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

Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 0. Collect hw2, return hw1, give out hw3. Project A competition. Bernoulli, binomial. Geometric random variable. Negative binomial E(X+Y) Harman / Negreanu , P(flop 2 pairs) Running it twice

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

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  1. Stat 35b: Introduction to Probability with Applications to Poker Outline for the day: 0. Collect hw2, return hw1, give out hw3. Project A competition. Bernoulli, binomial. Geometric random variable. Negative binomial E(X+Y) Harman / Negreanu , P(flop 2 pairs) Running it twice “Unbeatable” strategy Play (and lose) like the Pros P(rainbow flop) Nguyen vs. Szenkuti  u   u 

  2. 2. Bernoulli Random Variables. If X = 1 with probability p, and X = 0 otherwise, then X = Bernoulli (p). Probability mass function (pmf): P(X = 1) = p P(X = 0) = q, where p+q = 100%. If X is Bernoulli (p), then µ = E(X) = p, and s= √(pq). Binomial Random Variables. If X = # of times something with prob. p occurs, out of n independent trials Then X = Binomial (n.p). e.g. the number of pocket pairs, out of 10 hands. Key idea: X = Y1 + Y2 + … + Yn , where the Yi are independent and Bernoulli (p). pmf: P(X = k) = choose(n, k) * pk qn - k. If X is Binomial (n,p), then µ = np, and s = √(npq).

  3. 3. Geometric Random Variables. Suppose now X = # of trials until the first occurrence. (Again, each trial is independent, and each time the probability of an occurrence is p.) Then X = Geometric (p). e.g. the number of hands til you get your next pocket pair. [Including the hand where you get the pocket pair. If you get it right away, then X = 1.] Now X could be 1, 2, 3, …, up to ∞. pmf: P(X = k) = p1 qk - 1. e.g. say k=5: P(X = 5) = p1 q 4. Why? Must be 0 0 0 0 1. Prob. = q * q * q * q * p. If X is Geometric (p), then µ = 1/p, and s= (√q) ÷ p. e.g. Suppose X = the number of hands til your next pocket pair. P(X = 12)? E(X)? s? X = Geometric (5.88%). P(X = 12) = p1 q11 = 0.0588 * 0.9412 ^ 11 = 3.02%. E(X) = 1/p = 17.0. s= sqrt(0.9412) / 0.0588 = 16.5. So, you’d typically expect it to take 17 hands til your next pair, +/- around 16.5 hands.

  4. 4. Negative Binomial Random Variables. Suppose now X = # of trials until the rth occurrence. Then X = negative binomial (r,p). e.g. the number of hands you have to play til you’ve gotten r=3 pocket pairs. Now X could be 3, 4, 5, …, up to ∞. pmf: P(X = k) = choose(k-1, r-1) pr qk - r. e.g. say r=3 & k=7: P(X = 7) = choose(6,2) p3 q4. Why? Out of the first 6 hands, there must be exactly r-1 = 2 pairs. Then pair on 7th. P(exactly 2 pairs on first 6 hands) = choose(6,2) p2 q4. P(pair on 7th) = p. If X is negative binomial (r,p), then µ = r/p, and s= (√rq) ÷ p. e.g. Suppose X = the number of hands til your 12th pocket pair. P(X = 100)? E(X)? s? X = Negative Binomial (12, 5.88%). P(X = 100) = choose(99,11) p12 q88 = choose(99,11) * 0.0588 ^ 12 * 0.9412 ^ 88 = 0.104%. E(X) = r/p = 12/0.0588 = 204.1. s= sqrt(12*0.9412) / 0.0588 = 57.2. So, you’d typically expect it to take 204.1 hands til your 12th pair, +/- around 57.2 hands.

