STEP I 1998 question 14 solution - The Student Room
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STEP I 1998 question 14 solution

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To celebrate the opening of the financial year the finance minister of Genland flings a Slihing, a circular coin of radius a cm, where 0 < a < 1, onto a large board divided into squares by two sets of parallel lines 2 cm apart. If the coin does not cross any line, or if the coin covers an intersection, the tax on yaks remains unchanged. Otherwise the tax is doubled.
  1. Show that, in order to raise most tax, the value of a should be

    \left( 1+ \dfrac{\pi}{4} \right) ^{-1}

  2. If, indeed, a=\left( 1+ \frac{\pi}{4} \right) ^{-1} and the tax on yaks is 1 Slihing per yak this year, show that its expected value after n years will have passed is

    \left( \dfrac{8 + \pi}{4 + \pi} \right) ^n


Part 1
For this question we need only consider the centre of the coin. The coin will cover an intersection if and only if the centre of the coin is within a cm of an intersection. Likewise, the coin will not cross any line if and only if the centre of the coin is more than a cm away from any line. It may help if you draw a diagram similar to the (horrible MSPaint) one on the right. We are only considering one square in the grid as the situation is the same for all squares. In order to generate the most tax, we need to maximise the probability that the coin crosses a line but not an intersection and thus we want to maximise the shaded area.
The shaded area f(a) is given by:

f(a)= 2^2 -\pi a^2 - (2-2a)^2

f(a)= 8a -(4+ \pi)a^2

Finding a maximum by differentiation:

f'(a)= 8-2(4+ \pi)a =0
a= \left( 1+ \dfrac{\pi}{4} \right) ^{-1}

If you're worried about whether this stationary point is a maximum or a minimum, you can find the second derivative:

f''(a)= -2(4+ \pi) < 0, so the stationary point is indeed a maximum.


Part 2
First let us find the probability that the tax will double in any given year. We simply need to find the fraction of the area of the square that is shaded. The shaded area is given by

f((1+ \frac{\pi}{4})^{-1}) = 8(1+ \frac{\pi}{4})^{-1} -(4+ \pi)(1+ \frac{\pi}{4})^{-2}

f((1+ \frac{\pi}{4})^{-1}) = 4(1+ \frac{\pi}{4})^{-1}

Dividing this by 4 (the area of the square) gives us the probability p:

p= (1+ \frac{\pi}{4})^{-1}

Which is coincidentally (or not?) equal to a.
You may remember from S1 that the formula to find the expected value, E(X) of a finite discrete probability distribution is

E(X)= \displaystyle\sum_{i=0}^n x_ip_{x_i}

Where the random variable X has n different possible values. In a nutshell, you multiply each possible value with its respective probability and then add them all up.
In this question we are effectively measuring how many times the tax doubles and so the values of the random variable will be of the form 2^r. Their respective probabilities are equal to those of a standard binomial distribution: ^n \mathrm{C} _r p^r (1-p)^{n-r}.
Using our calculated value of p, the expected value is thus

\displaystyle\sum_{r=0}^n ^n \mathrm{C} _r ((1+ \frac{\pi}{4})^{-1})^r (1-(1+ \frac{\pi}{4})^{-1})^{n-r} 2^r

This eventually rearranges to give

\left(\dfrac{\pi}{4+\pi} \right)^n \displaystyle\sum_{r=0}^n ^n \mathrm{C} _r \left( \dfrac{8}{\pi} \right)^r

Considering that

(1+x)^n = \displaystyle\sum_{r=0}^n ^n \mathrm{C} _rx^r

We get

\left(\dfrac{\pi}{4+\pi} \right)^n \left( 1+ \dfrac{8}{\pi} \right)^n = \left( \dfrac{8 + \pi}{4 + \pi} \right) ^n


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