1

Coin Tossing Games

1) Game 1 of tossing one coin.

Keep tossing a coin until A or B wins. A wins on contiguous HH, and B wins on HT.

Q: [ ] [ ]?P Awins P Bwins= .

2) Game 2 of tossing one coin.

Keep tossing a coin until A or B wins. A wins on contiguous HH, and B wins on TH.

Q: [ ] [ ]?P Awins P Bwins=

3) Game of tossing own coin.

Each player has their own coin. A wins on contiguous HH, and B wins on HT.

Q: [ ] [ ]?P Awins P Bwins=

Solution of Game 3

Define

( ) : probability gets for the first time at the -th toss, 1,2,

( ) : probability gets for the first time at the -th toss, 1,2,A

B

P j A HH j j

P j B HT j j

==

[ ]

1 1

( ) ( ) and ( ) ( )

Then

( ) gets at or before the -th toss

( ) ( ) ( 1)

k k

A A B Bj j

A

A A A

Q k P j Q k P j

Q k P A HH k

P k Q k Q k

= =

== − −

[ ]Likewise

( ) gets at or before the -th toss

( ) ( ) ( 1)B

B B B

Q k P B HT k

P k Q k Q k

== − −

[ ] { } { }

[ ] { }

[ ]

1 1

1

1

Once ( ) and ( ) are found, the winning probabilities can be found from

( ) ( 1) ( 2) ( ) 1 ( )

( ) 1 ( )

( ) ( )

A B

A B B A Bk k

B Ak

A Bk

Q k Q k

P A wins P k P k P k P k Q k

P B wins P k Q k

P tie P k P k

∞ ∞

= =∞

=∞

=

= + + + + = −

= −

=

2

Find ( ) for 1, 2,3, 4.AP j j =

A general value of ( ) is difficult to find. We will find 1 ( ) instead.A AP k Q k−

( )paths to be counted for 2AP

H

T

( )paths to be counted for 3AP

H

T

( )paths to be counted for 4AP

H

T

3

Suppose has never got until the -th toss.

Define

( ) : the number of such sequences that end with a .

( ) : the number of such sequences that end with a .

Then

( ) ( )1 ( )

2

HA

TA

H TA A

A k

A HH k

u k Head

u k Tail

u k u kQ k

+− =

Likewise, suppose has never got until the -th toss.

Define

( ) : the number of such sequences that end with a .

( ) : the number of such sequences that end with a .

Then

( )1 ( )

HB

TB

H TB B

B

B HT k

u k Head

u k Tail

u k uQ k

+− = ( )

2kk

0222 2k

H

T

1 tossst 2nd

at least one 'HH' along the path

tossthk

no 'HH' along the pathending with 'H'

no 'HH' along the pathending with 'T'

( ) is the probability of a blue path.AQ kLeaf nodes form a simple sample space.

( )1 is the probability of a red path.AQ k−

4

Study of A, no HH up to the k-th toss

(1) { } 1 (1) { } 1

(2) { } 1 (2) { , } 2

(3) (2) (3) (2) (2)

( ) ( 1) for 2,3,

( ) ( 1) ( 1)

( 1) ( 2) for 3,4,

H TA A

H TA A

H T T H TA A A A A

H TA A

T T HA A A

T TA A

u H u T

u TH u HT TT

u u u u u

In general

u k u k k

u k u k u k

u k u k k

= = = =

= = = =

= = +

= − =

= − + −

= − + − =

Such a sequence is called the Fibonacci sequence.

1 11 1 5 1 5

( )2 25

k kTAu k

+ + + − = −

2 ( ) ( )( )

2

( ) ( ) ( 1)

k H TA A

A k

A A A

u k u kQ k

P k Q k Q k

− −=

= − −

Self Study.

{ }0 1

1 2

Derive the Fibonacci sequence 1,1,2,3,5,8, .

1.

for 2,3,n n n

x x

x x x n− −

= =

= + =

5

Study of B, no HT up to the k-th toss

(1) 1 (1) 1{ } { }

(2) 2 (2) 1{ , } { }

In general,

( ) ( ) 1

H TB B

H TB B

H TB B

u uH T

u uHH TH TT

u k k u k

= = = =

= = = =

= =

2 ( ) ( )( )

2

( ) ( ) ( 1)

k H TB B

B k

B B B

u k u kQ k

P k Q k Q k

− −=

= − −

Putting all together

[ ] { }1

39( ) 1 ( )

121A Bk

P A wins P k Q k∞

== − =

[ ] { }1

65( ) 1 ( )

121B Ak

P B wins P k Q k∞

== − =

[ ]

1

17( ) ( )

121A Bk

P tie P k P k∞

== =

6

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0 2 4 6 8 10 12

k

P[A win]

P[B win]