  5. 5) E(X+Y) = E(X) + E(Y).Whether X & Y are independent or not! Similarly, E(X + Y + Z + …) = E(X) + E(Y) + E(Z) + … And, if X & Y are independent, then V(X+Y) = V(X) + V(Y). so SD(X+Y) = √[SD(X)^2 + SD(Y)^2]. Example 1: Play 10 hands. X = your total number of pocket aces. What is E(X)? X is binomial (n,p) where n=10 and p = 0.00452, so E(X) = np = 0.0452. Alternatively, X = # of pocket aces on hand 1 + # of pocket aces on hand 2 + … So, E(X) = Expected # of AA on hand1 + Expected # of AA on hand2 + … Each term on the right = 1 * 0.00452 + 0 * 0.99548 = 0.00452. So E(X) = 0.00452 + 0.00452 + … + 0.00452 = 0.0452.

  6. E(X+Y) = E(X) + E(Y).Whether X & Y are independent or not! Similarly, E(X + Y + Z + …) = E(X) + E(Y) + E(Z) + … And, if X & Y are independent, then V(X+Y) = V(X) + V(Y). so SD(X+Y) = √[SD(X)^2 + SD(Y)^2]. Example 2: Play 1 hand, against 9 opponents. X = total # of people with pocket aces. Now what is E(X)? Note: not independent! If you have AA, then it’s unlikely anyone else does too. Nevertheless, Let X1 = 1 if player #1 has AA, and 0 otherwise. X2 = 1 if player #2 has AA, and 0 otherwise. etc. Then X = X1 + X2 + … + X10. So E(X) = E(X1) + E(X2) + … + E(X10) = 0.00452 + 0.00452 + … + 0.00452 = 0.0452. Harman vs. Negreanu….

  7. 6. Harman/Negreanu If you’re sure to be all-in next hand,what is P(you will flop two pairs)? Tough one. Don’t double-count (4 4 9 9 Q) and (9 9 4 4 Q)! There are choose(13,2) possibilities for the NUMBERS of the two pairs. For each such choice (such as 4 & 9), there are choose(4,2) choices for the suits of the lower pair, and same for the suits of the higher pair. So, choose(13,2) * choose(4,2) * choose(4,2) different possibilities for the two pairs. For each such choice, there are 44 [52 - 8 = 44] different possibilities for your fifth card, so that it’s not a full house but simply two pairs. So, P(flop two pairs) = choose(13,2) * choose(4,2) * choose(4,2) * 44 / choose(52,5) ~ 4.75%, or 1 in 21.

  8. 7) Running it twice. Harman has 10 7 . Negreanu has K Q . The flop is 10u 7 Ku . Harman’s all-in. $156,100 pot. P(Negreanu wins) = 29%. P(Harman wins) = 71%. Let X = amount Harman has after the hand. If they run it once, E(X) = $0 x 29% + $156,100 x 71% = $110,831. If they run it twice, what is E(X)? There’s some probability p1 that Harman wins both times ==> X = $156,100. There’s some probability p2 that they each win one ==> X = $78,050. There’s some probability p3 that Negreanu wins both ==> X = $0. E(X) = $156,100 x p1 + $78,050 x p2 + $0 x p3. If the different runs were independent, then p1 = P(Harman wins 1st run & 2nd run) would = P(Harman wins 1st run) x P(Harman wins 2nd run) = 71% x 71% = 50.4%. But, they’re not quite independent! Very hard to compute p1 and p2. However, you don’t need p1 and p2 ! X = the amount Harman gets from the 1st run + amount she gets from 2nd run, so E(X) = E(amount Harman gets from 1st run) + E(amount she gets from 2nd run) = $78,050 x P(Harman wins 1st run) + $0 x P(Harman loses first run) + $78,050 x P(Harman wins 2nd run) + $0 x P(Harman loses 2nd run) = $78,050 x 71% + $0 x 29% + $78,050 x 71% + $0 x 29% = $110,831.

