Some problems related to Dice

Debartha Paul

28 mins ago

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It was when I and one of my friends, Ahel, were playing a game of Ludo, that an idea struck both of our heads. Being final year undergraduate students of Statistics, both of us pondered upon the question, what if we can do some verification for some questions on dice. This brief spark in our brains led us to properly formulate the questions, think and write the solutions using mathematical concepts, and finally simulate them. Note that we did all the simulations using the R programming language. We also included the animation package in R for some visualisations of our graph.

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	rolls<-function() #function
	{
	x<-sample(1:6,1) #roll a die once
	i=1 #increase the counter
	while(x!=6) #until the roll is not a six
	{
	x<-sample(1:6,1) #continue rolling
	i=i+1 #and increasing the counter
	}
	return(i) #return the counter after the 6
	}

	rolls<-function() #function
	{
	j=0 #counter to check if the current roll is 6
	x<-sample(1:6,1) #roll once
	i=1 #increase the counter
	if(x==6) #if the roll is a six
	{
	j=1 #set this counter to 1
	x<-sample(1:6,1) #roll again
	i=i+1 #increase the counter
	if(x!=6) j=0 #if the roll is not 6, set j back to 0
	}
	while(j!=1) #while j is 0
	{
	x<-sample(1:6,1) #roll once
	i=i+1 #increase the roll counter
	if(x==6) #if the roll is a six
	{
	j=1 #set this counter to 1
	x<-sample(1:6,1) #roll again
	i=i+1 #increase the counter
	if(x!=6) j=0 #if the roll is not 6, set j back to 0
	}
	}
	return(i) #return the roll counter
	}

	rolls<-function()
	{
	j=0 #counter to check if the sequence is 65
	x<-sample(1:6,1) #roll once
	i=1 #increase the roll counter
	while(x==6) #while the roll results in a 6
	{
	j=1 #set this counter to 1
	x<-sample(1:6,1) #roll once again
	i=i+1 #increase the roll counter
	if(x!=5) j=0 #if that is not 5, replace j with 0
	}
	while(j!=1) #while j is not 1
	{
	x<-sample(1:6,1) #roll once
	i=i+1 #increase the roll counter
	while(x==6) #is the roll is 6
	{
	j=1 #set j as 1
	x<-sample(1:6,1) #roll again
	i=i+1 #if this roll is not 5
	if(x!=5) j=0 #set j back to 0
	}
	}
	return(i) #return the roll counter
	}

	rolls<-function() #function
	{
	x<-sample(1:6,1);y<-sample(1:6,1) #roll twice
	i=2 #set the roll counter
	while(abs(x–y)!=1) #while their absolute difference is not 1
	{
	x<-y;y<-sample(1:6,1) #replace the second roll with the first one and roll again
	i=i+1 #increase the counter
	}
	return(i) #return no. of rolls
	}

	rolls<-function(n) #function
	{
	x<-sample(1:6,n,replace=T) #roll n times
	if(length(unique(x))==6) return(1) #if the number of unique outcomes is 1
	#return 1
	else return(0) #else return 0
	}

	rolls<-function(n) #function
	{
	j=0 #set the counter for subsequence to 0
	x<-sample(1:6,n,replace=T) #roll n times
	y<-which(x==1) #see which rolls are 1
	for(i in y) #for every such rolls
	{
	if(n–i<5) break #if the index goes out of bounds, break
	else if((x[i+1]==2)&&(x[i+2]==3)&&(x[i+3]==4)&&(x[i+4]==5)&&(x[i+5]==6))
	#else check if the subsequence is 23456 or not
	{
	j=1 #if it is 23456, set j as 1
	break #and break from the loop
	}
	}
	return(j) #return the value of j
	}

	rolls<-function(n,m) #function
	{
	a<-sample(1:6,n,replace=T);b<-sample(1:6,m,replace=T) #roll the dice
	return(any(unique(a)%in%unique(b))) #check if any from the two set matches
	}

	mean_mat<-matrix(,nrow=12,ncol=12) #create a 12*12 matrix
	for(i in 1:12) #loop 12 times
	{
	for(j in 1:12) #loop 12 times again
	{
	x<-replicate(10000,rolls(i,j)) #repeat the function 10000 times
	mean_mat[i,j]<-mean(x) #add the mean of the above repitition to the matrix
	}
	}

	rolls<-function() #function
	{
	a<-sample(1:6,2);i<-1 #roll twice and set the roll counter as 1
	while(length(unique(a))!=6) #until all the faces have appeared
	{
	a<-c(a,sample(1:6,2,replace=T)) #continue rolling in pairs
	i=i+1 #and increase the counter
	}
	return(i) #return the number of rolls
	}

	x<-replicate(10000,rolls()) #repeat this 10000 times
	mean(x) #take the mean

	library(animation)
	set.seed(20)

	rolls<-function(n) #function
	{
	x<-sample(1:6,n,replace=T) #roll n times
	return(max(x)) #return the max face value obtained
	}

	mean_vec<-NULL #create a null vector
	for(i in 1:15) #loop 15 times
	{
	x<-replicate(10000,rolls(i)) #repeat 10000 times with i as n
	mean_vec<-c(mean_vec,mean(x)) #add the mean to the vector
	}

	rolls<-function(n) #function
	{
	x<-sample(1:6,n,replace=T) #roll n times
	return(all(x==sort(x))) #return if all the values are in a
	#non-decreasing sequence or not
	}

	n<-10 #take n as 10
	x<-replicate(1000000,rolls(n)) #repeat the function with n as 10
	mean(x) #take the mean

On average, how many times must a 6-sided die be rolled until a 6 turns up?

On average, how many times must a 6-sided die be rolled until a 6 turns up twice a row?

On average, how many times must a 6-sided die be rolled until the sequence 65 appears (that is a 6 followed by a 5)?

Note:

On average, how many times must a 6-sided die be rolled until two rolls in a row differ by 1 (such as a 2 followed by a 1 or 3, or a 6 followed by a 5)?

Bonus Question: What if we roll until two rolls in a row differ by no more than 1 (so we stop at a repeated roll, too)?

We roll a 6-sided die n times. What is the probability that all faces have appeared?

We roll a 6-sided die n times. What is the probability that all faces have appeared in order in six consecutive throws?

Person A rolls n dice and person B rolls m dice. What is the probability that they have a common face showing up?

On average, how many times must a pair of 6-sided dice be rolled until all sides appear at least once?

Suppose we can roll a 6-sided die up to n times. At any time we can stop, and that roll becomes our “score”. Our goal is to get the highest possible score, on average. How should we decide when to stop?

Suppose we roll a fair dice 10 times. What is the probability that the sequence of rolls is non-decreasing?