Random Walks, Gambler’s Ruin, and the Quest for Jack or Jillpersonal.psu.edu/dsr11/talks/cosmostalk.pdf · 2011. 4. 19. · Random Walks, Gambler’s Ruin,and the Quest for Jack

Random Walks, Gambler’s Ruin,and the Quest for Jack or Jill

Donald Richards

Penn State University

http://www.stat.psu.edu/∼richards

Random Walks, Gambler’s Ruin,and the Quest for Jack or Jill – p. 1/31

http://www.psu.edu/

http://www.stat.psu.edu/%7Erichards

Take a look around the room



Estimate the proportion of males-females in the audience




Most of you will probably guess: About 50-50





Guess the male-female proportion for newborns in the U.S.











Sex ratio :=The number of males

The number of females× 100








What is the sex ratio for U.S. newborns?








What is the sex ratio for U.S. newborns?

Would you guess: Approximately 100?


1940–2002: More boys than girls have been born in the U.S.


http://www.cdc.gov/nchs/data/nvsr/nvsr53/nvsr53_20.pdf


Here is a graph of the U.S. sex ratio from 1940–2002





In 2002, the U.S. sex ratio was 104.8






We estimate the probability that a newborn is male to be













104.8

104.8 + 100= 0.512, or 51.2%







104.8

104.8 + 100= 0.512, or 51.2%

Demographers study the consequences of “son preference”in many countries



Two childbearing schemes

Scheme #1:

Families continue to have children until a son arrives.

How many children will the “average” family have?

1

0.512, or 1.95

The average family will have about 1 boy and 0.95 girl


Scheme #2:

Families have children until #boys > #girls



Scheme #2:



1

(2 × 0.512) − 1, or 43


Scheme #2:



1

(2 × 0.512) − 1, or 43

The average family will have about 22 boys and 21 girls!


Scheme #2:



1

(2 × 0.512) − 1, or 43

The average family will have about 22 boys and 21 girls!

Why does a small difference in the childbearing schemeslead to a large difference in the average outcome?


An easy problem in basic probability

Annie tosses a penny repeatedly until the first head shows up

N : The number of tosses needed

Pr(N = 1) = Pr(H) =1

2

Pr(N = 2) = Pr(TH) =1

4

Pr(N = 3) = Pr(TTH) =1

8, etc.

If a billion people each toss a penny until head shows up then

About 1

2of them will need 1 toss

About 1

4of them will need 2 tosses

About 1

8of them will need 3 tosses, etc.


How many tosses will the “average” person need?

(1

2× 1

)+

(1

4× 2

)+

(1

8× 3) · · · = 2

Geometric series!∞∑

k=1

k ·(1

2

)k= 2

The expected value of N :

E(N) =∑

all k

k · Pr(N = k)

where the sum is over all possible values of N


Annie is given a new, unfair coin

Pr(H) = p, Pr(T ) = q where q = 1 − p

N = Number of tosses needed to get the first head

Pr(N = 1) = Pr(H) = p

Pr(N = 2) = Pr(TH) = qp

Pr(N = 3) = Pr(TTH) = q · q · p = q2p, etc.

In general,

Pr(N = k) = q · q · · · q︸︷︷︸

k−1

·p = qk−1p


The expected value of N :

E(N) =

∞∑

k=1

k · Pr(N = k)

=

∞∑

k=1

k · qk−1p =1

p

p = 1

2: E(N) = 2

p = 1

5: E(N) = 5

p = 1

10: E(N) = 10


Coin tossing and childbearing are similar probabilisticprocesses!

Coins and childbearing seem to have no memory

Childbearing Scheme #1:

Families continue to have children until a son arrives.How many children will the “average” family have?

