How can I calculate the probability of a increasingly likely positive outcome?

Question

I'm not even sure I'm phrasing the question properly, please let me know if there is any standard terms around this type of problem.

I'm trying to find an average number of attempts it would take to find a randomly placed pixel within a fixed size element. If a wrong pixel is chosen, it is removed from the choices.

Scenario:

There is a 100 x 100 pixel square. 1 pixel is the winning pixel, all others are losers. Remove each losing pixel as they are chosen.

A 100x100 square has a total of 10,000 pixels. Is the probability simply determined by multiplying 1/10000 * 1/9999 * 1/9998 * 1/9997 ....

How can I arrive at a number that gives me an idea of an average number of clicks it would take to find the winning pixel? I.e. can I run 1000 of these scenarios and get a fairly accurate estimate?

Answer 1

You're looking to calculate expected value of a variable $X$, which we define as number of clicks until success.

$$E[X] = \sum_{i=1}^{10,000} i \cdot Pr(X = i)$$

Note that for each possible $i \in \{1,2,...,10000\}$, $Pr(X=i) = \frac{1}{10,000}$. Thus,

$$E[X] = \sum_{i=1}^{10,000} \frac{i}{10,000} = 5000.5$$

Running the scenario 1000 times could get you close-ish to this answer. In R:

set.seed(1)
mean(replicate(1000, which(sample(1:10000, replace = F) == 1)))

Answer 2

Well, simply write the expected value formula, i.e. $\sum{xp(x)}$:

$$E[X]=1\times\frac{1}{10000}+2\times\frac{9999}{10000}\times\frac{1}{9999}+...=\frac{10000\times10001}{2}\times\frac{1}{10000}=\frac{10001}{2}$$

where $X$ is the number of throws needed.