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$A$ and $B$ play the following game: $A$ writes down either number $1$ or number $2$, and $B$ must guess which one. If the number that $A$ has written down is $i$ and $B$ has guessed correctly, $B$ receives $i$ units from $A$. If $B$ makes a wrong guess, $B$ pays $3/4$ unit to $A$. If $B$ randomizes his decision by guessing $1$ with probability $p$ and $2$ with probability $1 − p$, determine his expected gain if

$(a)$ $A$ has written down number $1$.

$(b)$ $A$ has written down number $2$.

What value of $p$ maximizes the minimum possible value of $B$’s expected gain, and what is this maximin value? (Note that $B$’s expected gain depends not only on $p$, but also on what $A$ does.)

Consider now player $A$. Suppose that she also randomizes her decision, writing down number $1$ with probability $q$. What is $A$’s expected loss if

$(c)$ $B$ chooses number $1$?

$(d)$ $B$ chooses number $2$?

What value of $q$ minimizes $A$’s maximum expected loss? Show that the minimum of $A$’s maximum expected loss is equal to the maximum of $B$’s minimum expected gain.

My attempt (just the results)

$(a)$: $\frac{1}{4}(7p-3)$.

$(b)$: $2-\frac{11}{4}p$.

$(c)$: $\frac{1}{4}(3-7q)$.

$(d)$: $2-\frac{5}{4}q$.

I don't understand the what the "maximize minimum" and "minimize maximum" part of the question means.

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  • $\begingroup$ When you are calculating expected loss in part (c), shouldn't you take loss as positive? Because you need to calculate expected loss (and not gain), when A gives the money, it is her loss, so it should be positive and when A gets money, it is her gain which should be her negative loss. So expected loss should be $1*q - 3(1-q)/4 = (7q-3)/4$. $\endgroup$
    – Ankit Seth
    Sep 11, 2021 at 6:40

2 Answers 2

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Maximizing the minimum possible value of B's expected gain means that A either chooses 1 or 2, and you take the one that is less favorable to B (minimum expected gain). Then you choose the p for which this minimum will be of maximal value.

Minimax means that both players act perfectly rationally, so your opponent always makes the best move (min for you), and you need to choose your best move accordingly (max for you).

I think the last question is about finding the Nash equilibrium.

(I wanted to comment but I didn't have 50 reputation.)

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  • $\begingroup$ I left out this part of the question: "This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. " Is that what this Nash equilibrium thing is? $\endgroup$
    – David
    Dec 11, 2016 at 22:57
  • $\begingroup$ I found it here: en.wikipedia.org/wiki/Minimax_theorem $\endgroup$
    – GregT
    Dec 11, 2016 at 23:09
  • $\begingroup$ Btw, Neumann was a real genius, with serious contributions to computers, game theory, the Manhattan Project and quantum mechanics. $\endgroup$
    – GregT
    Dec 11, 2016 at 23:12
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The way that I understand "maximizing the minimum" is as follows:

Let $y(p)=\frac{7}{4}p - \frac{3}{4}$ and $g(p)=2 - \frac{11}{4}p$. Let's plot the functions $y,g$ on a same graph and x axis being the value of $p \in [0,1]$. Because $y$ has positive slope and $g$ has negative slope, two lines most likely meet at a point $p \in [0,1]$. The meeting point is where the minimum of the two functions are maximized because, from that meeting point, if you select larger $p$, let's call it $p^*$ then $g(p^*)$ is going to be a new minimum that is smaller than we came from ($g(p) = y(p)$). Hope this helps!

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  • $\begingroup$ I applied the same strategy. It is like bias-variance tradeoff. $\endgroup$
    – Ankit Seth
    Sep 11, 2021 at 6:42

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