Question

1. Three players, A,B, and C, each flip their coins until one person has a different result from the others. The person having the different result wins.

(a) Using R, simulate this experiment 10000 times and give the resulting estimate of P(A wins).

Answer 1

# The possible outcomes when three coins are simultaneously flipped by A, B,C are :

Heads/Tails	HHH	HHT	HTH	HTT	THH	THT	TTH	TTT
1/0	111	110	101	100	011	010	001	000

# Let 1=Head=H and 0=Tail=T
#If anyone flips a coin, the probability that head(or tail) shows up is half(1/2).
#Therefore, it is like Bernoulli trial for each individual; with probability of success p=1/2.
# Also, each individual independently flips the coin.
# Bernoulli(p) is equivalent to Binomial(1,p)

p=1/2
A=rbinom(10000,1,p)
B=rbinom(10000,1,p)
C=rbinom(10000,1,p)

#data.frame(A,B,C)
M=cbind(A,B,C)
head(M)
A B C
[1,] 0 1 0
[2,] 1 0 1
[3,] 0 0 0
[4,] 1 1 1
[5,] 1 1 1
[6,] 1 0 1

M[which(A==B & B==C),] ## This will give positions(row number) where elements are '000' or '111'.
head(M[which(A==B & B==C),])
# A B C
#[1,] 0 0 0
#[2,] 1 1 1
#[3,] 1 1 1
#[4,] 1 1 1
#[5,] 0 0 0
#[6,] 0 0 0
length(which(A==B & B==C)) ## This total number of outcomes where nobody wins
#[1] 2505

# which(A!=B & B==C)
M[which(A!=B & B==C),] ## This will give positions(row number) where elements are '011' or '100'. i.e. Position when A wins.
unique(M[which(A!=B & B==C),])
# A B C
#[1,] 0 1 1
#[2,] 1 0 0
head(M[which(A!=B & B==C),])
# A B C
#[1,] 0 1 1
#[2,] 1 0 0
#[3,] 0 1 1
#[4,] 1 0 0
#[5,] 0 1 1
#[6,] 1 0 0
length(which(A!=B & B==C))
[1] 2542

# Estimate of Probability that A wins:
# Using fundamental definition of probability
# Probabilty= (number of favourable outcomes in an event) / (Total number of outcomes)
# Total outcomes=10000

PA=length(which(A!=B & B==C))/10000
PA
#[1] 0.2542

PB=length(which(B!=A & A==C))/10000
PB
#[1] 0.2504
PC=length(which(C!=A & A==B))/10000
PC
#[1] 0.2449
lenNO=length(which(A==B & B==C))
lenNO
#[1] 2505
#P(Nobody wins)
lenNO/10000
#[1] 0.2505
1-(PA+PB+PC)
#[1] 0.2505