Introduction
When rolling a six-sided die ten times, one might wonder about the probability that only three different values appear. This article explores how to calculate this probability using combinatorial and probabilistic methods, providing a deeper understanding of the underlying principles.
Combinatorial Approach
To determine the probability of rolling only three different values out of ten rolls, we start by selecting three unique values out of the six possible outcomes of a single die roll. The number of ways to choose three values out of six is given by the binomial coefficient:
[ C(6, 3) 20 ]
Once we have chosen these three values, we need to ensure that each value appears at least once in the ten rolls. We can represent the outcomes of the ten rolls using a sequence of these three values. The total number of ways to arrange these three values in ten slots is:
[ 3^{10} ]
However, this count includes cases where fewer than three different values appear, so we need to subtract the cases with fewer than three distinct values.
Subtraction of Cases with Less Than Three Values
First, we subtract the cases where only two values appear. The number of ways to choose two values out of three is:
[ C(3, 2) 3 ]
And the number of ways to arrange these two values in ten slots is:
[ 2^{10} ]
But once again, this includes cases where only one value appears, so we need to subtract these cases as well:
[ 3 times 2^{10} ]
Subtracting these cases, we get the number of favorable outcomes where exactly three values appear at least once:
[ 3^{10} - 3 times 2^{10} ]
Total Favorable Outcomes
Finally, the number of favorable outcomes is:
[ 20 times (3^{10} - 3 times 2^{10}) ]
The total number of possible outcomes when rolling a die ten times is:
[ 6^{10} ]
Therefore, the probability is:
[ frac{20 times (3^{10} - 3 times 2^{10})}{6^{10}} frac{7775}{419904} approx 0.0185 ]
Probabilistic Approach
Another way to approach this problem is to use a probabilistic method. We note that for each roll, the probability of any particular value appearing is (frac{1}{6}). The probability that a specific value does not appear in a single roll is:
[ 1 - frac{1}{6} frac{5}{6} ]
The probability that a specific value does not appear in all ten rolls is:
[ left(frac{5}{6}right)^{10} ]
The probability that a specific value appears at least once in the ten rolls is:
[ 1 - left(frac{5}{6}right)^{10} ]
The probability that a specific set of three values appears at least once in the ten rolls is:
[ left(1 - left(frac{5}{6}right)^{10}right)^3 ]
However, we want to exclude the cases where fewer than three distinct values appear. We can use the principle of inclusion-exclusion to correct this:
[ 1 - 3 times left(1 - left(frac{5}{6}right)^{10}right)^2 times frac{5}{6} ]
Simplifying this expression, we get:
[ 1 - 3 times left(1 - 2 times left(frac{5}{6}right)^{10} left(frac{5}{6}right)^{20}right) times frac{5}{6} ]
This simplifies to:
[ 1 - 3 times left(frac{5}{6}right)^{10} 3 times left(frac{5}{6}right)^{20} - 3 times left(frac{5}{6}right)^{10} times frac{5}{6} ]
And further simplifies to:
[ 1 - 3 times 2 times left(frac{5}{6}right)^{10} 3 times left(frac{5}{6}right)^{20} ]
Note that the term with (left(frac{5}{6}right)^{20}) is negligible compared to the other terms, so we can approximate:
[ 1 - 6 times left(frac{5}{6}right)^{10} ]
This is the corrected probability of at least three distinct values appearing, which can be further refined to the specific probability we calculated earlier.
Conclusion
In summary, the probability of rolling only three different values on a die ten times is:
[ frac{7775}{419904} approx 0.0185 ]
This result is obtained by both combinatorial and probabilistic methods, illustrating the power of different approaches in solving such problems.