Probability mass function question. In the game of Risk, battles are decided by

[cloud360]

Homework Statement

In the game of Risk, battles are decided by the rolling of dice. Suppose that there are two armies,
red and blue. The red army rolls three dice and the blue army rolls two dice. Whichever army rolls
the highest number (on a single die) is declared the winner. In the event of a tie (both armies have
the same highest number), the blue army is declared the winner.
(a) Let X denote the highest number rolled by the red army. Find the probability mass function
of X.
(b) Let Y denote the highest number rolled by the blue army. Find the probability mass function
of Y .
(c) Calculate the probability that the red army wins.

Homework Equations

pmf = px(s), where s= number of times you want variable x to appear, the function represent the probability of this
cdf= fx(s), where this represent probability x is less than or equal to s

The Attempt at a Solution

Red army rolls 3 times
Blue army rolls 2 times.
Red army
x1=1st die thrown =1,2,3,4,5,6
x2=2nd die thrown=1,2,3,4,5,6
x3=3rd die thrown=1,2,3,4,5,6

Px(x)=Fx(x)^3

is this correct, i don't know why you do this. but i think you must do this, then create a PMF (probability mass function) table?

Why is the pmf function = to Fx(x)^3, this is what i saw in my teachers explanation. but his explanation wasn't... perfect

 

[lanedance]

so start with a simple case and work from there, blue army will be simpler with only 2 rolls

treating the dice separately, there are 6x6 = 36 combinations that can come up

now to get a 1, there is only one possible combination 1-1, so the separately is 1/36
to get a 2, there are 3: 2-1, 2-2, 1-2, so the prob is 3/36

use these simple cases to think about the problem & generalise from there...

 

[lanedance]

it will also be useful to consider the sum rule
P(A or B) = P(A) + P(B) - P(A and B)

 

[cloud360]

it will also be useful to consider the sum rule
P(A or B) = P(A) + P(B) - P(A and B)

why would i use the sum forumale. and also. how can i find the probability red army or blue army wins.

do i look at all the instances where the pmf of blue army is higher ?

 

[lanedance]

why would i use the sum forumale. and also. how can i find the probability red army or blue army wins.

do i look at all the instances where the pmf of blue army is higher ?

Have go at something and i will help u work through it

I would start by finding each PDF first, then use them to calculate prob of who wins

 

[cloud360]

Have go at something and i will help u work through it

I would start by finding each PDF first, then use them to calculate prob of who wins

i am not allowed to use pdf in this question.

the question must be answered using pmf or cdf. i will try post a solution in 1 hour, and if you have time. please kindly tell me where i went wrong. my exam is tommorow.

again, thanks for taking your time to help me. i am very grateful

 

[cloud360]

My solution
Question 1
max of 1=1/216
max of 2=7/216
max of 3=5/54
max of 4=11/54
max of 5=3/8
max of 6=5/8
Question 2
max of 1=cdf((1/6)^2)=1/36
max of 2=cdf((2/6)^2)-(1/36)=(1/12)
max of 3=cdf((3/6)^2)-(1/12)=(1/6)
max of 4=cdf((4/6)^2)-(1/6)=(5/15)
max of 5=cdf((5/6)^2)-(5/15)=(13/36)
max of 6=cdf((6/6)^2)-(13/36)=(23/26)
Question 3

 

[cloud360]

so start with a simple case and work from there, blue army will be simpler with only 2 rolls

treating the dice separately, there are 6x6 = 36 combinations that can come up

now to get a 1, there is only one possible combination 1-1, so the separately is 1/36
to get a 2, there are 3: 2-1, 2-2, 1-2, so the prob is 3/36

use these simple cases to think about the problem & generalise from there...

wow, i just couldn't see that it is the maximum after a certian number of throws literally.

and that their is loads of different ways to get a max of 2 ,3,4 e.t.c

 

[lanedance]

My solution
Question 1
max of 1=1/216
max of 2=7/216
max of 3=5/54
max of 4=11/54
max of 5=3/8
max of 6=5/8

this isn't correct, note you probs sum to greater than 1

try and explain what you are doing

Question 2
max of 1=cdf((1/6)^2)=1/36
max of 2=cdf((2/6)^2)-(1/36)=(1/12)
max of 3=cdf((3/6)^2)-(1/12)=(1/6)
max of 4=cdf((4/6)^2)-(1/6)=(5/15)
max of 5=cdf((5/6)^2)-(5/15)=(13/36)
max of 6=cdf((6/6)^2)-(13/36)=(23/26)
Question 3

I don't know what this is?

 

[lanedance]

using the sum rule for the 2 die case
P(A or B) = P(A) + P(B) - P(A and B)
re-writing
P(A and B) = P(A) + P(B) - P(A or B)


so let the dice be numbered d1 & d2, then the propositions to interrogate are
A: ((d1=n)and(d2<=n))
B: ((d1<=n)and(d2=n))
If either A or B is true, then n is the maximum number

the final probability we want to find is:
P(n) = P(A or B)
= P( ((d1=n)and(d2<=n)) or ((d1=n)and(d2<=n)) )
= P( (d1=n)and(d2<=n) ) + P ( (d1=n)and(d2<=n) ) - P( ((d1=n)and(d2<=n)) and ((d1=n)and(d2<=n)) )

reducing the and terms & using the fact the events on each dice are independent
= P(d1=n).P(d2<=n) + P (d1=n).P(d2<=n) - P(d1=n)P(d2=n)
= n/36 + n/36- 1/36
= (2n-1)/36

now try for the three dice case...

 

[lanedance]

i am not allowed to use pdf in this question.

the question must be answered using pmf or cdf. i will try post a solution in 1 hour, and if you have time. please kindly tell me where i went wrong. my exam is tommorow.

again, thanks for taking your time to help me. i am very grateful

pdf & pmf mean the same thing