Transcript E(X+Y)

Stat 35b: Introduction to Probability with Applications to Poker
Outline for the day:
1. E(X+Y)
2. Negreanu / Harman, and “running it twice”
3. Helmuth’s strategy
4. “Unbeatable” strategy
5. WPT example
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1) 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.
Example 2: Play 1 hand, against 9 opponents. X = total # of pocket aces at the table.
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.
2) Harman / Negreanu, and 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.
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.
3) 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.
4) “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.
11/4/05, 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.