Here's the question:
Four numbers are drawn at random from a box of ten numbers 0, 1, ..., 9. Find the probability that the largest number drawn is a six if the draws are made with replacement. The answer, evidently, is
$$\left(\frac{7}{10}\right)^4-\left(\frac{6}{10}\right)^4$$
I understand this approach; since there are 7 possible numbers that are $\le6$ and 6 possible numbers that are $\le5$ (and since $\le5 \Rightarrow \le6)$, one can use the Difference Rule to obtain the probability that all numbers drawn are $\le6$ and at least one is $\gt5$, thereby ensuring that the largest draw is a six.
What I do not understand--and what is puzzling me quite a bit--is why the below approach is not equivalent:
$$\frac{K\cdot1\cdot7^3}{10^4} \tag*{for some natural number $K$}$$
I thought that I could use this method to tabulate the number of equally likely outcomes and divide by the number of all possible outcomes, but equivalence does not hold for any natural number $K$ (I first thought to try permutations but later realized the no $K$ works). And, if there is some error in my reasoning for this second approach, is there a similar method that is equivalent to the first, correct approach?
Also, to make sure I understand, suppose I instead wanted to calculate the probability of a six, two fives, and a one (in any order). Would that just be
$$\frac{4!}{10^4}$$
What about the probability of a six, two numbers in {3, 4, 5}, and a number in {1, 2}?
Thanks in advance. This has me questioning my basic understanding of probability.
Your initial calculation shows there are $7^4-6^4=1105$ combinations with largest number $6$. This is not divisible by $7^3=343$, not even by $7$. The error is that combinations with more than one $6$ get double counted in your approach. The logical value for $K$ is $4$, which selects which die is the $6$. $4 \cdot 343=1372$ But you are counting the ones with exactly two sixes, ${4 \choose 2}6^2=216$ twice, the ones with exactly three sixes ${4 \choose 3}6=24$ three times and the one with four sixes four times. Subtracting off the extras gives $1372-216\cdot 1-24\cdot 2-1\cdot 3=1105$. When a combination has more than one six, each six can be the distinguished one in your expression.