Problems

Solution:

+ - ↺

Solution: After \(5\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 1 \text { - go positive every time}\\ 4 & 0 \\ 3 & \binom{5}{1} \text { - go positive every time but 1} \\ 2 &0 \\ \hline & 6 \end{array} Therefore there are \(\frac{6}{2^5} = \frac{3}{16}\) ways to get to \(\{2,3,4,5\}\) After \(6\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 0 \\ 4 & \binom{6}{1} \text { - go positive every time but 1}\\ 3 & 0 \\ 2 & \binom{6}{2} - \text{ - go positive every time but 2} \\ \hline & 21 \end{array} Therefore there are \(\frac{21}{2^6} = \frac{21}{64}\) ways to get to \(\{2,3,4,5\}\) After \(2k\) steps we can reach \(2\) or \(4\). To get to \(2\) we must take \(k+1\) positive steps and \(k-1\) negative steps, ie \(\binom{2k}{k-1}\). To get to \(4\) we must take \(k+2\) positive steps and \(k-2\) negative steps, ie \(\binom{2k}{k-2}\) Therefore there are \(\binom{2k+1}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k}} \binom{2k+1}{k-1}\) After \(2k+1\) steps we can reach \(3\) or \(5\). To get to \(3\) we must take \(k+2\) positive steps and \(k-1\) negative steps, ie \(\binom{2k+1}{k-1}\). To get to \(5\) we must take \(k+3\) positive steps and \(k-2\) negative steps, ie \(\binom{2k+1}{k-2}\) Therefore there are \(\binom{2k+2}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k+1}} \binom{2k+2}{k-1}\) To find the maximum of \(P(s)\) notice that \begin{align*} && \frac{P(2k+1)}{P(2k)} &= \frac12 \frac{\binom{2k+2}{k-1}}{\binom{2k+1}{k-1}} \\ &&&= \frac12 \frac{(2k+2)!(k-1)!(k+2)!}{(2k+1)!(k-1)!(k+3)!} \\ &&&= \frac12 \frac{2k+2}{k+3} = \frac{k+1}{k+3} < 1 \end{align*} So we should only look at the even terms. \begin{align*} && \frac{P(2k+2)}{P(2k)} &= \frac14 \frac{\binom{2k+3}{k}}{\binom{2k+1}{k-1}} \\ &&&= \frac14 \frac{(2k+3)!(k-1)!(k+2)!}{(2k+1)!k!(k+3)!} \\ &&&= \frac14 \frac{(2k+3)(2k+2)}{k(k+3)} \\ &&&= \frac{(2k+3)(k+1)}{2k(k+3)} \geq 1 \\ \Leftrightarrow && (2k+3)(k+1) &\geq 2k(k+3) \\ \Leftrightarrow && 2k^2+5k+3 &\geq 2k^2+6k \\ \Leftrightarrow && 3 &\geq k \\ \end{align*} Therefore the maximum is when \(s = 2\cdot 3\) or \(s = 2\cdot 4\) which we computed earlier to be \(\frac{21}{64}\)

Solution: \begin{align*} \mathbb{P}(\text{run out reading for toffee on } N\text{th occassion}) &= \binom{N-1}{n}p^nq^{N-1-n}p \end{align*} Since out of the first \(N-1\) times, we need to choose toffee \(n\) times, and then choose it again for the \(N\)th time. Therefore: \begin{align*} \mathbb{P}(\text{reaching for toffee} | \text{run out on }N\text{th occassion}) &= \frac{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion})}{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion}) + \mathbb{P}(\text{reaching for mint and run out on }N\text{th occassion})} \\ &= \frac{ \binom{N-1}{n}p^nq^{N-1-n}p}{ \binom{N-1}{n}p^nq^{N-1-n}p + \binom{N-1}{m}q^mp^{N-1-m}q} \\ &= \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} \l \frac{q}{p} \r^{m+ n+ 2-N}} \end{align*} Some conclusions we can draw from this are: As \(p \to 1, q \to 0\), the probability they were reaching for a Toffee tends to \(1\). (And vice versa). If \(p = q\), then the probability is: \begin{align*} \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} } \end{align*} Since \(n+1 \leq N \leq n+m+1\) where \(n \geq m\) we can notice that: \begin{align*} \text{if } N = m + n + 1 && \binom{m+n+1 - 1}{n} &= \binom{m+n+1 - 1}{m} & \text{ so } \mathbb{P} = \frac12 \\ \text{if } N = n+k && \binom{n+k-1}{n} &< \binom{n+k-1}{m} & \text{ so } \mathbb{P} < \frac12 \\ \end{align*}