Interpreting Game Theory Questions
Matthew Barrera 
I have come across two game theory questions, and got stuck in a similar stage in both.
Problem 1:
There are two boxes, correctly labeled as 4 candies and 6 candies according to their contents. You and your partner are famously perfectly rational. The game proceeds by each of you deciding independently which box to choose. If you both pick the same box, you split its' contents (2 or 3 candies each, depending on the box chosen). If you pick different boxes, you split the total (i.e. 5 candies each). What is the optimal strategy for you?
Solution of Problem 1:I got stuck with too many undetermined probability values:
$$p_{4}^\text{me}, (p_{6}^\text{me} = 1-p_{4}^\text{me}),p_{4}^\text{partner},(p_{6}^\text{partner}=1-p_{4}^\text{partner}).$$
I couldn't see how to decide on my probabilities, as I don't know my partner's.Apparently it is ok to assume that my partner's probabilities equal mine. From there, it is simple to get the expected payoff and maximize it as a single variable function.
Problem 2: (Mark Joshi et al, Quant Job Interview Questions and Answers)
We play a game: I pick a number $n$ from 1 to 100. If you guess correctly, I pay you \$$n$ and zero otherwise. How much would you pay to play this game?
My Attempt at Problem 2:Let the choice of $n$ by the dealer be given by the probabilities$$p_n^\text{delaer},\quad n \in \{1,2,\dots,100\}. $$Here is where I found myself stuck again. How can I decide on a strategy when I don't know the dealer's probabilities?
The solution in the book reads
Note that this is really a game theory question. Each player has to pick a strategy. In this case, an optimal strategy will be one in which the expected winnings is independent of the strategy the opponent picks.... The solution is to pick $k$ with probability proportional to $1/k$. We therefore pick $k$ with probability $$\frac{1}{k} \left( \sum_{j=1}^{100} \frac{1}{j} \right)^{-1}$$.
I can't see how the author got around the unknown dealer probabilities, and found this strategy.
My questions are:
- In problem 1, is it supposed to be obvious that my probabilities are equal to my partner's? Does that follow from our perfect rationality?
- How does one end up with the $C/k$ probabilities given in the solution? How do they follow from the question? It seems to me that more wording is necessary for that, isn't it?
- Are there some general rules when interpreting game theory problems as equations which I am missing?
Thank you!
$\endgroup$ 11 Answer
$\begingroup$For problem 1, when using mixed strategies we must use one player’s payoffs to solve for the player’s strategy. This is because the point of my mixed strategy is to make you indifferent, and vice versa. Suppose that Row is going to randomize. Then Row’s payoffs must be equal for all strategies that Row plays with positive probability. But that equality in Row’s payoffs doesn’t determine the probabilities with which Row plays the various rows. Instead, that equality in Row’s payoffs will determine the probabilities with which Column plays the various columns. The reason is that it is Column’s probabilities that determine the expected payoffs for Row; if Row is going to randomize, then Column’s probabilities must be such that Row is willing to randomize.
Problem 1 gives the game\begin{array}{c|lcr} I/II & \text{Box4} & \text{Box6} \\ \hline \text{Box4} & (2,2) & (5,5) \\ \text{Box6} & (5,5) & (3,3) \\ \end{array}
If player I best responds with a mixed strategy player II must make him indifferent between the choice of box. So we want the expected payoff between choosing Box4 and Box6 to be equal for player I. This gives
$$2p_2 + 5(1 - p_2) = 5p_2 + 3(1 - p_2)$$$$p_2 = \frac{2}{5}$$
This means the only way player I can be indifferent to choosing either box is if player II chooses Box4 with probability $\frac{2}{5}$ and Box6 with probability $1 - \frac{2}{5} = \frac{3}{5}$. To find the probabilities for player I we use the same process.
It ends up being that in mixed strategies the Nash Equilibrium is player I: $(\frac{2}{5}, \frac{3}{5})$ and player II: $(\frac{2}{5},\frac{3}{5})$. In addition there are two pure equilibriums (Box4, Box6) and (Box6, Box4).
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