(a)

library(epiDisplay)
GambRuinFunc=function(A,B,p)
{
  Rec=c(A)
  while(A>0) {
    ProbVal=runif(1)
    if(ProbVal <= p)
    {
      B = A+B
      A = (B+A)*0.1
    }else{
      B = B-A
      A = A
      }
    Rec = c(Rec,A)
    if(B<A){break}
  }
  Duration = length(Rec)
  return(Duration)
}
set.seed(1)
Durations = numeric()
M = 5000
for(i in 1:M)
{
  NextDuration = GambRuinFunc(A=100,B=900,p=18/38)
  Durations = c(Durations,NextDuration)
}
prob100<-length(which(Durations<=100))/5000
prob100

## [1] 0.173

The probability that you have “busted” (lost all your money) by the time you have placed your one hundredth bet is 0.173.

(b)

prob500<-length(which(Durations<=500))/5000
prob500

## [1] 0.63

The probability that you have busted by the time you have placed your five hundredth bet by simulating the outcome directly is 0.6296.

(c)

d5000<-Durations[which(Durations<=5000)]
mean(d5000)

## [1] 505

the mean time you go bust, given that you go bust within the first 5000 hand is 505.1114.

(d)

library(epiDisplay)
GambRuinFunc1=function(A,B,p)
{
  Rec=c(A)
  while(A>0) {
    ProbVal=runif(1)
    if(ProbVal <= p)
    {
      B = A+B
      A = (B+A)*0.1
    }else{
      B = B-A
      A = A
      }
    Rec = c(Rec,A)
    if(length(Rec)==100){break}
    if(B<A){break}
  }
  Duration = length(Rec)
  return(B)
}
set.seed(1)
m100 = numeric()
M = 5000
for(i in 1:M)
{
  nextm = GambRuinFunc1(A=100,B=900,p=18/38)
  m100 = c(m100,nextm)
}
mean(m100)

## [1] 469

var(m100)

## [1] 849960

The mean of your bankroll after 100 hands (including busts) is 469.1183; The variance of your bankroll after 100 hands (including busts) is 849960.

(e)

library(epiDisplay)
GambRuinFunc2=function(A,B,p)
{
  Rec=c(A)
  while(A>0) {
    ProbVal=runif(1)
    if(ProbVal <= p)
    {
      B = A+B
      A = (B+A)*0.1
    }else{
      B = B-A
      A = A
      }
    Rec = c(Rec,A)
    if(length(Rec)==500){break}
    if(B<A){break}
  }
  Duration = length(Rec)
  return(B)
}
set.seed(1)
m500 = numeric()
M = 5000
for(i in 1:M)
{
  nextm = GambRuinFunc2(A=100,B=900,p=18/38)
  m500 = c(m500,nextm)
}
mean(m500)

## [1] 32.9

var(m500)

## [1] 95787

The mean of your bankroll after 500 hands (including busts) is 32.88478; The variance of your bankroll after 500 hands (including busts) is 95786.5.

(a)

Is this a fair game? Simulate one hundred thousand sequential hands to determine the size of your return. Then repeat this simulation 99 more times to get a range of values to calculate the expectation.

MCFunc=function(A,B,p)
{
  k=0
  prob<-p
  while(A>0) {
    ProbVal=runif(1)
    if(ProbVal <= p)
    {
      B = B+A
      p = p+0.01
    }else{
      B = B-A
      p = prob
    }
    k=k+1
    if(k==100000){break}
    if(B<A){break}
  }
  return(B)
}

size = numeric()
M = 100
for(i in 1:M)
{
  nextb = MCFunc(A=100,B=900,p=0.48)
  size = c(size,nextb)
}
mean(size)

## [1] 0

The expectation of the size of return is 0, which means one will go busted after one hundred thousand sequential hands, so the game is not fair.

(b)

Repeat this process but change the starting probability to a new value within 2% either way. Get the expected return after 100 repetitions. Keep exploring until you have a return value that is as fair as you can make it. Can you do this automatically?

set.seed(1)
prop = numeric()
for (j in 1:20) {
size = numeric()
M = 100
for(i in 1:M)
{
  nextb = MCFunc(A=100,B=900,p=0.46+0.002*j)
  size = c(size,nextb)
}
nextprop = length(which(size==0))/100
prop<-c(prop,nextprop)
}
prop

##  [1] 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
## [15] 0.96 0.94 0.88 0.80 0.71 0.67

We found that if we choose the starting probability as 0.5, the proportion of size of return equals to 0 is closest to 0.5, which means the game is fairest under the starting probability of 0.5.

(c)

Repeat again, keeping the initial probability at 48%, but this time change the probability increment to a value different from 1%. Get the expected return after 100 repetitions. Keep changing this value until you have a return value that is as fair as you can make it.

MCFunc1=function(A,B,p,step)
{
  k=0
  prob<-p
  while(A>0) {
    ProbVal=runif(1)
    if(ProbVal <= p)
    {
      B = B+A
      p = p+step
    }else{
      B = B-A
      p = prob
    }
    k=k+1
    if(k==100000){break}
    if(B<A){break}
  }
  return(B)
}

set.seed(1)
prop = numeric()
for (j in 1:20) {
size = numeric()
M = 100
for(i in 1:M)
{
  nextb = MCFunc1(A=100,B=900,p=0.48,step=0.002*j)
  size = c(size,nextb)
}
nextprop = length(which(size==0))/100
prop<-c(prop,nextprop)
}
prop

##  [1] 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 0.94 0.92 0.74 0.70 0.67 0.60
## [15] 0.57 0.57 0.50 0.39 0.38 0.34

0.002*17

## [1] 0.034

We found that if we choose the probability increment to 0.034, the proportion of size of return equals to 0 is 0.5, which means the game is fair under the probability increment of 0.034.

For Game 1 (Change the starting probability)

bestp1<-0.5
size1 = numeric()
M = 100
for(i in 1:M)
{
  nextb = MCFunc(A=100,B=900,p=bestp1)
  size1 = c(size1,nextb)
}
###Bootstrap
set.seed(1)
bs1<-sample(size1,10000,replace = TRUE)
mean(bs1)

## [1] 56836

var(bs1)

## [1] 10785467594

For Game 2 (Change the probability increment)

bestp2<-0.034
size2 = numeric()
M = 100
for(i in 1:M)
{
  nextb = MCFunc1(A=100,B=900,p=0.48,step=bestp2)
  size2 = c(size2,nextb)
}
set.seed(1)
bs2<-sample(size2,10000,replace = TRUE)
mean(bs2)

## [1] 5449959

var(bs2)

## [1] 2.47e+13

var(bs1)/var(bs2)

## [1] 0.000437

The variance for game1 is 10785467594, the variance for game1 is 2.469009e+13. The proportion between the variance of two games’ sizes of return (0.0004368338) shows that the first game (Change the starting probability) has the smaller variance in returns.