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[ A wins ] = P[ B wins ]? .
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
.
3) Game of tossing own coin.
Each player has their own coin. A wins on HH and B wins on HT (or TH)
Solution of Game (3)
Define
PA ( j ) : probability that A gets HH for the first time at the j -th toss, j = 1, 2,
PB ( j ) : probability that B gets HT for the first time at the j -th toss, j = 1, 2,
QA (k ) 
k
 PA ( j )
and
j =1
QB (k ) 
k
 PB ( j )
j =1
Then
QA (k ) = P [ A gets HH at or before the k -th toss ]
PA (k ) = QA (k ) − QA (k − 1)
Once QA (k ) and QB (k ) are found, the winning probabilities can be found from
P [ A wins ] =
P [ B wins ] =
∞
∞
∞
k =1
∞
k =1
 PA (k ) {PB (k + 1) + PB (k + 2) +  } =  PA (k ) {1 − QB (k )}
 PB (k ) {1 − QA (k )}
k =1
P [tie] =  PA (k ) PB (k )
k =1
2
Q: Find PA ( j ) for j = 1, 2,3, 4.
Sequences to be counted in PA ( 2 )
Sequences to be counted in PA ( 3)
H
T
H
T
PA (k ) is difficult to find, We will find 1 − QA (k ) first.
1st toss
k th toss
2nd
no 'HH' along the path
ending with 'H'
H
T
at least one 'HH' along the path
no 'HH' along the path
ending with 'T'
21
2
2 leaf nodes
Simple Sample Space
Q A ( k ) is the probability of a blue path.
1 − Q A ( k ) is the probability of a red path.
2k
3
Suppose A has never got HH until the k -th toss.
Define
u AH (k ) : the number of such sequences that end with a Head .
uTA (k ) : the number of such sequences that end with a Tail.
Then
1 − QA ( k ) =
u AH (k ) + uTA (k )
2k
Likewise, suppose B has never got HT until the k -th toss.
Define
uBH (k ) : the number of such sequences that end with a Head .
uTB (k ) : the number of such sequences that end with a Tail.
Then
1 − QB (k ) =
u BH (k ) + uTB (k )
2k
Study of A, winning on HH
u AH (1) = {H } = 1
uTA (1) = {T } = 1
u AH (2) = {TH } = 1
uTA (2) = {HT , TT } = 2
u AH (3) = uTA (2)
uTA (3) = u AH (2) + uTA (2)
In general
u AH ( k ) = uTA ( k − 1)
for k = 2,3,
uTA ( k ) = uTA ( k − 1) + u AH ( k − 1)
= uTA ( k − 1) + uTA ( k − 2) for k = 3,4,
uTA (k ) is called the Fibonacci sequence.
uTA (k ) =
k +1
k +1

1− 5  
1  1 + 5 
− 


 
2
5  2 

 

4
Self Study.
Derive the Fibonacci sequence {1,1, 2,3,5,8,} .
x0 = x1 = 1.
xn = xn −1 + xn − 2
for n = 2,3,
Study of B, wining on HT
uBH (1) = {H } = 1
uTB (1) = {T } = 1
uBH (2) = {HH , TH } = 2
uTB (2) = {TT } = 1
In general,
uBH ( k ) = k
QB (k ) =
uTB ( k ) = 1
2k − u BH (k ) − uTB (k )
2k
PB (k ) = QB (k ) − QB (k − 1)
Putting all together
∞
P [ A wins ] =  PA (k ) {1 − QB (k )} =
k =1
P [ B wins ] =
P [tie ] =
∞
∞
 PB (k ) {1 − QA (k )}
k =1
17
 PA (k ) PB (k ) = 121
k =1
=
39
121
65
121
5
0.2
0.18
0.16
0.14
0.12
0.1
P[A win]
0.08
P[B win]
0.06
0.04
0.02
0
0
2
4
6
k
8
10
12