Stat 35b: Introduction to Probability with Applications to - PowerPoint Presentation

Slide1

Stat 35b: Introduction to Probability with Applications to Poker

Outline for the day:E(X+Y) = E(X) + E(Y).Harman/Negreanu.Running a hand multiple times, expected value and variance.Geometric random variables.Negative binomial random variables.Midterm is Feb 21, in class. 50 min. Open book plus one page of notes, double sided. Bring a calculator!





u







u

Slide2

1

) E(X+Y) = E(X) + E(Y). pp126-127. A fact given in ch.7 is that E(X+Y) = E(X) + E(Y), for any random variables X and Y, whether X & Y are independent or not, as long as E(X) and E(Y) are finite. 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. E(X) = ?

Note: not independent! If you have AA, then it’s unlikely anyone else does too.

Nevertheless, Let X

1

= 1 if player #1 has AA, and 0 otherwise.

X

2

= 1 if player #2 has AA, and 0 otherwise, and so on.

Then X = X

1

+ X

2

+ … + X

10

. So E(X) = E(X

1

) + E(X

2

) + … + E(X

10

)

= 0.00452 + 0.00452 + … + 0.00452 = 0.0452.Slide3

Deal the cards face up, without reshuffling.

Let Z = the number of cards til the 2nd king. What is E(Z)? I’ll answer this next time. The solution uses the fact E(X+Y+Z + ...) = E(X) + E(Y) + E(Z) + ....  u  



u

Slide4

2) Harman and Negreanu,

3) Running a hand multiple times, E(X) and V(X). Farha vs. Antonius too. E(X+Y) = E(X) + E(Y). Whether X & Y are independent or not! And, if X & Y are independent, then V(X+Y) = V(X) + V(Y). so SD(X+Y) = √[SD(X)^2 + SD(Y)^2]. Also, if Y = 9X, then E(Y) = 9E(Y), and SD(Y) = 9SD(X). V(Y) = 81V(X).Farha vs. Antonius. Running it 4 times. Let X = chips you have after the hand. Let p be the prob. you win.X = X1 + X2 + X3 + X4, where X

1

= chips won from the first “run”, etc.

E(X) = E(X

1

) + E(X

2

) + E(X

3

) + E(X

4

)

= 1/4 pot (p) + 1/4 pot (p) + 1/4 pot (p) + 1/4 pot (p)

= pot (p)

= same as E(Y), where Y = chips you have after the hand if you ran it once!

But the SD is smaller: clearly X

1

= Y/4, so SD(X

1

) = SD(Y)/4. So, V(X

1

) = V(Y)/16.

V(X) ~ V(X

1

) + V(X

2

) + V(X

3

) + V(X

4

),

= 4 V(X

1

)

= 4 V(Y) / 16

= V(Y) / 4.

So SD(X) = SD(Y) / 2. Slide5

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) = 28.69%. P(Harman wins) = 71.31%. Let X = amount Harman has after the hand.If they run it once, E(X) = $0 x 29% + $156,100 x 71.31% = $111,314.90.

If they run it twice, what is E(X)?

There’s some probability p

1

that Harman wins both times ==> X = $156,100.

There’s some probability p

2

that they each win one ==> X = $78,050.

There’s some probability p

3

that Negreanu wins both ==> X = $0.

E(X) = $156,100 x p

1

+ $78,050 x p

2

+ $0 x p

3

.

If the different runs were

independent,

then p

1

= P(Harman wins 1st run & 2nd run)

would = P(Harman wins 1st run) x P(Harman wins 2nd run) = 71.31% x 71.31% ~ 50.85%.

But, they’re not quite independent! Very hard to compute p

1

and p

2

.

However, you don’t need p

1

and p

2

!

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 71.31% + $0 x 28.69% + $78,050 x 71.31% + $0 x 28.69% =

$111,314.90.Slide6

HAND RECAP: Harman 10

 7 . Negreanu K Q . Flop 10u 7 Ku . Harman’s all-in. $156,100 pot. P(Negreanu wins) = 28.69%.P(Harman wins) = 71.31%. ---------------------------------------------------------------------------

The standard deviation (SD) changes a lot!

Say they run it once

.

(see p127.)

V(X) = E(X

2

) - µ

2

.

µ = $111,314.9, so µ

2

~ $12.3 billion.

E(X

2

) = ($156,100

2

)(71.31%) + (0

2

)(28.69%) = $17.3 billion.

V(X) = $17.3 billion - $12.3 bill. = $5.09 billion. SD

s

= sqrt($5.09 billion)~$71,400.

So if they run it once, Harman expects to get back about $111,314.9

+/- $71,400

.

If they run it twice?

Hard to compute, but approximately, if each run were independent, then V(X

1

+X

2

) = V(X

1

) + V(X

2

),

so if X

1

= amount she gets back on 1st run, and X

2

= amount she gets from 2nd run,

then V(X

1

+X

2

) ~ V(X

1

) + V(X

2

) ~ $1.25 billion + $1.25 billion = $2.5 billion,

The standard deviation

s

= sqrt($2.5 billion) ~ $50,000.

So if they run it twice, Harman expects to get back about $111,314.9

+/- $50,000.Slide7

4. Geometric Random Variables

, ch 5.3.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) = p

1

q

11

= 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.