How to Win Games Part 4: Counter Prudential Strategies

Last week, we talked about how to counter someone who plays spitefully by using your prudential strategy. However, what if you know someone is playing prudentially?

This week, we will conclude the four game theory posts by learning how to determine your counter prudential strategy, which does exactly what it says: counters the prudential strategy.

First, you will have to figure out their prudential strategy though. Let's go back to the police-criminal game from last week.

	Crime	Lay Low
Patrol	3, -5	0, 1
Donuts	-2, 3	2, 0

We had determined that the police's prudential strategy is 4/7 patrol and 3/7 donuts. So, what do the criminals do?

Well, this isn't their mixed-strategy equilibrium. Remember that the mixed-strategy equilibrium is a way to make the other player's payoffs equal so that they don't have an advantage by playing one way or another.

Since this is the prudential strategy, and therefore not the mixed-strategy equilibrium, the criminals do have an advantage by playing one way or another. So, let's determine the expected payoff of playing either strategy.

For committing crime, they will get -5 4/7 of the time and 3 3/7 of the time. So, we do:

-5(4/7) +3(3/7) = -11/7

For laying low, we do the same thing:

1(4/7) + 0(3/7) = 4/7

Since 4/7 is greater than -11/7, the criminals should lay low every single time for their counter prudential strategy.

Though this type of thing isn't a proof or pattern like I normally post about, I find it really cool that you can analyze games just like you analyze math. If you have a lot more time, you can analyze games like chess, poker, black jack, and even sports.

Answer: Here is the answer to July's problem of the week. Make sure you do last week's as well!

Easy:
e = 24
h = 30
m = 12
n = 18
A = 81

Hard:
t = 45
z = 4.5
s = 50
a = 1
b = -1250
c = 390625
x = 625
P = 50π

For the rope problem from the hard problem of Monday, you must first set both ends of one rope on fire and set only one end of the other rope. Half an hour later, your first rope has burned completely leaving your second rope with 30 minutes left. Now, set the other end, and put your plant over the flame for its fifteen minute cooking.

Problem of the Week Day 5: Week of 8/20/12 - 8/24/12

Today is the final day of the problem of the week. The answers will be going up in late September.

Easy: Take a trapezoid with an area of n and bases of lengths t and m^3. What is the height h of this trapezoid?

h =

Hard: Take a circle with radius a centimeters and a trapezoid with bases s and k centimeters and a height of (2n-8y-4)/5 millimeters. Find the perimeter of the shape with the bigger area in centimeters. Round to the nearest tenth of a centimeter.

p =

Problem of the Week Day 4: Week of 8/20/12 - 8/24/12

Today is day four of the problem of the week. Good luck!

Easy: Solve for the next number n in this sequence:

x, sm^2, t, sgm, b, y, n

n =

Hard: Find the value of n in this sequence:

y, s-1, z, l, n, k-a-5,...

n =

Problem of the Week Day 3: Week of 8/20/12 - 8/24/12

Today is day three of the problem of the week. Unlike normal Wednesdays, I have decided to give a few probability questions. The easy one you should be able to figure out with some simple math, and the hard one is explained in a previous post on the math behind a card trick.

Easy: You and your friend decide to play a game where you take a b card deck and deal down a card. On each turn, you both bet a dollar out of the two twenties that you brought with you. If the card dealt is a face card, you win the money and if it is a number card, your friend wins the money. Whoever runs out of money first loses.

Before the game, you rigged the deck so that y% of the cards are face cards and x% of the cards are number cards, making it much easier for you to win. Based on this cheat, determine how many turns it will take before your friend runs out of money.

t =

If you can't figure out a way to solve this problem, you probably have the values of x and y wrong. If you do, go back and make sure you have checked for dominant strategies, and then a mixed strategy equilibrium.

Hard: First off, solve the following problem:

2g - s = k
k =

Your friend bets you f dollars that you can't win a game he made up, which you accept. The game is that he takes a k card deck with numbers from 1 to s written on them. You will randomly guess a number from 1 to s, and then see if the next card has that number on it. If you fall into a conscious pattern, your friend automatically gets the money. Your goal is to get through the whole deck and never guess a card correctly. Determine the odds that your friend will get your money and you will therefore lose. Round to the nearest full percent.

l = ___%

This problem seems pretty daunting, but we did go over it in a blog post (not a problem of the week). If you can find this post and your calculator, the problem won't be too difficult. Good luck.

Problem of the Week Day 2: Week of 8/20/12 - 8/24/12

Today is day 2 of the problem of the week. Remember to use yesterday's answers in today's problem.

Easy: For this problem, you will need to bring up yesterday's problem as well. Look at the matrix from yesterday and find the highest number there. Let's call that number g.

g =

Now, find the second highest number in the matrix. Let's call that number s.

s =

Finally, find the average of all of the numbers in the matrix. Round that to the nearest hundredth. That number will be called m.

m =

Now that we have those, we can begin the problem. Take a right triangle with the following side lengths. All of the measurements are in millimeters.

a = sgm + x
b = ___
c = y

Try to determine the length of b.

b =

Hard: This problem is trigonometric, but is also a real world problem. For my science fair experiment in 2011, I had to solve almost the same exact problem and it was actually really cool to see the real world application of trigonometry. I hope you find the applications of trigonometry as cool as I did while you complete this word problem. Before you begin, complete these two calculations:

f = x + (a + b)/(y + z)
f =

g = x + a + b
g =

Now for the fun part. Say you need to create a wooden ramp that is f meters long and is propped at an a° angle. You will prop it up with another piece of wood g centimeters up from the bottom and you want the support to make it an a° angle ramp. How many centimeters long should your support be to achieve this angle? Round to the nearest centimeter.

s =

Good luck, and remember to save your answers for tomorrow.