User:Trettels:Session 1 - 16JAN

Session 2: 16 JAN 2013
Code for Question 2 (written in MATLAB):
Group: TrettelS, TSanders, LIan, RCardenas

Part 1:

clear all; close all; clc;

range = [1 exp(1) pi 4 5 6 exp(pi)]; % the value weights for each side

range=cumsum(range/sum(range)); % builds our numberline

% roll the first die...

t_base = rand(1,1e6);

t1(t_base <=range(1))=1;

t1(t_base>range(1) & t_base <= range(2))=2;

t1(t_base>range(2) & t_base <= range(3))=3;

t1(t_base>range(3) & t_base <= range(4))=4;

t1(t_base>range(4) & t_base <= range(5))=5;

t1(t_base>range(5) & t_base <= range(6))=6;

t1(t_base>range(6) & t_base <= range(7))=7;

% and roll the second

t_base=rand(1,1e6);

t2(t_base <=range(1))=1;

t2(t_base>range(1) & t_base <= range(2))=2;

t2(t_base>range(2) & t_base <= range(3))=3;

t2(t_base>range(3) & t_base <= range(4))=4;

t2(t_base>range(4) & t_base <= range(5))=5;

t2(t_base>range(5) & t_base <= range(6))=6;

t2(t_base>range(6) & t_base <= range(7))=7;

%find out how many of our results add to 8

result1=sum(t1+t2==8)

Result was 62239

Part 2: What is the analytic expectation for 1,000,000 rolls?

We will take advantage of the fact that for $n$ rolls, $n*E[$ one roll sums to 8$]=n*\sum _{k=1}^{k_{max}}p(k)*p(k_{max}-k)$, where $k_{max}=7$.

range2 = [1 exp(1) pi 4 5 6 exp(pi)];

range2 = range2/sum(range2);

p_8 = 1e6 *sum(range2(1:7).*range2(7:-1:1))

Result is (approximately) 62377.04196435287

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