a horse race has 10 entries and one person owns 5 of those horses. assuming that there are no ties, what is the probability that those 5 horses finish 1,2,3, 4th and 5th regardless of the order?
.0040
.0009
N=10C5=10!/(5!5!)=252
A horse race has 14 entries and one person owns 5 of those horses...
To find the probability of those 5 horses finishing 1st, 2nd, 3rd, 4th, and 5th regardless of the order, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
The total number of possible outcomes can be calculated using the concept of permutations. Since there are 10 entries in the race, the total number of possible outcomes is 10 factorial (10!). This can be written as:
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Now, let's calculate the number of favorable outcomes. Since one person owns 5 horses, the 5 horses owned by that person must finish among the top 5 positions. So, we fix the positions of those 5 horses and calculate the number of ways the remaining 5 horses can fill the remaining positions.
The number of ways the remaining 5 horses can fill the remaining positions can be calculated using the concept of permutations as well. However, since order does not matter for the remaining horses, we need to use combinations instead of permutations.
The number of combinations can be calculated using the formula:
nCr = n! / (r! * (n - r)!)
In this case, we need to calculate the combination of 5 horses out of the remaining 5 horses, which can be written as:
5C5 = 5! / (5! * (5 - 5)!)
Simplifying this, we get:
5C5 = 5! / (5! * 0!) = 1
So, there is only 1 way for the remaining 5 horses to fill the remaining positions.
Now, let's multiply the number of favorable outcomes by the number of possible outcomes to find the probability:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes) = (1) / (10!)
Calculating the total number of possible outcomes:
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800
Therefore, the probability that those 5 horses finish 1st, 2nd, 3rd, 4th, and 5th regardless of the order is:
Probability = 1 / 3,628,800
So, the probability is a very small fraction, approximately 2.76 × 10^(-7).
Assume
There are
N=10C5 ways of randomly choosing 5 horses out of 10.
where
N=10!/(5!5!)=25
We assume that
1. order does not count,
2. winning probabilities are equal among horses,
3. results are completely random.
Then the owner has one chance in 252 that his 5 horses are placed 1st to 5th.