Given 8 independent choices of a random number between 1 and 100, what is the probability of any given choice being the highest number?

Question

Given 8 people choosing a random number between 1 and 100, with repetition allowed, what's the chance of being the one to choose the highest number? Assuming it is truly random.

Obviously the chance of choosing one particular number is $ 1/100 $

If rank didn't come into play, I would think it would simply be $ 1/8 $

However, even with the choices being independent, the results rely on all the others, so would they technically be dependent?

That and the rank of the numbers coming into play is what's confusing. It's been several years since I touched a probability textbook and have forgotten most of all of it.

Now, assuming that implies they are dependent, how does one chain 8 conditional probabilities into one calculation?

My best guess for notation is $ P((A | B)|C...) $ but I'm not sure where I would start with calculations.

Answer 1

The probability that everyone else choose a number at most as great as you is:

$$\sum\limits_{k=1}^{100} \dfrac{k^{7}}{100^8}=\dfrac{520233321667}{4000000000000}$$

The probability that everyone else choose a number less that you did is:

$$\sum\limits_{k=2}^{100} \dfrac{(k-1)^{7}}{100^8}=\dfrac{480233321667}{4000000000000}$$

Neither is exactly $1/8$ because of the non-zero probabilities of ties.

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PS: This of course presumes choices are made with independent uniform discrete integer distribution.   Check these assumptions.

Answer 2

If I am interpreting your question correctly, you want to know the distribution of $N_{max}$ where $N_{max}$ is the number of the person with maximum choice. Let $X_i$ be the i-th person's choice.

$P(N_{max}=r)=P(X_r\geq X_i\forall i\neq r)=\sum_{x=1}^{100}P(X_r=x, X_j\leq x\forall j\neq r)=\sum_{x=1}^{100}P(X_r=x)\prod_{j\neq r}P(X_j\leq x)=\sum_{x=1}^{100}(\dfrac{1}{100})(\dfrac{x}{100})^{r-1}$

This holds for $r=1,2,...,8$.