  9. HAND RECAP: Harman 10 7 . Negreanu K Q . Flop 10u 7 Ku . Harman’s all-in. $156,100 pot. P(Negreanu wins) = 29%. P(Harman wins) = 71%. --------------------------------------------------------------------------- The standard deviation changes a lot! Say they run it once. V(X) = E(X2) - µ2. µ = $110,831, so µ2 ~ $12.3 billion. E(X2) = ($156,1002)(71%) + (02)(29%) = $17.3 billion. So V(X) = $17.3 billion - $12.3 billion = $5 billion. The standard deviation s = sqrt($5 billion) ~ $71,000. So if they run it once, Harman expects to get back about $110,831 +/- $71,000. If they run it twice? Hard to compute, but approximately, if each run were independent, then V(X1+X2) = V(X1) + V(X2), so if X1 = amount she gets back on 1st run, and X2 = amount she gets from 2nd run, then V(X1+X2) ~ V(X1) + V(X2) ~ $1.3 billion + $1.3 billion = $2.6 billion, The standard deviation s = sqrt($2.6 billion) ~ $51,000. So if they run it twice, Harman expects to get back about $110,831 +/- $51,000.

  10. 8) “Unbeatable Texas Holdem Strategy” http://www.freepokerstrategy.com : all in with AK-AT or any pair. P(getting such a hand) = 4 x [16/1326] + 13 x [6/1326] = 4 x 1.2% + 13 x 0.45% = 10.7%. Say you’re dealt 100 hands. Pay ~11 blinds = $55. Expect 10.7 (~ 11) such good hands. Say you’re called by 88-AA, and AK, for $100 on avg. P(player 1 has one of these) = 7 x 0.45% + 1.2% = 4.4%. P(of 8 opponents, someone has one of these) ~ 1 - (95.6%)8 = 30%. So, you win pre-flop 70% of the time. (Say $10 on avg.) = 11 x 70% x $10 = $77 profit. Other 30%, you’re on avg about a 65-35 underdog, so you win 11 x 30% x 35% x $100 = $115.50 lose 11 x 30% x 65% x $100 = $214.50. Total: exp. to win $77 + $115.50 - $55 - $214.50 = -$77/ 100 hands.

  11. 9) Phil Helmuth, Play Poker Like the Pros, Collins, 2003. Strategy for beginners: AA, KK, QQ, or AK. P(getting one of these hands)? 3 x choose(4,2)/choose(52,2) + 4x4/choose(52,2) = 3 x 6/1326 + 16/1326 = 3 x 0.45% + 1.21% = 2.56% = 1 in 39. Say you play $100 NL, table of 9, blinds 2/3, for 39x9 = 351 hands. Pay 5 x 39 = 195 dollars in blinds. Expect to play 9 hands. Say P(win preflop) ~ 50%, and in those hands you win ~ $8. Other 50%, always vs. 1 opponent, 60% to win $100. So, expected winnings after 351 hands = -$195 + 9 x 50% x $8 + 9 x 50% x 60% x $100 + 9 x 50% x 40% x -$100 = -$69. That is, you lose $69 every 351 hands on average = $20 per 100 hands.

  12. 10) Rainbow flop = all 3 cards are of different suits. 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%.

  13. 11) 11/4/05, Travel Channel, World Poker Tour, $1 million Bay 101 Shooting Star. 4 players left, blinds $20,000 / $40,000, with $5,000 antes. Avg stack = $1.1 mil. 1st to act: Danny Nguyen, A 7. All in for $545,000. Next to act: Shandor Szentkuti, A K. Call. Others (Gus Hansen & Jay Martens) fold. (66% - 29%). Flop: 5 K 5 . (tv 99.5%; cardplayer.com: 99.4% - 0.6%). P(tie) = P(55 or A5) = (1 + 2*2) ÷ choose(45,2) = 0.505%. 1 in 198. P(Nguyen wins) = P(77) = choose(3,2) ÷ choose(45,2) = 0.30%. 1 in 330. [Note: tv said “odds of running 7’s on the turn and river are 274:1.” Given Hansen/Martens’ cards, choose(3,2) ÷ choose(41,2) = 1 in 273.3.] Turn: 7. River: 7! * Szentkuti was eliminated next hand, in 4th place. Nguyen went on to win it all.

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