Csdrhrt

I'm trying to calculate the probability of rolling exactly 1 four with 2 die.
Using probability,
$$
P(text{exactly 1 four}) = frac{1}{6}*frac{5}{6} * 2! = frac{10}{36}
$$

If I try to use combinatorics (specifically the 'stars and bars' approach) to solve this problem, I find that the total number of combinations is 21 or $binom{6 + 2 - 1}{2}$. There's 5 ways to select exactly 1 four (I think) where it's $binom{1}{1}*binom{5}{1}$

Is this an incorrect application of combinatorics? Why? How come the probability is $frac{10}{36}$ using probability but $frac{5}{21}$ the other way?

Please let me know if I have made some error in calculations, it's been a long time since I have been in school.

The reason for the discrepancy is that outcomes in which the dice display two different numbers are twice as likely to occur as those outcomes in which they display the same number.

The $21$ possible outcomes you counted consist of $6$ outcomes in which both six-sided dice display the same number and $15$ outcomes in which the two six-sided dice display different numbers.
$$binom{6}{1} + binom{6}{2} = 21$$
This is the number of distinguishable outcomes you would see if you had two indistinguishable six-sided dice, for instance, if both the dice were white with black dots.

However, these outcomes are not equally likely. Imagine that we now have one blue six-sided die and one red six-sided die. There are now $36$ possible outcomes since for each of the six possible outcomes on the blue die, there are six possible outcomes on the red die.

There are still six outcomes in which the two dice display the same number. However, there are now $30$ outcomes in which the two dice display different numbers since for each of the six possible outcomes on the blue die, there are five ways to obtain a different number on the red die.

Assuming the dice are fair, among the $21$ outcomes you counted, the $15$ outcomes in which the dice display different numbers are twice as likely to occur as those in which they display the same number. For instance, the outcome $(3, 3)$ on the white dice can only occur as $(color{blue}{3}, color{red}{3})$ on the colored dice, while the outcome $(2, 5)$ on the white dice can occur as $(color{blue}{2}, color{red}{5})$ or $(color{blue}{5}, color{red}{2})$ on the colored dice, where $(color{blue}{b}, color{red}{r})$ represents the ordered pair of outcomes on the blue die and red die, respectively.

If we make the assumption that a six-sided die is fair, then the probability that a 4 is obtained when a six-sided die is thrown is $1/6$. Since this is true for each roll, we can use the binomial distribution, which states that the probability that a success occurs exactly $k$ times in $n$ trials, each of which has probability $p$ of success, is
$$Pr(X = k) = binom{n}{k}p^k(1 - p)^{n - k}$$
where $binom{n}{k}$ is the number of ways exactly $k$ successes can occur in $n$ trials, $p^k$ is the probability of $k$ successes, and $(1 - p)^{n - k}$ is the probability of $n - k$ failures.

Here the trials are the outcomes on the blue die and the red die. Hence, the probability of obtaining exactly one 4 when two fair six-sided dice are thrown is
$$Pr(X = 1) = binom{2}{1}left(frac{1}{6}right)left(frac{5}{6}right)$$
as your first calculation showed.

After spending additional time looking around on the internet, it appears my combinations are not equiprobable. Probability of rolling 6 when throwing k dice one by one vs all at once.