Problem source

A taxi driver keeps a packet of toffees and a packet of mints in her taxi. From time to time she takes either a toffee (with probability $p$) or mint (with probability $q=1-p$). At the beginning of the week she has $n$ toffees and $m$ mints in the packets. On the $N$th
occasion that she reaches for a sweet, she discovers (for the first time) that she has run out of that kind of sweet. What is the probability that she was reaching for a toffee?

Solution source

\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*}