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

I am a muslim and there is ramadan going on. The last 10 days of ramadan are really special, and in one of these 10 days there is one special night that no one knows about.
The special day is more likely to happen on the odd nights, but it can be on any day.

How can the probability of a given day being the special day be modelled statistically?

My logic: Since we know it needs to fall in one of the 10 days and these all equally likely to land on any day. So the probability of that happening is 10%. I would think the binomial theorem would play a role because either it happens that day or it doesn't. But when I calculated it all the way to n =9 (0 included) my probability got really low as I approached the last day.

• Do you know/can assume there is only one “special night” or could there be several? – Gala Jul 31 '13 at 6:29
• there is only one in those 10 days. 100%. like I want to know what is the prob itll be today. if not today what is the prob having it tomm. – MathGeek Jul 31 '13 at 6:34
• It seems that what you are looking for might be the hypergeometric distribution. – Gala Jul 31 '13 at 6:45
• The meaning of "probability" depends on context. Could you please clarify, then, what is known at the point you wish to compute probabilities? For instance, do you wish to model the probability that tomorrow is the special day given that the special day has not yet occurred? – whuber Jul 31 '13 at 15:20

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 $1 \le t \le n$.
• $\rm{Pr}_n(T\le t)$, 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 $1 \le 2 \not\le 1$.

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.

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

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

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

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.