Stat 35b: Introduction to Probability with Applications to Poker Outline for the day:

1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1. Review list Bayes’ Rule example CLT example Other examples * Final is Thursday, 12/11/07, 3-6am, in Geology 3656. * Open note. * Bring a calculator and a pen or pencil. * about 20-25 multiple choice + 2-3 open-answers + an extra-credit with 2 open-answer parts.  u   u 

2. 1. Review List Basic principles of counting Permutations & combinations Addition rule Conditional probability Multiplication rule Independence Odds ratios Random variables (RVs) Discrete RVs, and probability mass function (pmf) Expected value Pot odds calculations Deal-making Variance and SD Bernoulli RV [0-1. µ = p, s= √(pq). ] Binomial RV [# of successes, out of n tries. µ = np, s= √(npq).] Geometric RV [# of tries til 1st success. µ = 1/p, s= (√q) / p. ] Negative binomial RV [# of tries til rth success. µ = r/p, s= (√rq) / p. ] E(X+Y), V(X+Y)

3. 1. Review List, continued Continuous RVs Probability density function (pdf) Uniform RV Normal RV Law of Large Numbers (LLN) Central Limit Theorem (CLT) Bayes’ Rule

4. 2. Bayes’ Rule Example If B1 , …, Bn are disjoint events with P(B1 or … or Bn) = 1, then P(Bi | A) = P(A | Bi ) * P(Bi) ÷ [ ∑P(A | Bj)P(Bj)]. Q. Suppose your opponent looks at her cards and then looks at her chips immediately. You know that she’d do that with 100% probability if she had AA or KK, and with 50% probability if she had AK. Given only this information, what is the probability that she has AA? A. We want P(AA | looked at chips). But we know P(looked at chips | AA) = 100%. P(AA | looked at chips) = P(looked at chips | AA) P(AA) ÷ [P(looked at chips | AA)P(AA) +P(looked at chips | KK)P(KK) +P(looked at chips|AK)P(AK)] = 100% x P(AA) ÷ [100% x P(AA) + 100% x P(KK) + 50% x P(AK)] = 100% x 6/1326 ÷ [100% x 6/1326 + 100% x 6/1326 + 50% x 16/1326] = 6/1326 ÷ [20/1326] = 30.0%.

5. 3. CLT Example • Central Limit Theorem (CLT): if X1 , X2 …, Xn are iid with mean µ& SD s, then • (X - µ) ÷ (s/√n) ---> Standard Normal. (mean 0, SD 1). • In other words, X is like a draw from a normal distribution • with mean µ and standard deviation ofs÷√n. That is, 95% of the time, X is in the interval µ +/- 1.96 (s/√n). Q. Suppose you average $5 profit per hour, with a SD of $60 per hour. If you play 1600 hours, let Y be your average profit over those 1600 hours. What is range where Y is 95% likely to fall? A. We want µ +/- 1.96 (s/√n), where µ = $5, s = $60, and n=1600. So the answer is $5 +/- 1.96 x $60 / √(1600) = $5 +/- $2.94, or the range [$2.06, $7.94].

6. 4. Rainbow flops. P(Rainbow flop) = choose(4,3) * 13 * 13 * 13 ÷ choose(52,3) choices for the 3 suits numbers on the 3 cards possible flops ~ 39.76%. Alternative way: conceptually, order the flop cards. No matter what flop card #1 is, P(suit of #2 ≠ suit of #1 & suit of #3 ≠ suits of #1 and #2) = P(suit #2 ≠ suit #1) * P(suit #3 ≠ suits #1 and #2 | suit #2 ≠ suit #1) = 39/51 * 26/50 ~ 39.76%. Q: Out of 100 hands, expected number of rainbow flops? +/- what? X = Binomial (n,p), with n = 100, p = 39.76%, q = 60.24%. E(X) = np = 100 * 0.3976 = 39.76 SD(X) = √(npq) = sqrt(23.95) = 4.89. So, expected around 39.76 +/- 4.89 rainbow flops, out of 100 hands.

7. More poker examples. Suppose you’re all-in next hand, no matter what. What is the probability that you will eventually make 4 of a kind? (In other words, what is the probability that, after all 5 board cards are dealt, you will have 4 of a kind? Note that this includes the possibility that there is 4 of a kind on the board.) Answer: Forget about which 2 cards are yours! 13 * choose(48,3) / choose(52,7) = .00168, or about 1 in 595.

8. 4. Poker Examples, continued Given that both of your hole cards are the same suit, what is the probability that you will eventually make a flush? (Note that technically this includes the possibility that the 5 board cards are all the same suit, even if this is not the same suit as your two cards!) P(exactly 3 of your suit or exactly 4 of your suit or 5 of your suit or 5 of other suit) = [choose(11,3)*choose(39,2) + choose(11,4)*39 + choose(11,5) + 3*choose(13,5)] choose(50,5) ~ 6.58% (iii) (The worst possible beat.) You have pocket aces, and your opponent has 6 2. The first two cards of the flop are revealed, and they are both aces! At this point, what is your opponent’s probability of winning the hand? 6 cards out. 46 cards left in the deck. The last 3 board cards must come: 3 4 5 , not nec. in that order. P(opponent wins) = 1 / choose(46,3) = 0.0000659 , or 1 in 15,180.

9. 4. Poker Examples, continued The definition of a jackpot hand is a hand that meets the following 2 conditions: (a) You have a full house with 3 aces, or better. That is, you have a royal flush, straight flush, 4 of a kind, or a full house with 3 aces. (b) Your best 5-card hand must use both of your hole cards. [Ignore the possibility of ambiguity as to whether your hole card plays, such as where you have A5 and the board is AA553.] (iv) Given that you make a hand satisfying Condition (a), what is the probability that you satisfy Condition (b)? Imagine ordering the 7 cards so that the first 5 are the ones actually used. What’s the chance that your 2 cards are in those first 5? = P(your first card is in those 5 AND your 2nd card is in those 5) = 5/7 * 4/6 = 47.6% Many 2-card hands that seem weak nevertheless have some chance of being two-player jackpot hands. If you have 94, it is possible for you and your opponent to both make jackpot hands, if for instance the board comes something like 44477, and your opponent has 77. Are there any 2-card hands you could have such that, no matter what your opponent has and no matter what the board is, you and your opponent cannot possibly both have jackpot hands? List them. Only 32 offsuit. To satisfy (b), the board must be 22233 or 33322….