Outline: 1) Odds ratios, continued. 2) Expected value revisited, Harrington’s strategy 3) Pot odds 4) Examples

1) Odds ratios, revisited: Odds ratio of A = P(A)/P(A c ) Odds against A = Odds ratio of A c = P(A c )/P(A). An advantage of probability over odds ratios is the mult. rule: P(A & B) = P(A) x P(B|A), but can’t multiply odds ratios. Example: Last week’s High Stakes Poker, 1/25/07, on GSN: Blinds $300/$600 + $1200 “straddle” + eight $100 antes = $2900. Gabe Kaplan: K  K , raises to $4200. Chen folds, Doyle folds, Matusow calls with A  9 . Gold folds, Alaie folds, Sheikhan folds, Ramdin folds. Flop: T  8  J . Kaplan checks, Matusow bets $8000, Kaplan calls. Turn: A . (both players check) River: 6 . Kaplan bets $16000, Matusow folds! (pot $43,300)

Gabe Kaplan: K  K . Mike Matusow: A  9 . Flop: T  8  J . To win, Matusow needs a 7, Q, or A. (But not Q9 or QA, and they tie with 79…. And not two more  s. ) And, Matusow can also win with two more  s. Given this, what’s the probability that Matusow makes a flush? P(  on turn AND  on river) = P(  turn) x P(  river |  turn) = 9/45 x 8/44 = 3.6%. Odds against it? 96.3% ÷ 3.6% = : 1. Odds against  on turn = 4 : 1. Odds against  on river |  on turn = 4.5 : 1. Multiplying, you’d get (4 x 4.5) = 18. Moral: you can’t multiply odds ratios!

3) Expected Value. For a discrete random variable X with pmf f(b), the expected value of X is the sum ∑ b f(b). = b 1 f(b 1 ) + b 2 f(b 2 ) + … b m f(b m ) +

Dan Harrington says that, with a hand like AA, you really want to be one-on-one. True or false? * Best possible pre-flop situation is to be all in with AA vs A8, where the 8 is the same suit as one of your aces, in which case you're about 94% to win. (the 8 could equivalently be a 6,7, or 9.) If you are all in for $100, then your expected holdings afterwards are $188. 1) In a more typical situation: you have AA and are up against TT. You're 80% to win, so your expected value is $160. 2) Suppose that, after the hand vs TT, you get QQ and get up against someone with A9 who has more chips than you do. The chance of you winning this hand is 72%, and the chance of you winning both this hand and the hand above is 58%, so your expected holdings after both hands are $232: you have 58% chance of having $400, and 42% chance to have $0. 3) Now suppose instead that you have AA and are all in against 3 callers with A8, KJ suited, and 44. Now you're 58.4% to quadruple up. So your expected holdings after the hand are $234, and the situation is just like (actually slightly better than) #1 and #2 combined: 58.4% chance to hold $400, and 41.6% chance for $0. * So, being all-in with AA against 3 players is much better than being all-in with AA against one player: in fact, it's about like having two of these lucky one-on-one situations.

What about with a low pair? a) You have $100 and 55 and are up against A9. You are 56% to win, so if X = your holdings after the hand, then the expected value E(X) = ($200 x 56%) + ($0 x 44%) = $112. b) You have $100 and 55 and are up against A9, KJ, and QJ suited. Seems pretty terrible, doesn't it? But your odds are 27.3% to quadruple, so your expected value E(X) = ($400 x 27.3%) + ($0 x 72.7%) = $109. About the same!

a) You have $100 and KK and are all-in against TT. You're 81% to double up, so your expected number of chips after the hand is $162. b) You have $100 and KK and are all-in against A9 and TT. You're 58% to have $300, so your expected value is $300 x 58% = $174. So, if you have KK and an opponent with TT has already called you, and another who has A9 is thinking about whether to call you too, you actually want A9 to call! Given this, one may question Harrington's suggested strategy of raising huge in order to isolate yourself against one player. What about with KK?

3) POT ODDS. Suppose someone bets (or raises) you, going all-in. What should your chances of winning be in order for you to correctly call? Let B = the amount BET to you, i.e. the additional amount you'd need to put in if you want to call. So, for instance, if you bet 100 & your opponent with 800 left went all-in, B = 700. Let T = the TOTAL amount in the pot right now (including your opponent's bet). Let p = your Probability of winning the hand if you call. So prob. of losing = 1-p. Let C = the number of CHIPS you have right now. Let X = the number of chips you have after the hand is over. If you call, then E[X] = (C - B)(1-p) + (C + T)(p) = C(1-p+p) - B(1-p) + Tp = C - B + Bp + Tp If you fold, then E[X] = C. You want your expected number of chips to be maximized, so it's worth calling if C - B + Bp + Tp > C, that is, if p > B / (B+T). Amount bet ÷ Amount you’d win.

Example: "Poker Superstars Invitational Tournament", FSN, October Ted Forrest: 1 million chips Freddy Deeb: 825,000Blinds: 15,000 / 30,000 Cindy Violette: 650,000 Elie Elezra: 575,000(pot = 45,000) 1) Elezra raises to 100,000 (pot = 145,000) 2) Forrest folds. 3) Deeb, the small blind with A  8 , folds. 4) Violette, the big blind with K J, calls.(pot = 215,000) 5) The flop is:2 7  A 6) Violette bets 100,000.(pot = 315,000) 7) 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) = 32%.

Q: Based on expected value, should she have called? Her chances must be at least 375,000 / (790, ,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 your opponent is bluffing. 38% seems the most plausible. 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 7 3  ! Her chances were 51%. Bad fold. How many outs did she have? 8 s, plus 3 kings and 3 jacks = 14. And 2 chances. So a very rough guess is 28% + 28% = 56%. A better guess is 14/ /45 - (14/45)(13/44) = 53.0%. Must subtract from this the prob. that she hits a K or J but Elezra hits a non- 7 or 3 = 6/45 x 4/44 + 4/45 x 6/44 = 2.4%. Also, subtract the chance that she gets the 3 but Elezra makes a full house = 1/45 x 4/44 + 4/45 x 1/44 = 0.4% So you get 50.2%.