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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: Uniform random variables and R Kaplan vs. Gazes, More counting problems Express, implied and reverse implied odds Yang vs. Kravchenko. u u . Uniform Random Variables and R
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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: Uniform random variables and R Kaplan vs. Gazes, More counting problems Express, implied and reverse implied odds Yang vs. Kravchenko u u
Uniform Random Variables and R Continuous random variables are often characterized by their probability density functions (pdf, or density): a function f(x) such that P{X is in B} = ∫B f(x) dx . Uniform: f(x) = c, for x in (a, b). = 0, for all other x. [Note: c must = 1/(b-a), so that ∫ab f(x) dx = P{X is in (a,b)} = 1.] (For teams & examples for hw & computer projects, see http://www.stat.ucla.edu/~frederic/35b/F09/diamond1.txt )
2) Poker After Dark: Kaplan vs. Gazes 3) More counting problems -- How likely is it to make 4-of-a-kind? 1 in ___ ? -- What about the probability of flopping 4-of-a-kind? -- What about the prob. of flopping 4-of-a-kind, given that you have a pocket pair? -- Is it less likely than … * flopping an ace-high flush? * flopping a straight-flush?
Suppose you’re all in next hand, no matter what cards you get. P(eventually make 4-of-a-kind)? [including case where all 4 are on board] Trick: 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. P(flop 4-of-a-kind) = 13*48 / choose(52,5) = 0.024% = 1 in 4165. P(flop 4-of-a-kind | pocket pair)? No matter which pocket pair you have, there are choose(50,3) possible flops, each equally likely, and how many of them give you 4-of-a-kind? 48. (e.g. if you have 7 7, then need to flop 7u 7 x, & there are 48 choices for x) So P(flop 4-of-a-kind | pp) = 48/choose(50,3) = 0.245% = 1 in 408.
P(flop an ace high flush)? [where the ace might be on the board] -- 4 suits -- one of the cards must be an ace. choose(12,4) possibilities for the others. So P(flop ace high flush) = 4 * choose(12,4) / choose(52,5) = 0.0762%, or 1 in 1313. P(flop a straight flush)? -- 4 suits -- 10 different straight-flushes in each suit. (5 high, 6 high, …, Ace high) So P(flop straight flush) = 4 * 10 / choose(52,5) = 0.00154%, or 1 in 64974.
4) Express, implied, and reverse implied odds. From 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.
5. 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.
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.
