stat 35b: introduction to probability with applications to poker poker code competition: all-in or...

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Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.

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Page 1: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

Stat 35b: Introduction to Probability with Applications to Poker

Poker Code competition: all-in or fold.

Page 2: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

5) P(SOMEONE has AA, given you have KK)? Out of your 8 opponents?Note that given that you have KK, P(player 2 has AA & player 3 has AA)

= P(player 2 has AA) x P(player 3 has AA | player 2 has AA) = choose(4,2) / choose(50,2) x 1/choose(48,2)= 0.0000043, or 1 in 230,000.

So, very little overlap! Given you have KK,P(someone has AA) = P(player2 has AA or player3 has AA or … or pl.9 has AA)~ P(player2 has AA) + P(player3 has AA) + … + P(player9 has AA)= 8 x choose(4,2) / choose(50,2) = 3.9%, or 1 in 26. -----------What is exactly P(SOMEONE has an Ace, given you have KK)? (8 opponents)

(or more than one ace)P(SOMEONE has an Ace) = 100% - P(nobody has an Ace).P(nobody has an Ace) = P(pl2 doesn’t have one & pl.3 doesn’t & … & pl.9 doesn’t) = P(pl.2 doesn’t) x P(pl.3 doesn’t | pl.2 doesn’t) x … x P(pl.9 doesn’t | 2-8 don’t) = choose(46,2)/choose(50,2) x choose(44,2)/choose(50,2) x … x ch(32,2)/ch(50,2) = 1.6%. So P(SOMEONE has an Ace) = 98.4%.

River: 7!

Page 3: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

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 or 5A)

= (2/45 x 1/44) + (2/45 x 2/44) + (2/45 x 2/44) = 0.505%. 1 in 198. P(Nguyen wins) = P(77) = 3/45 x 2/44 = 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, 3/41 x 2/40 = 1 in 273.3). ]

* Szentkuti was eliminated next hand, in 4th place. Nguyen went on to win it all.

Turn: 7. River: 7!

Page 4: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

11/4/05, Travel Channel, World Poker Tour, $1 million Bay 101 Shooting Star.

3 players left, blinds $20,000 / $40,000, with $5,000 antes. Avg stack = $1.4 mil.

(pot = $75,000)

1st to act: Gus Hansen, K 9. Raises to $110,000. (pot = $185,000)

Small blind: Dr. Jay Martens, A Q. Re-raises to $310,000. (pot = $475,000)

Big blind: Danny Nguyen, 7 3. Folds.

Hansen calls. (tv: 63%-36%.) (pot = $675,000)

Flop: 4 9 6. (tv: 77%-23%; cardplayer.com: 77.9%-22.1%)

P(no A nor Q on next 2 cards) = 37/43 x 36/42 = 73.8%

P(AK or A9 or QK or Q9) = (9+6+9+6) ÷ (43 choose 2) = 3.3%

So P(Hansen wins) = 73.8% + 3.3% = 77.1%. P(Martens wins) = 22.9%.

Page 5: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

1st to act: Gus Hansen, K 9. Raises to $110,000. (pot = $185,000)

Small blind: Dr. Jay Martens, A Q. Re-raises to $310,000. (pot = $475,000)

Hansen calls. (pot = $675,000)

Flop: 4 9 6. P(Hansen wins) = 77.1%. P(Martens wins) = 22.9%.

Martens checks. Hansen all-in for $800,000 more. (pot = $1,475,000)

Martens calls. (pot = $2,275,000)

Vince Van Patten: “The doctor making the wrong move at this point. He still can get lucky of course.”

Was it the wrong move?

His prob. of winning should be ≥ $800,000 ÷ $2,275,000 = 35.2%.

Here it was 22.9%. So, if Martens knew what cards Hansen had, he’d be making the wrong move. But given all the possibilities, it seems very reasonable to assume he had a 35.2% chance to win. (Harrington: 10%!)

* Turn: A! River: 2.

* Hansen was eliminated 2 hands later, in 3rd place. Martens then lost to Nguyen.

Page 6: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

LLNCLT, Variance and prop bets.

Rainbow board.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%.

Page 7: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

Expected value (µ) = ∑ y P(y)Sample mean (X) = ∑Xi / nSample standard deviation = √[∑(Xi - X)2 / (n-1)]iid: independent and identically distributed.

Suppose X1, X2 , etc. are iid with expected value µ and sd ,

LAW OF LARGE NUMBERS: X ---> µ .

CENTRAL LIMIT THEOREM:(X - µ) ÷ (/√n) ---> Standard Normal.

Page 8: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

95% between -1.96 and 1.96

Page 9: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

Truth: -49 to 51, exp. value = 1.0

Page 10: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

Truth: -49 to 51, exp. value = 1.0Estimated as X +/- 1.96 /√n = .95 +/- 0.28

Page 11: Stat 35b: Introduction to Probability with Applications to Poker Poker Code competition: all-in or fold.   u

* Poker has high standard deviation. Important to keep track of results.

* Don’t just track ∑Xi.

Track X +/- 1.96 /√n .

Make sure it’s converging to something positive.