The problem asks for the probability of being a Winner in a lottery where you select a sequence of four different numbers from 0 through 62, and the order is important.

Total Possible Outcomes:

You are selecting 4 different numbers from 0 to 62, which gives you 63 total options (since 0 is included). Since the order of the numbers matters, we are dealing with a permutation.

The number of ways to choose and order 4 numbers from a set of 63 is calculated using the permutation formula:

$P(n, k) = \frac{n!}{(n - k)!}$

where $n$ is the total number of choices (63), and $k$ is the number of selections (4).

$P(63, 4) = \frac{63!}{(63-4)!} = \frac{63!}{59!} = 63 \times 62 \times 61 \times 60$

Let's compute this:

$P(63, 4) = 63 \times 62 \times 61 \times 60 = 897,480$

So, there are 897,480 possible sequences of four different numbers.

Favorable Outcome:

There is exactly one correct sequence that matches the drawn numbers in the exact order. Therefore, the number of favorable outcomes is 1.

Probability of Being a Winner:

The probability of being a Winner is the ratio of favorable outcomes to total possible outcomes:

$\text{Probability of being a Winner} = \frac{1}{897,480}$

Thus, the probability of being a Winner is:

$\boxed{\frac{1}{897,480}}$

Let me know if you want further clarification or a more detailed explanation!

Related Questions:

1. What is the probability of winning the Booby Prize?
2. How would the probability change if you could select the same number more than once?
3. What if the range of numbers increased to 100 instead of 62? How would that affect the probability?
4. How many different ways are there to select the numbers without considering the order?
5. What if only 3 numbers were drawn? How does this impact the total number of possible outcomes?

Tip:

Always pay attention to whether order matters in probability problems! If it does, permutations are the key. If not, combinations are used.