In this problem, you are looking for the probability of winning a bet by correctly selecting the three horses that finish in the top three positions, without needing to specify the exact order.

Step 1: Total Number of Possible Outcomes

First, let's calculate the total number of ways to choose 3 horses from a field of 12 horses. Since the order doesn't matter, we can use combinations to find this.

The formula for combinations is: $C(n, r) = \frac{n!}{r!(n-r)!}$ where $n$ is the total number of horses, and $r$ is the number of horses we want to choose. In this case, $n = 12$ and $r = 3$.

$C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$ This simplifies to: $C(12, 3) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

Thus, there are 220 possible ways to choose any 3 horses from 12.

Step 2: Number of Successful Outcomes

There is exactly 1 successful outcome: choosing the 3 horses that actually finish first, second, and third.

Step 3: Probability of Winning

The probability is the ratio of successful outcomes to total possible outcomes, which is: $\text{Probability} = \frac{1}{220}$

Therefore, the probability of winning the bet is: $\frac{1}{220}$

Final Answer:

The probability of randomly selecting the top 3 horses is $\frac{1}{220}$.

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Would you like further details on any of these steps? Here are some related questions to expand on:

1. How would the probability change if the order of horses mattered?
2. What is the formula for permutations, and how does it differ from combinations?
3. If there were more than 12 horses, how would the calculation be adjusted?
4. How can we calculate the odds in favor of winning versus losing this bet?
5. What is the expected number of tries to win this bet?

Tip: For problems involving choosing items where order does not matter, always use combinations, whereas for problems involving order, use permutations.