: 54% T9: 87% 73: 50% Harrington's principle: always assume at least a 10% chance that opponent is bluffing. Bayesian approach: average all possibilities, weighting them by their likelihood. Maybe she's conservative.... but then why play the hand at all? Reality: Elezra had 7u 3. Her chances were 51%. Bad fold. What was her prob. of winning (given just her cards and Elezra’s, and the flop)? Of choose(45,2) = 990 combinations for the turn & river, how many give her the win? First, how many outs did she have? eight us + 3 kings + 3 jacks = 14. She wins with (out, out) or (out, nonout) or (non-u Q, non-u T) choose(14,2) + 14 x 31 + 3 * 3 = 534 but not (k or j, 7 or non-u 3) and not (3u , 7 or non-u 3) - 6 * 4 - 1 * 4 = 506. So the answer is 506 / 990 = 51.1%.

4. Yang / Kravchenko. Yang A 10u. Pot is 19million. Bet is 8.55 million. Needs P(win) > 8.55 ÷ (8.55 + 19) = 31%. vs. AA: 8.5%. AJ-AK: 25-27%. KK-TT: 29%. 99-22: 44-48%. KQs: 56%. Bayesian method: average these probabilities, weighting each by its likelihood. See p49-53.

4. Yang / Kravchenko. Yang A 10u. Pot is 19.0 million. Bet is 8.55 million. Suppose that, averaging the different probabilities, P(Yang wins) = 30%. And say Yang calls. Let X = the number of chips Kravchenko has after the hand. What is E(X)? [Note, if Yang folds, then X = 19.0 million for sure.] E(X) = ∑ [k * P(X=k)] = [0 * 30%] + [27.55 million * 70%] = 19.285 million.

5. Deal-making. (Expected value, game theory) Game-theory: For a symmetric-game tournament, the probability of winning is approx. optimized by the myopic rule (in each hand, maximize your expected number of chips), and P(you win) = your proportion of chips (Theorems 7.6.6 and 7.6.7 on pp 151-152). For a fair deal, the amount you win = the expected value of the amount you will win. See p61.

For instance, suppose a tournament is winner-take-all, for $8600. With 6 players left, you have 1/4 of the chips left. An EVEN SPLIT would give you $8600 ÷ 6 = $1433. A PROPORTIONAL SPLIT would give you $8600 x (your fraction of chips) = $8600 x (1/4) = $2150. A FAIR DEAL would give you the expected value of the amount you will win = $8600 x P(you get 1st place) = $2150. But suppose the tournament is not winner-take-all, but pays $3800 for 1st, $2000 for 2nd, $1200 for 3rd, $700 for 4th, $500 for 5th, $400 for 6th. Then a FAIR DEAL would give you $3800 x P(1st place) + $2000 x P(2nd) +$1200 x P(3rd)+$700xP(4th) +$500xP(5th) +$400xP(6th). Hard to determine these probabilities. But, P(1st) = 25%, and you might roughly estimate the others as P(2nd) ~ 20%, P(3rd) ~ 20%, P(4th) ~ 15%, P(5th) ~10%, P(6th) ~ 10%, and get $3800 x 25% + $2000 x 25% +$1200 x 20% + $700x 15% + $500x 10% +$400x 5% = $1865. If you have 40% of the chips in play, then: EVEN SPLIT = $1433. PROPORTIONAL SPLIT = $3440. FAIR DEAL ~ $2500!

Another example. Before the Wasicka/Binger/Gold hand, Gold had 60M, Wasicka 18M, Binger 11M. Payouts: 1st place $12M, 2nd place $6.1M, 3rd place $4.1M. Proportional split: of the total prize pool left, you get your proportion of chips in play. e.g. $22.2M left, so Gold gets 60M/(60M+18M+11M) x $22.2M ~ $15.0M. A fair deal would give you P(you get 1st place) x $12M + P(you get 2nd place) x $6.1M + P(3rd pl.) x $4.1M . *Even split: Gold $7.4M, Wasicka $7.4M, Binger $7.4M. *Proportional split: Gold $15.0M, Wasicka $4.5M, Binger $2.7M. *Fair split: Gold $10M, Wasicka $6.5M, Binger $5.7M. *End result: Gold $12M, Wasicka $6.1M, Binger $4.1M.

Luck vs. skill. pp 71-79. Any thoughts? name1 = c("gravity","tommy","ursula","timemachine","vera","william","xena”) decision1 = list(gravity, tommy, ursula, timemachine, vera, william, xena) tourn1(name1, decision1, myfast1 = 2) ## run quickly tourn1(name1, decision1, myfast1 = 0) ## run slowly, showing key hands