Solved – How to model the probability of an event occurring on a given day when it must occur once in a set of 10 days

As @whuber notes, there are one or two ambiguities in the question.

I would think the binomial theorem would play a role because either it happens that day or it doesn't

First of all, the binomial distribution isn't relevant, so far as I can see. The binomial distribution could be used to model the number of special days in the last 10 days, if there were a fixed, and independent, probability, of each day being special. But we know that there's exactly one special day, so the binomial distribution does not apply.

Second, there seem to be two principal quantities of interest:

• $rm{Pr}_n(T=t)$, the probability of day $t$ being the special day (given it hasn't happened already), with $n$ days left, where $1 le t le n$.
• $rm{Pr}_n(Tle t)$, the probability of the special day occurring on or before day $t$ (given it hasn't happened already), with $n$ days left, where once again $1 le t le n$. This probability is known as the "cumulative" probability.

Given these two quantities, we can, for example, work out the probability that tomorrow is the special day (given it hasn't happened already) by evaluating $rm{Pr}_n(T=2)$.

In the simple case, $T$ is uniformly distributed, so $rm{Pr}_n(T=t) = 1/n$, for $1 le t le n$, or $0$ otherwise. With $n=10$, as in your example, $rm{Pr}_{10}(T=2) = 1/10$, simply by substituting $t=2$. However, with only 1 day left, $rm{Pr}_1(T=2) = 0$, because $1 le 2 notle 1$.

my probability got really low as I approached the last day.

Here, I think you are interested in the cumulative probability, the probability that the special day will occur on or before day $t$. Obviously this should $increase$ as $t$ gets closer to $n$. The way to calculate the cumulative probability is to calculate the probability of the special day not occurring by day $t$, and subtract this from 1 to get the probability of the special day occurring. We calculate the probability of the special day not occurring by day $t$ by multiplying together the chance of it not occurring on the first day (with $n$ days left) with the chance of it not occurring on the second day (i.e. on the first day of the remaining $n-1$ days left), and so on, resulting in $t$ multiplications altogether.

$$begin{eqnarray*} rm{Pr}_n(Tle 1) &=& 1 – (1-rm{Pr}_n(T=1)) \ &=& 1 – frac{n-1}{n} \ &=& frac{1}{n} \ rm{Pr}_n(Tle 2) &=& 1 – (1-rm{Pr}_n(T=1)) times (1-rm{Pr}_{n-1}(T=1)) \ &=& 1 – frac{n-1}{n} times frac{n-2}{n-1} \ &=& 1 – frac{n-2}{n} \ &=& frac{2}{n} \ rm{Pr}_n(Tle 3) &=& 1 – (1-rm{Pr}_n(T=1)) times (1-rm{Pr}_{n-1}(T=1)) times (1-rm{Pr}_{n-2}(T=1)) \ &=& 1 – frac{n-1}{n} times frac{n-2}{n-1} times frac{n-3}{n-2} \ &=& 1 – frac{n-3}{n} \ &=& frac{3}{n} ;;mbox{and in general, for}; t le n\ rm{Pr}_n(Tle t) &=& frac{t}{n} end{eqnarray*}$$

To make things more explicit, in your case, with $n=10$, we have

$$begin{eqnarray*} rm{Pr}_{10}(Tle 1) &=& 1 – (1-rm{Pr}_{10}(T=1)) \ &=& 1 – frac{9}{10} \ &=& frac{1}{10} \ rm{Pr}_{10}(Tle 2) &=& 1 – (1-rm{Pr}_{10}(T=1)) times (1-rm{Pr}_9(T=1)) \ &=& 1 – frac{9}{10} times frac{8}{9} \ &=& 1 – frac{8}{10} \ &=& frac{2}{10} \ rm{Pr}_{10}(Tle 3) &=& 1 – (1-rm{Pr}_{10}(T=1)) times (1-rm{Pr}_{9}(T=1)) times (1-rm{Pr}_{8}(T=1)) \ &=& 1 – frac{9}{10} times frac{8}{9} times frac{7}{8} \ &=& 1 – frac{7}{10} \ &=& frac{3}{10} ;;mbox{and in general, for}; t le 10\ rm{Pr}_n(Tle t) &=& frac{t}{10} end{eqnarray*}$$

Of course, this gives the probability of the special day happening by day 10 as 100%, which is reassuring!

The special day is more likely to happen on the odd nights, but it can be on any day.

You can generalize the argument to calculate cumulative probabilities in the case where $T$ is not uniform, but you won't end up with such a neat solution.