For U.S. families in 2002, p = 0.512. Therefore

E(N) =1

0.512= 1.95

The average family will have about 1 boy and 0.95 girl


Childbearing Scheme #2:



N = Number of children in a randomly chosen family

N is an odd number

B = G + 1 so N = G + B = G + G + 1 = 2G + 1

In each family, the last two children are boys

For simplicity, suppose that Pr(B) = Pr(G) = 1/2


Pr(N = 1) = Pr(B) =1

2

Pr(N = 3) = Pr(GBB) =1

8

Pr(N = 5) = Pr(GGBBB or GBGBB)

=(1

2

)5+

(1

2

)5= 2 ·

(1

2

)5

What about Pr(N = 7)?

Suppose a family has 4 boys and 3 girls in the sequenceGGBBGBB

We describe this using a diagram


• • • • •

• • • • •

• • • • •

• • • • •

• • • • •

x (girls)

y (boys)

- -

6

6-

6

6

��

��

��

��

��

��

��

�

A family with outcome GGBBGBB

There are 5 ways to walk from (0,0) to (3,3) insidethe region x ≥ y ≥ 0.


Therefore Pr(N = 7) = 5 ·(

1

2

)7

W (j, k): The number of ways to walk from (0, 0)to (j, k) inside the region x ≥ y ≥ 0

W (3, 3) = 5

We need a general formula for W (k, k)

W (1, 0) = 1, W (1, 1) = 1

W (2, 2) = W (2, 1), W (3, 3) = W (3, 2)

A pattern emerges: W (k, k) = W (k, k − 1)


How about W (j, k)?

W (2, 1) = W (2, 0) + W (1, 1)

In general,

W (j, k) = W (j, k − 1) + W (j − 1, k), j > k

Solve these recurrence relations using mathematical induction:

W (k, k) =(2k)!

(k + 1)! k!

Check: W (3, 3) =6!

4! 3!=

720

24 × 6= 5


Proceed as before:

Pr(N = 2k + 1) =1

2· · ·

1

2︸︷︷︸

k+1 B′s

·1

2· · ·

1

2︸︷︷︸

k G′s

·W (k, k)

Conclusion:

Pr(N = 2k + 1) =(1

2

)2k+1W (k, k)


The expected family size:

E(N) =

∞∑

k=0

(2k + 1) · Pr(N = 2k + 1)

=

∞∑

k=0

(2k + 1) ·(1

2

)2k+1W (k, k)

Bad news: The terms in the series increase very quickly,and the series “blows up” (diverges)

E(N) = ∞

If Pr(H) ≤ 1

2then the average family size is infinite!


Suppose Pr(Success) = p

p Game0.512 Male baby in 20020.5 Coin tossing0.492 Craps0.438 Roulette (always betting on red)0.25 Keno (single number bet)

1

146×106 Pennsylvania Powerball Jackpot

www.palottery.state.pa.us/lottery

To win the PA Powerball Jackpot more times than lost,an average player must buy infinitely many tickets!


http://www.palottery.state.pa.us/lottery/cwp/view.asp?q=457089

The moral of the story

Avoid any casino game if Pr(Success) ≤ 1

2

What if Pr(Success) > 1

2?

Recall that

Pr(N = 2k + 1) = p · · · p︸︷︷︸

k+1 wins

q · · · q︸︷︷︸

k losses

·W (k, k)

= pk+1qk W (k, k)

E(N) =

∞∑

k=0

(2k + 1) · Pr(N = 2k + 1)


Using college-level mathematics, we discover that:

If p > 1

2then

E(N) =1

2p − 1

If p = 0.512 then

E(N) =1

(2 × 0.512) − 1≃ 43

If U.S. families decide to have more boys than girls thenthe average family will have about 22 boys and 21 girls!


Is the World a Giant Gambling Machine?

