help me please

[EDIT: Fri Jun 24 23:04:46 UTC 2011 - Reformat - MKF]

Simulation of tossing a pair of fair dice can be done using the following two MATLAB lines

X=ceil(6*rand(1,n));
Y=ceil(6*rand(1,n));

where n is the number of tosses. Write a MATLAB script to determine the probability P[X+Y=7]. Use a “for” loop to run simulation one hundred times with n = 10000. Plot the 100 estimated probability values along with the theoretical result of P[X+Y=7].

That what i have done.

n=10000
m=100
B=0
for k=1:n
    P=0
X=ceil(6*rand(1,n));
Y=ceil(6*rand(1,n));
Z=X+Y;

for i=1:m
     if Z(i)==7;
     P=P+1;
end
end

disp('trial');
k
disp('success rate');
P=P/n;
 Xaxis(k) = k;
    Yaxis(k) = P;
    B=B+P;
end    

disp('Total Average: ')
B/m

plot(Xaxis,Yaxis);

clear;

______________________________________________________________

Write a MATLAB script to do the following a. Create 10000 random variables uniformly distributed between 2 and 4. b. Create a histogram to approximate the actual probability density function. c. Superimpose the actual probability density function to the above histogram.

[Matt Fig]

What have you done so far?

See this guide:

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Walter Roberson]

If the simulation is being run 50 times, then what are the 100 values that are to be plotted ?

Answer 1

[Matt Fig]

Thanks for showing some work. Here is how one could approach the first task:

n = 10000; 
m = 100;
PROB = zeros(1,m);
for ii = 1:m
    X = ceil(6*rand(1,n));
    Y = ceil(6*rand(1,n));
    Z = X+Y;
    PROB(ii) = sum(Z==7)/n;
end

plot(1:m,PROB,'b*')
hold on
line([1 m],mean(PROB)*ones(1,2),'color','r')
line([1 m],[1 1]./6,'color','k')
legend({'Trial Values';'Trials Mean';'Theory'})

The second task uses some skill learned in the first, so try to understand what I did and what went wrong in what you did. Also, look at the MATLAB HIST and HISTC functions.

Answer 2

[Sean de Wolski]

X = rand(10000,50,6);
[v,v] = sort(X,3);
plot([sum(sum(v(:,:,1:2),3)==7);repmat(1/5*10000,1,50)]');
legend('Experimental','Ideal');

For loops are boring

[Sean de Wolski]

Must be the Fridays; Walter, I learned this engine from you!!

(The sort operation index)

[Walter Roberson]

Ah, I didn't notice it was the index you were pulling out.

Very well, carry on. ;-)

[Walter Roberson]

Ideal should be 1/6 not 1/5:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

That's 6 possibilities out of (6*6), so probability is 1/6

[Sean de Wolski]

I disagree. You are guaranteed to get a number 1:6 (the first number); one of the next five numbers you choose will add to seven with the first.

[Sean de Wolski]

Ahh you're right. I'm thinking exclusive. Doh!

[Walter Roberson]

No.

>> mean(mean((ceil(6*rand(5000,10000)) + ceil(6*rand(5000,10000))) == 7,2))

ans =

0.1666153

You generate the first number; it is certain to be in range. 7 minus that number is going to be a single number in the range 1 to 6, which is exactly the possibilities for a die. There is therefore a 1 in 6 probability that the second dice will come out equal to the value needed to make the total be 7.

[Walter Roberson]

Your pub or mine, Sean?

[Sean de Wolski]

Yours, I'll be there in an hour!

plot(sum(sum(ceil(rand(10000,50,2)*6),3)==7))

[southie]

could you show me using the for loop the professor specify that we use for loop

[Matt Fig]

percy, show what you have done so far! Don't put it in a comment or a new answer, edit your original post by clicking on the Edit link, then show your code.

[Walter Roberson]

for K = 1:1

data = sum(sum(ceil(rand(10000,50,2)*6),3)==7);

end

plot(data)

Answer 3

[Walter Roberson]

Without the ideal part:

plot(mean((ceil(6*rand(50,10000)) + ceil(6*rand(50,10000))) == 7,2))