Probability in the Game of Risk  Problem Three
This post looks at other ways to solve the problem when two
attacking armies defend against one defending army. I had broached this situation in a previous
post.
Probability in the Game of Risk: Problem Set Two
Questions:
What is the probability that the maximum of two dice rolls
from the attacking army is 1, 2, 3, 4, 5, or 6?
Given these probabilities what is the probability that the
defending army is wiped out? What is
the probability that the defending army survives and the attacking army loses one
army?
Analysis:
What is the
probability that the maximum of two dice rolls from the attacking army is 1, 2,
3, 4, 5, or 6?
I will find the probability the max of two dice rolls equals
k for k= 1 to 6 two different ways.
First way:
Denote FR= first roll and SR = Second roll
Step one: Find the
probability that FR>=k or SR>=k.
There are a number of ways to do this.
I use the formula for the probability of the union of the two events.
P(FR>=k or SR>=k) = P(FR>=k) + P (SR>=k) – P(FR>k)*P(SR>=k)
Step two: Note that
PR(FR=k or SR=k) =
PR(FR>=k or SR>=k) – PR(FR>=k+1 or SR>=k+1)
Calculating the probability of the highest of two dice rolls is
equal to k


k

p(x>=k)

P(FR>=k or SR>=K)

Prob the highest of FR or
SR is equal to k

1

1.000

1.000

0.028

2

0.833

0.972

0.083

3

0.667

0.889

0.139

4

0.500

0.750

0.194

5

0.333

0.556

0.250

6

0.167

0.306

0.306

1.000

Second Way:
Step One: List all 36
outcomes of the two dice rolls.
Step Two: Take the maximum of the two dice rolls for all 36
observations.
Step Three: Add the
number of times max of two dice rolls is equal to k for the six values of k.
Step Four: Divide the
sum above by 36.
A second way to calculate the prob the max of two dice roll is
equal to k


FD

SD

max(FD,SD)

max=1

max=2

max=3

max=4

max=5

max=6

1

1

1

1

0

0

0

0

0

2

1

2

0

1

0

0

0

0

3

1

3

0

0

1

0

0

0

4

1

4

0

0

0

1

0

0

5

1

5

0

0

0

0

1

0

6

1

6

0

0

0

0

0

1

1

2

2

0

1

0

0

0

0

2

2

2

0

1

0

0

0

0

3

2

3

0

0

1

0

0

0

4

2

4

0

0

0

1

0

0

5

2

5

0

0

0

0

1

0

6

2

6

0

0

0

0

0

1

1

3

3

0

0

1

0

0

0

2

3

3

0

0

1

0

0

0

3

3

3

0

0

1

0

0

0

4

3

4

0

0

0

1

0

0

5

3

5

0

0

0

0

1

0

6

3

6

0

0

0

0

0

1

1

4

4

0

0

0

1

0

0

2

4

4

0

0

0

1

0

0

3

4

4

0

0

0

1

0

0

4

4

4

0

0

0

1

0

0

5

4

5

0

0

0

0

1

0

6

4

6

0

0

0

0

0

1

1

5

5

0

0

0

0

1

0

2

5

5

0

0

0

0

1

0

3

5

5

0

0

0

0

1

0

4

5

5

0

0

0

0

1

0

5

5

5

0

0

0

0

1

0

6

5

6

0

0

0

0

0

1

1

6

6

0

0

0

0

0

1

2

6

6

0

0

0

0

0

1

3

6

6

0

0

0

0

0

1

4

6

6

0

0

0

0

0

1

5

6

6

0

0

0

0

0

1

6

6

6

0

0

0

0

0

1

1

3

5

7

9

11


0.028

0.083

0.139

0.194

0.250

0.306

Note the estimate of the P(Max(FD,SD)=k) on the bottom of
the table above are identical to the estimates obtained using the first method.
This is a big relief.
What is the
probability that the defending army is wiped out? What is the probability the defending army
survives and takes one attacking army?
Again, there are at least two ways to solve this
problem.
The First
Method:
The most direct brute
force way to solve this problem is to list all 216 outcomes and count the
number of outcomes where the attacking nation wins and the number of outcomes
where the defending nation wins and divide by 216 the total number of outcomes.
I did this in a previous post:
Probability in the Game of Risk: Problem Set Two
I found that when the attacking nation uses two armies and
the defending nation uses one army there is a 0.579 probability of the
attacking nation winning and a 0.421 probability of the defending nation
winning.
The Second Method:
The second approach involves understanding the concept of
independent events and conditional probabilities.
There are six outcomes for the defending army each with a
probability of 1/6.
For each of the defending army outcomes I calculate the
probability the attacking nation wins.
Remember in Risk the tie goes to the defending army. If k =1 the probability of the attacking army
winning is 1 0.028. If k=2 the probability of the attacking army
winning is 1 (0.028+.083)
And so on.
Probability Attacking
Nation Wins for Each Defending Army Outcomes


Defending Army Event

Probability of Defending
Army Event

Prob the highest of FR or
SR is equal to k

Probability Attacking
Nation Wins Given Defending Army Result

1

0.167

0.028

0.972

2

0.167

0.083

0.889

3

0.167

0.139

0.750

4

0.167

0.194

0.556

5

0.167

0.250

0.306

6

0.167

0.306

0.000

The probability the defending army has outcome k is 1/6 for
each value of the dice.
The probability the attacking nation wins for each value of
k is presented in the fourth column of the table above.
The probability the attacking army wins is the sum of the
products the probability the defending army has outcome k and the probability
the attacking nation wins given the defending army has outcome k.
The probability the attacking nation wins is the sum product
of the second column and the fourth column in the table above.
The event defending nation wins is the complement of the
event the attacking nation wins. The probability of the complement of an event
is 1 – probability of the event.
Probabilities of
Victories for Attacking and Defending Nations


Probability the attacking
army wins

0.579

Probability the Defending
Army Wins

0.421

Note the answer from the second method is identical to the
answer from the first method.
This makes me happy!!
Authors Note: Around 10 probability problems using the game
of risk will be posted on my blog.
These problems will be a chapter or two in my next book on probability
and statistics.
In 1997, I wrote a short book called Statistical
Applications of Baseball. The book was
favorably reviews by the JASA journal Chance.
It is available on Kindle.
I have been very busy on my policy blog, my finance blog,
and my politics blog. Interested readers
might look at the following material.
Thoughts on a new
mandatory pension:
Post describes a policy proposal designed to create a new
mandatory pension. Is such a proposal
an effective way to increase pension coverage?
Should this proposal be tied to a plan to improve the solvency of Social
Security?
Analyzing a Stock
Fund that Invests in Six Other Stock Funds:
Wall Street is coming up with a lot of innovative mutual
funds and ETFs. My finance blog is
looking at these developments.
My Experience at the
Colorado Caucus:
I am intensely watching this election. Go to 2016 Memos for my insights.
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