Outline for the day: 1.Discuss handout / get new handout. 2.Teams 3.Example projects 4.Expected value 5.Pot odds calculations 6.Hansen / Negreanu 7.P(4.

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Stat 35b: Introduction to Probability with Applications to Poker
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Stat 35b: Introduction to Probability with Applications to Poker
Stat 35b: Introduction to Probability with Applications to Poker
Stat 35b: Introduction to Probability with Applications to Poker
Stat 35b: Introduction to Probability with Applications to Poker
Stat 35b: Introduction to Probability with Applications to Poker
Stat 35b: Introduction to Probability with Applications to Poker
Presentation transcript:

Outline for the day: 1.Discuss handout / get new handout. 2.Teams 3.Example projects 4.Expected value 5.Pot odds calculations 6.Hansen / Negreanu 7.P(4 of a kind) and other calculations Project: (Due by on Sunday, Nov 15, 8:00pm).   u 

Project: (Due by on Sunday, Nov 15, 8:00pm). Teams.txt ## MAKE TEAMS: name1 = c("a","b","c","d","e","f","g","h","i") x = c("RobertC", "IrisC", "AlexaG", "EvanL", "PeterM", "KaitlinM", "WilliamM", "AramN", "BrianO", "PeterQ", "MarcR", "JayveerS", "SaraS", "DanielS", "TannazT", "NhatT", "JamieT", "NicolasW", "BrianW") n = length(x) teams = function(){ y = x[sample(n)] for(i in 1:8){ cat("\n", "team", name1[i], y[(2*i-1):(2*i)]) } i = 9 cat("\n", "team", name1[i], y[(2*i-1):(2*i+1)]) } teams() team a WilliamM PeterQ team b AramN RobertC team c BrianO AlexaG team d JayveerS SaraS team e PeterM TannazT team f EvanL NicolasW team g DanielS NhatT team h BrianW KaitlinM team i IrisC JamieT MarcR   u 

## Examples at unbeatable1 = function(numattable1, crds1, board1, round1, currentbet, mychips1, pot1, roundbets, blinds1, chips1, ind1, dealer1, tablesleft){ ## any pair, or AT-AK, a1 = 0 if((crds1[1,1] == crds1[2,1]) || ((crds1[1,1] > 13.5) && (crds1[2,1]>9.5))) a1 = mychips1 a1 } ## end of unbeatable version2 = function(numattable1, crds1, board1, round1, currentbet, mychips1, pot1,roundbets, blinds1, chips1, ind1, dealer1, tablesleft){ ## any pair, or AT-AK, or suited connectors a1 = 0 if((crds1[1,1] == crds1[2,1]) || ((crds1[1,1] > 13.5) && (crds1[2,1]>9.5)) || ((crds1[1,1]-crds1[2,1]==1) && (crds1[1,2] == crds1[2,2]))) a1 = mychips1 a1 } ## end of unbeatable

Expected Value. For a discrete random variable X, the probability mass function f(b) = P(X=b). The expected value of X = ∑ b f(b). The sum is over all possible values of b. (continuous random variables later…) The expected value is also called the mean and denoted E(X) or . Ex: 2 cards are dealt to you. X = 1 if pair, 0 otherwise. P(X is 1) ~ 5.9%, P(X is 0) ~ 94.1%. E(X) = (1 x 5.9%) + (0 x 94.1%) = 5.9%, or Ex. Coin, X=20 if heads, X=10 if tails. E(X) = (20x50%) + (10x50%) = 15. Ex. Lotto ticket. f($10million) = 1/choose(52,6) = 1/20million, f($0) = 1-1/20mil. E(X) = ($10mil x 1/20million) = $0.50. The expected value of X represents a best guess at X.

Some reasons why Expected Value applies to poker: Tournaments: some game theory results suggest that, in symmetric, winner- take-all games, the optimal strategy is the one which uses the myopic rule: that is, given any choice of options, always choose the one that maximizes your expected value. Laws of large numbers: Some statistical theory indicates that, if you repeat an experiment over and over repeatedly, your long-term average will ultimately converge to the expected value. So again, it makes sense to try to maximize expected value when playing poker (or making deals). Checking results: A great way to check whether you are a long-term winning or losing player, or to verify if a certain strategy works or not, is to check whether the sample mean is positive and to see if it has converged to the expected value.

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, if you bet 100 & your opponent with 800 left went all-in, B = 700. Let POT = the 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 CHIPS = the number of chips you have right now. If you call, then E[your chips at end] = (CHIPS - B)(1-p) + (CHIPS + POT)(p) = CHIPS(1-p+p) - B(1-p) + POT(p) = CHIPS - B + Bp + POTp If you fold, then E[your chips at end] = CHIPS. You want your expected number of chips to be maximized, so it's worth calling if -B + Bp + POTp > 0, i.e. if p > B / (B+POT).

Pot odds and expected value, continued. 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. Daniel Negreanu Gus Hansen High Stakes Poker.

-- Which is more likely: * flopping a full house, or * eventually making 4 of a kind?

Suppose you’re all in next hand, no matter what cards you get. P(flop a full house)? Key idea: forget order! Consider all combinations of your 2 cards and the flop. Sets of 5 cards. Any such combo is equally likely! choose(52,5) different ones. P(flop full house) = # of different full houses / choose(52,5) How many different full houses are possible? 13 * choose(4,3) different choices for the triple. For each such choice, there are 12 * choose(4,2) choices left for the pair. So, P(flop full house) = 13 * choose(4,3) * 12 * choose(4,2) / choose(52,5) ~ 0.144%, or 1 in 694.

P(eventually make 4-of-a-kind)? [including case where all 4 are on board] Again, just forget card order, and consider all collections of 7 cards. Out of choose(52,7) different combinations, each equally likely, how many of them involve 4-of-a-kind? 13 choices for the 4-of-a-kind. For each such choice, there are choose(48,3) possibilities for the other 3 cards. So, P(4-of-a-kind) = 13 * choose(48,3) / choose(52,7) ~ 0.168%, or 1 in 595. Uniform random variables in R. Teams for Project A.