In a game, I toss a coin repeatedly. The probability, \(p\), that the coin shows Heads on any given toss is given by
\[ p= \frac N{N+1} \,, \]
where \(N\) is a positive integer. The outcomes of any two tosses are independent.
The game has two versions. In each version, I can choose to stop playing after any number of tosses, in which case I win £\(H\), where \(H\) is the number of Heads I have tossed. However, the game may end before that, in which case I win nothing.
In version 1, the game ends when the coin first shows Tails (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Find, in terms of \(h\) and \(p\), an expression for my expected winnings and show that I can maximise my expected winnings by choosing \(h=N\).
In version 2, the game ends when the coin shows Tails on two consecutive tosses (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Show that my expected winnings are
\[ \frac{ hN^h (N+2)^h}{(N+1)^{2h}} \,.\]
In the case \(N=2\,\), use the approximation \(\log_3 2 \approx 0.63\) to show that the maximum value of my expected winnings is approximately £3.
Solution:
Since we either win \(h\) or \(0\), to calculate the expected winnings we just need to calculate the probability that we get \(h\) consecutive heads, therefore:
\begin{align*}
&& \mathbb{E}(\text{winnings}) &= E_h \\
&&&= h \cdot \left ( \frac{N}{N+1} \right)^h \\
&& \frac{E_{h+1}}{E_h} &= \frac{h+1}{h }\left ( \frac{N}{N+1} \right)
\end{align*}
Therefore \(E_h\) is increasing if \(h \leq N\), so we can maximise our winnings by taking \(h = N\). (In fact, we could take \(h = N\) or \(h = N+1\), but arguably \(h = N\) is better as we have the same expected value but lower variance).
We can have up to \(h\) tails appearing (if we imagine slots for tails of the form \(\underbrace{\_H\_H\_H\_\cdots\_H}_{h\text{ spaces and }h\, H}\) so, we have
\begin{align*}
&& \mathbb{P}(\text{wins}) &= \sum_{t = 0}^h \mathbb{P}(\text{wins and } t\text{ tails}) \\
&&&= \sum_{t = 0}^h\binom{h}{t} \left ( \frac{N}{N+1} \right)^h\left ( \frac{1}{N+1} \right)^t \\
&&&= \left ( \frac{N}{N+1} \right)^h \sum_{t = 0}^h\binom{h}{t}\left ( \frac{1}{N+1} \right)^t \cdot 1^{h-t} \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( 1 + \left ( \frac{1}{N+1} \right) \right)^h \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( \frac{N+2}{N+1}\right)^h \\
&&&= \frac{N^h(N+2)^h}{(N+1)^{2h}} \\
\Rightarrow && \E(\text{winnings}) &= h \cdot \frac{N^h(N+2)^h}{(N+1)^{2h}}
\end{align*}
If \(N = 2\), we have
\begin{align*}
&& \E(\text{winnings}) &= E_h \\
&&&= h \cdot \frac{2^h\cdot2^{2h}}{3^{2h}}\\
&&&= h \cdot \frac{2^{3h}}{3^{2h}} \\
\Rightarrow && \frac{E_{h+1}}{E_h} &= \frac{h+1}{h} \frac{8}{9} \\
\end{align*}
Therefore to maximise the winnings we should take \(h = 8\), and the expected winnings will be:
\begin{align*}
&& E_8 &= 8 \cdot \frac{2^{24}}{3^{16}} \\
\Rightarrow && \log_3 E_8 &= 27 \log_3 2 - 16 \\
&&&\approx 24 \cdot 0.63 - 16 \\
&&&\approx 17 - 16 \\
&&&\approx 1 \\
\Rightarrow && E_8 &\approx 3
\end{align*}