Craps Casino poker Wheel of FortuneRoulette Keno Slot machines

Spanish 21 Bingo Texas Hold ’EmBlackjack Faro Stud pokerBaccarat Let it Ride Video pokerBoxing Basketball Horse racingFootball Hockey Dog racingSoccer Cockfights Jai AlaiStocks Currencies BondsOptions Futures Derivatives

Mutual funds ETFs MortgagesPork bellies FCOJ Heating oilElectricity Cocoa Natural gasLive cattle Cotton GasolineSoybeans Coffee Wheat

Sugar Oats PalladiumCrude oil Corn PlatinumCopper Gold SilverPropane Rice Jet fuel


The Importance of Expected Value

Roulette wheel: 0, 00, 1, 2, 3, 4, 5, . . . , 34, 35, 36

A gambler spends a day betting $1 on 6 (what a dummy)

Result Win Lose

Probability 1/38 37/38

Profit $35 -$1

Expected profit =(35 × 1

38

)+

(− 1 × 37

38

)= − 2

38= −0.053

The average gambler loses 5.3% of his money per bet on 6


Better luck with craps?

Result Win Lose

Probability 0.492 0.507

Profit $1 -$1

Expected profit =(1 × 0.492) +

(− 1 × 0.507

)= −0.014

The average gambler loses 1.4% of his money per game


The 13 most important words in casino gambling

All Casino Games HaveNegative Expected Value∗

∗And Don’t You Ever Forget It!

See Robert de Niro in the movie “Casino”

“Expected value” applies to many people playing once orto one person playing often

Perhaps, one person can play often and avoid going bust?


The Gambler’s Ruin Problem

Annie has $a and Bobbie has $b. They decide to bet onrepeated tosses of a coin.

On each toss, Annie collects $1 from Bobbie if the coin showsheads. If the coin shows tails then Annie pays $1 to Bobbie.

What is the probability that Annie goes bust?

James Bernoulli (1654–1705), a brilliant thinker

ca := Pr(Annie goes bust)

Bernoulli worked out recurrence relations for ca


Let p = Pr(Heads), q = Pr(Tails)

Bernoulli proved that if p 6= 1

2then

Pr(Annie goes bust) =1 −

(pq

)b

1 −(

pq

)a+b

If p = 1

2then

Pr(Annie goes bust) =b

a + b


Game p

Coin tossing 0.5

Craps 0.492

Roulette (red) 0.448

PA Powerball Jackpot1

146 × 106≃ 0


Annie goes to Las Vegas

You are Annie; your dad gave you $106 to play craps

a =$106 (One Million Dollars!)

Bobbie is Las Vegas; she has $109

b =$109 (One Billion Dollars!)

Pr(You go bust eventually) =1 −

(.492.507

)109

1 −(

.492

.507

)106+109≃ 100%

In every casino game,

Pr(You go bust eventually from repeated play) ≃ 100%


Annie vs. The Rest of the World

Annie: A hedge fund with a weakness for rapid trading in theGiant Gambling Machine

She has $109 to trade anything and everything

Annie’s probability of winning on each trade is (at most!) .492

Bobbie: The Rest of the World; she has $1014

Pr(Annie goes bust eventually) =1 −

(.492.507

)1014

1 −(

.492

.507

)109+1014≃ 100%

Conclude: Lots of rapid-trading hedge funds will go busteventually

Their weakness: An addiction to high-speed trading


Jack Binion, owner of Binion’s Horseshoe Casino:



“As long as you have no weaknesses,Las Vegas is an easy place to live. Butif you have a weakness, we’ll find it.” ∗



“As long as you have no weaknesses,Las Vegas is an easy place to live. Butif you have a weakness, we’ll find it.” ∗

∗ And when we find it, ...


People you need to thank!



Abby Thompson



Abby Thompson

Jennifer Judkins



Abby Thompson

Jennifer Judkins

The COSMOS faculty, staff, and RAs



Abby Thompson

Jennifer Judkins

The COSMOS faculty, staff, and RAs

Your parents (Go call your mother!)


Documents

Random Walks, Gambler’s Ruin, and the Quest for Jack or Jillpersonal.psu.edu/dsr11/talks/cosmostalk.pdf · 2011. 4. 19. · Random Walks, Gambler’s Ruin,and the Quest for Jack