Spinner of Probability

Let the probability of getting a different number for each spin of $n$ spins, where $n \le x$ , be $p(n)$ . Then for the first spin, $p(1) = \dfrac xx$ , since any number can be the first different number. For the second spin, since 1 number of been chosen, there are remaining $x-1$ numbers to be chosen, then $p(2) = \dfrac xx \times \dfrac {x-1}x$ . Similarly, $p(3) = \dfrac xx \times \dfrac {x-1}x \times \dfrac {x-2}x$ and so on. And $p(x) = \dfrac xx \times \dfrac {x-1}x \times \dfrac {x-2}x \cdots \dfrac 3x \times \dfrac 2x \times \dfrac 1x$ $= \boxed{\dfrac {x!}{x^x}}$ .

The second spin has a $\frac{x-1}{x}$ probability of giving a result different from the first.

If the second spin gives a result different from the first, the third spin has a $\frac{x-2}{x}$ probability of being different from the first two.

This pattern continues all the way to the last spin, which has a $\frac{1}{x}$ probability of being different from all the other spins if they all yielded different results.

To find the probability that all spins yield different numbers, multiply all the probabilities together: $\frac{x-1}{x} \cdot \frac{x-2}{x} \cdot ... \frac{1}{x}$ . This simplifies to $\frac{(x-1)!}{x^{x-1}}$ , because in the numerator, we start with $x-1$ and multiply all the way down to 1, and in the denominator, we multiply $x$ by itself $x-1$ times. We can multiply the numerator and denominator by $x$ , to achieve a result of $\frac{x!}{x^x}$ .