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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Midterms 2.Flushes 3.Hellmuth vs. Farha 4.Worst possible beat 5.Jackpot.

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Presentation on theme: "Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Midterms 2.Flushes 3.Hellmuth vs. Farha 4.Worst possible beat 5.Jackpot."— Presentation transcript:

1 Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 1.Midterms 2.Flushes 3.Hellmuth vs. Farha 4.Worst possible beat 5.Jackpot hands. 6.Random Walks. * Computer Project B will be due on Monday, Nov 30, at 8:00pm. * No class Thursday, Dec 3. Last class is Dec 1. * Final is Thursday, Dec 10, 3-6pm, in class. * Open note. * Bring a calculator and a pen or pencil. * about multiple choice open-answers + an extra-credit with 2 open-answer parts. u u

2 2. Flushes. a) 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%. b) Given that your hole cards are different suits, what is P(flush)? = P(4 of suit A or 5 of suit A or 4 of suit B or 5 of suit B or 5 of suit C or 5 of suit D) = = [C(12,4)*38 + C(12,5) + C(12,4)*38 + C(12,5) + C(13,5) + C(13,5)] choose(50,5) ~ 1.97%. 3. Hellmuth vs. Farha.

3 4. 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 opponents 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) = , or 1 in 15,180.

4 5. Jackpot hands 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. Whats 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%

5 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. (v)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 or 33322….

6 6. Random walks. Suppose that X 1, X 2, …, are iid and that Y k = X 1 + … + X k for k = 1, 2, …. Y 0 = 0. Then the totals {Y 1, Y 2, …} is a random walk. The classical example is when each X i is 1 or -1 with probability ½ each. For the classical random walk,. P(Y k = j) = Choose{k, (k+j)/2} (1/2) k..If j and k are > 0, then the number of paths from (0,j) to (n,k) that touch the x-axis = the number of paths from (0,-j) to (n,y). [reflection principle].In an election, if candidate X gets x votes, and candidate Y gets y votes, where x > y, then the probability that X always leads Y throughout the counting is (x-y) / (x+y). [ballot theorem].P(Y 1 0, Y 2 0, …, Y 2n 0) = P(Y 2n = 0).

7 Ballot theorem. Suppose that in an election, candidate X gets x votes, and candidate Y gets y votes, where x > y. The probability that X always leads Y throughout the counting is (x-y) / (x+y). Proof. We know that, after counting n = x+y votes, the total difference in votes is x-y = a. We want to count the number of paths from (1,1) to (n,a) that do not touch the x-axis. By the reflection principle, the number of paths from (1,1) to (n,a) that do touch the x-axis equals the total number of paths from (1,-1) to (n,a). So the number of paths from (1,1) to (n,a) that do not touch the x-axis equals the number of paths from (1,1) to (n,a) minus the number of paths from (1,-1) to (n,a) = Choose(n-1,x-1) – Choose(n-1,x) = (n-1)! / [(x-1)! (n-x)!] – (n-1)! / [x! (n-x-1)!] = (n-1)! x / [x! (n-x)!] – (n-1)! (n-x) / [x! (n-x)!] = {n! / [x! (n-x)!]} (x/n) – {n! / [x! (n-x)!]} (n-x)/n = (x – y) / (x+y) {n! / [x! (n-x)!]} = (x – y) / (x+y) {Number of paths from (0,0) to (n,a)}. And each path is equally likely.

8 For a classical random walk, P(Y 1 0, Y 2 0, …, Y 2n 0) = P(Y 2n = 0). Proof. The number of paths from (0,0) to (2n, 2r) that dont touch the x-axis at positive times = the number of paths from (1,1) to (2n,2r) that dont touch the x-axis at positive times = paths from (1,1) to (2n,2r) - paths from (1,-1) to (2n,2r) by reflection principle = N 2n-1,2r-1 – N 2n-1,2r+1 Let p n,x = P(Y n = x). P(Y 1 > 0, Y 2 > 0, …, Y 2n-1 > 0, Y 2n = 2r) = ½[p 2n-1,2r-1 – p 2n-1,2r+1 ]. Summing from r = 1 to, P(Y 1 > 0, Y 2 > 0, …, Y 2n-1 > 0, Y 2n > 0) = ½[p 2n-1,1 – p 2n-1,3 ] + ½[p 2n-1,3 – p 2n-1,5 ] + ½[p 2n-1,5 – p 2n-1,7 ] + … = (1/2) p 2n-1,1 = (1/2) P(Y 2n = 0). By symmetry, the same is true for P(Y 1 < 0, Y 2 < 0, …, Y 2n-1 < 0, Y 2n < 0). So, P(Y 1 0, Y 2 0, …, Y 2n 0) = P(Y 2n = 0).


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