Game Theory (Part 14)

Here is the first test in our game theory course. If you’re following along on this blog, you can do it and then look at the answers below.

Definitions

1. Define a Nash equilibrium for a 2-player normal form
game.

2. Define the expected value of some function with respect to some probability distribution.

Proof

3. Suppose Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view. are the payoff matrices for 2-player normal form game. Prove that if Image may be NSFW.
Clik here to view. (i,j) is a Nash equilibrium, there cannot exist a choice Image may be NSFW.
Clik here to view. for player A that strictly dominates choice Image may be NSFW.
Clik here to view.

2-Player normal form games

Consider this 2-player normal form game:

Image may be NSFW.
Clik here to view. $\begin{array}{rrr} (-2,1) & (4,2) & (2,-5) \\ (2,3) & (-6,2) & (5,3) \\ (1,0) & (2,-4) & (4,-3) \end{array}$

4. Find all the Nash equilibria. Draw a box around each Nash
equilibrium.

For problems 5-8 do not simplify your answers by working out the binomial coefficients, etc.

Probabilities

5. If you draw 3 cards from a well-shuffled standard deck,
what is the probability that at least 2 are hearts?

6. If you flip 4 fair and independent coins, what is the probability that exactly 2 land heads up?

Expected values

7. Suppose you pay $2 to enter a lottery. Suppose you have a 1% chance of winning $100, and otherwise you win nothing. What is the expected value of your payoff, including your winnings but also the money you paid?

8. Suppose you draw two cards from a well-shuffled standard deck. Suppose you win $100 if you get two aces, $10 if you get one ace, and nothing if you get no aces. What is your expected payoff?

Extra credit

About how many ways are there to choose 3 atoms from all the atoms in the observable universe? Since this question is for extra credit, I’ll make it hard: I’ll only accept answers written in scientific notation, for example Image may be NSFW.
Clik here to view. $2 \times 10^{50}.$

And here are the answers to the first test.

Definitions

1. Given a 2-player normal form game where A’s
payoff is Image may be NSFW.
Clik here to view. $A_{ij}$ and B’s payoff is Image may be NSFW.
Clik here to view. $B_{ij}$ , a pair of choices Image may be NSFW.
Clik here to view. (i,j) is a Nash equilibrium if:

1) For all Image may be NSFW.
Clik here to view. $1 \le i' \le m,$ Image may be NSFW.
Clik here to view. $A_{i'j} \le A_{ij}.$

2) For all Image may be NSFW.
Clik here to view. $1 \le j' \le n,$ Image may be NSFW.
Clik here to view. $B_{ij'} \le B_{ij}.$

2. The expected value of a function Image may be NSFW.
Clik here to view. $f : X \to \mathbb{R}$ with respect to a probability distribution Image may be NSFW.
Clik here to view. on the finite set Image may be NSFW.
Clik here to view. is

Image may be NSFW.
Clik here to view. $\sum_{i \in X} f(i) p_i$

Note that a good definition makes it clear what term is being defined, by writing it in boldface or underlining it. Also, it’s best if all variables used in the definition are explained: here they are Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view.

Proof

3. Theorem. Suppose Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view. are the payoff matrices for 2-player normal form game. If Image may be NSFW.
Clik here to view. (i,j) is a Nash equilibrium, there cannot exist a choice Image may be NSFW.
Clik here to view. for player A that strictly dominates choice Image may be NSFW.
Clik here to view..

Proof. Suppose that Image may be NSFW.
Clik here to view. (i,j) is a Nash equilibrium. Then

Image may be NSFW.
Clik here to view. $A_{ij} \ge A_{i' j}$

for any choice Image may be NSFW.
Clik here to view. for player A. On the other hand, if choice Image may be NSFW.
Clik here to view. strictly dominates choice Image may be NSFW.
Clik here to view., then

Image may be NSFW.
Clik here to view. $A_{i'j} > A_{i j}$

This contradicts the previous inequality, so there cannot exist
a choice Image may be NSFW.
Clik here to view. for player A that strictly dominates choice Image may be NSFW.
Clik here to view.. █

Note that the really good way to write a proof involves:

• First writing “Theorem” and stating the theorem.
• Saying “Proof” at the start of the proof.
• Giving an argument that starts with the hypotheses and leads to the conclusion.
• Marking the end of the proof with “Q.E.D.” or “█” or something similar.

2-Player normal form games

4. In this 2-player normal form game, the three Nash equilibria are marked with boxes:

Image may be NSFW.
Clik here to view. $\begin{array}{rrr} (-2,1) & \boxed{(4,2)} & (2,-5) \\ \boxed{(2,3)} & (-6,2) & \boxed{(5,3)} \\ (1,0) & (2,-4) & (4,-3) \end{array}$

Probabilities

5. If you draw 3 cards from a well-shuffled standard deck,
what is the probability that at least 2 are hearts?

Answer. One correct answer is

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{13}{2} \binom{39}{1}}{\binom{52}{3}} + \frac{\binom{13}{3} \binom{39}{0}}{\binom{52}{3}} }$

since there are:

• Image may be NSFW.
Clik here to view. $\binom{52}{3}$ ways to choose 3 cards, all equally likely,

• Image may be NSFW.
Clik here to view. $\binom{13}{2}$ ways to choose 2 hearts and Image may be NSFW.
Clik here to view. $\binom{39}{1}$ ways to chose 1 non-heart, and

• Image may be NSFW.
Clik here to view. $\binom{13}{3}$ ways to choose 2 hearts and Image may be NSFW.
Clik here to view. $\binom{39}{0}$ ways to chose 0 non-hearts.

