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# Expected value - PowerPoint PPT Presentation

Expected value (µ) = ∑ y P(y). Sample mean ( X ) = ∑X i / n. Sample standard deviation = √[∑(X i - X ) 2 / (n-1)]. iid: independent and identically distributed. Suppose X 1 , X 2 , etc. are iid with expected value µ and sd s ,. LAW OF LARGE NUMBERS : X ---> µ .

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Sample mean (X) = ∑Xi / n

Sample standard deviation = √[∑(Xi - X)2 / (n-1)]

iid: independent and identically distributed.

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

LAW OF LARGE NUMBERS:

X ---> µ .

CENTRAL LIMIT THEOREM:

(X - µ) ÷ (s/√n) ---> Standard Normal.

Estimated as X +/- 1.96 s/√n = .95 +/- 0.28

Important to keep track of results.

* Don’t just track ∑Xi.

Track X +/- 1.96 s/√n .

Make sure it’s converging to something positive.

strategy for beginners: AA, KK, QQ, or AK.

P(getting one of these hands)?

3(4/52)(3/51) + 2/13(4/51) = 1.36% + 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.

“Unbeatable Texas Holdem Strategy”: all in with AK-AT or pair.

P(getting such a hand) = 4 x [16/(52 choose 2)] + 13 x [6/(52 chs 2)]

= 4 x 1.2% + 13 x 0.45%

= 10.7%.

Play 100 times. Expect ~ 11 hands. Pay ~11 blinds = \$55.

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(someone of 8 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.

11/4/05, 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). ]

Turn: 7.

River: 7!

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

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%.

1st to act: Gus Hansen, K 101 Shooting Star. 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%!)

River: 2.

* Turn: A!

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

•   