Optimal strategies for this card game? I recently had an interesting card game for an interview.
The interviewer and I have two cards each, one with a '5' painted on it and one with a '10'. We pick a card each and show it at the same time. If we picked the same card, I receive nothing, while if we picked different cards, the interviewer pays me the number he picked in dollars. What is the interviewer most likely to do if he intends to minimise my payoff, and what should my strategy be to counter this?
Consideration 1: I pick '5' all the time, but if the interviewer knows I do this he would pick '5' as well so that I receive nothing and my expected payoff would just be zero.
Consideration 2: I assign a probability $p_1$ of picking a '5' (and $1-p_1$ of picking a '10'), and the interviewer assigns a probability $p_2$ of picking a '5' (and $1-p_2$ of picking a '10'). My payoff would then be
$$\mathbb{E}=10p_1(1-p_2)+5(1-p_1)p_2.$$
I was thinking of differentiation somehow, but $p_1$ aims to maximise $\mathbb{E}$ while $p_2$ aims to minimise $\mathbb{E}$. Is this even the correct strategy for both of us?
Consideration 3: A friend suggested assigning probabilities of $\frac13$ and $\frac23$ to the cards respectively for a Nash equilibrium since it is a symmetric game. Where does this come from intuitively? Do these probabilities match with the above equation?
Any help/comments on the considerations is greatly appreciated, cheers!
 A: The logic of equilibrium here is: you should choose your strategy (i.e. determine your probabilities of picking $5$ and $10$) in such a way that, however your opponent chooses his strategy, your expected gain is not influenced.
Let's say you pick $5$ with probability $p_1$ and your opponent picks $5$ with probability $p_2$. Then, once you fixed your choice of $p_1$, the expectation $\mathbb{E}$ is a linear function in $p_2$: $$\mathbb{E} = 10p_1 + (5 - 15p_1)p_2.$$
In this situation, you see that choosing $p_1 = 1/3$ will guarantee an expectation of $10/3$, regardless of $p_2$.
If, let's say, you choose $p_1 < 1/3$. Then there is the "risk" that your opponent somehow knows your strategy (e.g. via statistics from multiple rounds of the game), and then chooses $p_2 = 0$. You then have $\mathbb{E} = 10p_1 < 10/3$.
If, in the other direction, you choose $p_1 > 1/3$. Then again assume your opponent gets to know your strategy, and then chooses $p_2 = 1$. You then have $\mathbb{E} = 5 - 5p_1 < 10/3$.
Therefore, you see that the "safest" way of playing is to choose $p_1 = 1/3$. It is in that sense that we say it's the "optimal strategy".
I hope this whole logic makes sense to you.
And as an exercise, you may try to solve the general situation: say the two cards are labelled A and B, and your gain has $4$ possibilities: $a, b, c, d$, which correspond to the $4$ combinations AA, AB, BA, BB of your cards. Then what is your optimal strategy?
