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Suppose that A and B play a strange game. At the beginning of the game A can choose one real number - $x$ - between $0$ and $60$. Each round of the game goes like this:

  1. B names random real number $b$ between $0$ and $60$ with uniform probability.

  2. If $b > x$ then A pays $3(b - x)$ dollars. If $b <= x$ then A pays $x - b$ dollars.

How should A choose $x$ so that the expected payment is minimal?

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Consider that for a guess from A of $x$ that, over time, they expect to pay out the area of the graph:

$\int_0^x (x-b) db + \int_x^{60} 3(b-x) db$

Note that when $b=x$ then the payout is zero in both integrals, and so we do not need to worry about what happens here.

The integrals evaluate to:

$\left(xb-\frac{b^2}{2}\right)\mid_0^x + \left(\frac{3b^2}{2}-3xb\right)\mid_x^{60}$
$=(x^2-\frac{x^2}{2}-0 + 0)+(5400-180x-\frac{3x^2}{2}+3x^2)$
$=2x^2-180x+5400$

To optimize A's choice of $x$, we need to minimize this function, and so we take the differential:

$\frac{d}{dx}=4x-180$

and set to zero to find

$x=45$

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45
Think marginal risk.
Consider that B is chosing $b' = (b-x)$. The value $b'$ is then chosen uniformly between $-x$ and $60-x$. The payout is $3 b'$ if $b>0$, or $-b'$ if $b'<0$.
If A moves $x$ by a small amount, then the payout is the same everywhere, it only changes by moving the range of $b'$. When increasing $x$ a little bit, A adds a risk of a $3b' = 3(60-x)$ payout near the high end and avoids a risk of a $-b' = +x$ payout near the lower end.
A should therefore increase $x$ if $x < 3(60-x)$ i.e. if $x < 45$. A should lower $x$ if $x>45$. In short, the best is $x = 45$.

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