# II_Question12_Prob

I have attempted a full solution below (spoilers, you should try the question first before looking at the answer!!)

i)

Winnings can either be £0, or £h \Rightarrow expected winnings are given by \mathrm{f}(h) = h\mathrm{P}(H=h).

In order to toss h heads, there must be a string of h heads, else the game ends. Hence \mathrm{P}(H=h)=p^h \Rightarrow \mathrm{f}(h) = hp^h .

\mathrm{f}'(h)=p^h + hp^h\ln p, and so at stationary values of \mathrm{f}(h), we have \mathrm{f}'(h) = 0 \Rightarrow 1 + h\ln p= 0 \Rightarrow h = -\dfrac{1}{\ln p}.

Using p=\dfrac{N}{N+1}, we have h=-\dfrac{1}{\ln\left(\frac{N}{N+1}\right)} = \dfrac{1}{\ln\left(1+N^{-1}\right)}.

The Taylor series about x=0, the Maclaurin series, of \ln(1+x), begins x - \dfrac{1}{2}x^2 + \dfrac{1}{3}x^3, and so h=\dfrac{1}{N^{-1} - \frac{1}{2}N^{-2} + \frac{1}{3}N^{-3} - \dotsb} \approx \dfrac{1}{N^{-1}} = N.

Now, note that h - N = \dfrac{\frac{1}{2}N^{-1} - \frac{1}{3}N^{-2} + \dotsb}{N^{-1} - \frac{1}{2}N^{-2} + \frac{1}{3}N^{-3} - \dotsb} = \dfrac{\frac{1}{2} - \frac{1}{3}N^{-1} + \dotsb}{1 - \frac{1}{2}N^{-1} + \frac{1}{3}N^{-2} - \dotsb}.

Hence \lim\limits_{N \to \infty}(h - N) = \dfrac{1}{2}, and so N < h < N + \dfrac{1}{2}, since approximation \ln\left(1+N^{-1}\right) \approx N^{-1} is an over-estimate (can be seen graphically for \ln(1+x) \approx x), so we infer that it’s reciprocal as an approximation for h is an underestimate.

Since \mathrm{f}(h) will increase up to a maximum, and then decrease; and is not symmetric about this maximum point, we have N and N+1 contending to be the value maximising \mathrm{f}(h).

We have \dfrac{\mathrm{f}(N)}{\mathrm{f}(N+1)} = \dfrac{Np^N}{(N+1)p^{N+1}} = \dfrac{N}{N+1} \cdot \dfrac{1}{p} = 1 . That is both N and N+1 will maximise \mathrm{f}(h).

Therefore, choosing h = N will maximise \mathrm{f}(h).

ii)

Similarly, the expected winnings are \mathrm{g}(h) = h\mathrm{P}(H = h).

If we repeatedly have H or TH for h times, we will have all possible sequences that win with h heads.

So \mathrm{P}(H = h) = \left(p + (1-p)p\right)^h = \left(1 - (1-p)^2\right)^h.

Using p = \dfrac{N}{N+1} gives \mathrm{P}(H = h) = \dfrac{N^h(N+2)^h}{(N+1)^{2h}} \Rightarrow \mathrm{g}(h) = \dfrac{hN^h(N+2)^h}{(N+1)^{2h}}.

When N=2, \mathrm{g}(h) = h\left(\dfrac{8}{9}\right)^h. We have \mathrm{g}'(h) = \left(\dfrac{8}{9}\right)^h + h \left(\dfrac{8}{9}\right)^h \ln \dfrac{8}{9}, and so for stationary \mathrm{g}(h), h = \dfrac{1}{\ln \frac{9}{8}}.

Apply logarithm laws gives h = \dfrac{1}{2\ln 3 - 3\ln 2} = \dfrac{1/\ln 3}{2\log_33 - 3\log_32} \approx \dfrac{1/\ln 3}{2 - 3\times 0.63} = \dfrac{1/\ln 3}{0.11} = \dfrac{100}{11\ln 3}. Approximately, 11 \approx 12.5 and \ln 3 \approx 1, respectively over- and under-approximating, so 11 \ln 3 \approx 12.5 \Rightarrow h \approx 8.

Therefore, \mathrm{g}(h) \approx 8\left(\dfrac{8}{9}\right)^8 = \dfrac{8^9}{9^8} = \dfrac{(9-1)^9}{9^8} = \dfrac{9^9(1 - \frac{1}{9})^9}{9^8} = 9(1 - \frac{1}{9})^9.

\mathrm{g}(h) \approx 9\left(1-9 \cdot \dfrac{1}{9} + \dfrac{9\cdot 8}{2}\cdot \dfrac{1}{9^2} - \dfrac{9\cdot 8 \cdot 7}{6} \cdot \dfrac{1}{9^3}\right) = 4 - \dfrac{28}{27} \approx 3.

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