## Sunday, April 3, 2016

### Probability in the Game of Risk -- Problem Three

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.