Another correct answer is

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{13}{2} \cdot 39 }{\binom{52}{3}} + \frac{\binom{13}{3}} {\binom{52}{3}}}$

This is equal since Image may be NSFW.
Clik here to view. $\binom{39}{0} = 1$ and Image may be NSFW.
Clik here to view. $\binom{39}{1} = 39.$

6. If you flip 4 fair and independent coins, what is the probability that exactly 2 land heads up?

Answer. The probability

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{4}{2}}{2^4} }$

since there are Image may be NSFW.
Clik here to view. 2^4 possible ways the coins can land, all
equally likely, and Image may be NSFW.
Clik here to view. $\binom{4}{2}$ ways to choose 2 of the 4 coins to land heads up.

Expected values

Answer. One correct answer is

Image may be NSFW.
Clik here to view. $0.01 \cdot \$98 \quad + \quad 0.99 \cdot (-\$ 2)$

Another correct answer is

Image may be NSFW.
Clik here to view. $0.01 \cdot \$100 \quad + \quad 0.99 \cdot \$0 \quad - \quad \$2$

Of course these are equal.

8. Suppose you draw two cards from a well-shuffled standard deck. Suppose you win $100 if you get two aces, $10 if you get one ace, and nothing if you get no aces. What is your expected payoff?

Answer. One correct answer is

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{4}{2} \binom{48}{0}}{\binom{52}{2}} \cdot \$100 \quad + \quad \frac{\binom{4}{1} \binom{48}{1}}{\binom{52}{2}} \cdot \$10 \quad + \quad \frac{\binom{4}{0} \binom{48}{2}}{\binom{52}{2}} \cdot \$ 0 }$

since Image may be NSFW.
Clik here to view. $\binom{4}{n} \binom{48}{3-n}$ is the number of ways to pick Image may be NSFW.
Clik here to view. aces and Image may be NSFW.
Clik here to view. 3-n non-aces. Of course we can also leave off the last term, which is zero:

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{4}{2} \binom{48}{0}}{\binom{52}{2}} \cdot \$100 \quad + \quad \frac{\binom{4}{1} \binom{48}{1}}{\binom{52}{2}} \cdot \$10 }$

Since Image may be NSFW.
Clik here to view. $\binom{48}{0} = 1$ we can also write this as

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{4}{2}}{\binom{52}{2}} \cdot \$100 \quad + \quad \frac{\binom{4}{1} \binom{48}{1}}{\binom{52}{2}} \cdot \$10 }$

Or, since Image may be NSFW.
Clik here to view. $\binom{4}{1} = 4$ and Image may be NSFW.
Clik here to view. $\binom{48}{1} = 48$ we can write this as

Image may be NSFW.
Clik here to view. $\displaystyle{ \frac{\binom{4}{2}}{\binom{52}{2}} \cdot \$100 + \frac{4 \cdot 48}{\binom{52}{2}} \cdot \$10 }$

But I said not to bother simplifying the binomial coefficents.

Extra credit

Answer. In class I said the number of atoms in the
observable universe is about Image may be NSFW.
Clik here to view. $10^{80},$ and I said I might put this on the test. So, the answer is

Image may be NSFW.
Clik here to view. $\begin{array}{ccl} \displaystyle{ \binom{10^{80}}{3}} &=& \displaystyle{ \frac{10^{80}(10^{80} - 1)(10^{80} - 2)}{3 \cdot 2 \cdot 1} } \\ \\ &\approx& \displaystyle{ \frac{10^{80} \cdot 10^{80} \cdot 10^{80}}{6} } \\ \\ &=& \displaystyle{ \frac{10^{240}}{6} } \\ \\ &\approx & 1.667 \times 10^{239} \end{array}$

Note that the question asked for an approximate answer, since we don’t know exactly how many atoms there are in the observable universe. The right answer to a question like this gives no more decimal places than we have in our data, so Image may be NSFW.
Clik here to view. $1.667 \times 10^{239}$ is actually too precise! You only have one digit in the data I gave you, so a better answer is

Image may be NSFW.
Clik here to view. $\mathbf{ 2 \times 10^{239} }$

Since the figure Image may be NSFW.
Clik here to view. $10^{80}$ is very approximate, another correct answer is

Image may be NSFW.
Clik here to view. $\mathbf{ 10^{239} }$

Game Theory (Part 14)

Definitions

Proof

2-Player normal form games

Probabilities

Expected values

Extra credit

Definitions

Proof

2-Player normal form games

Probabilities

Expected values

Extra credit

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