Binomial probability doesn't give the correct number?

If you roll five dice your chances of getting all five matching is $\frac{1}{{6}^{5}}=\frac{1}{7776}\approx 0.08\mathrm{\%}$.

I recently learned the binomial probability formula: $P=nCr\times {p}^{r}\times (1-p{)}^{n-r}$.

- $nCr=\frac{n!}{r!(n-r)!}$ << The different possible combinations

- r is the number of dice you need to match

- n is the number of dice you're rolling

I use $\frac{1}{6}$ in the formula because that's dice probability

Because I am rolling 5 dice and need 5 matches, I set $r=5$ and $n=5$ and came up with this equation: $P=\frac{5!}{(5-5)!5!}\times (\frac{1}{6}{)}^{5}\times (1-\frac{1}{6}{)}^{5-5}\approx 0.012\mathrm{\%}$

But when I set $r=4$, $n=4$ with the formula below I do get the desired result. $P=\frac{4!}{(4-4)!4!}\times (\frac{1}{6}{)}^{4}\times (1-\frac{1}{6}{)}^{4-4}\approx 0.08\mathrm{\%}$

Why is this? Shouldn't r and n be 5 because I'm rolling 5 dice and need 5 matches?

If you roll five dice your chances of getting all five matching is $\frac{1}{{6}^{5}}=\frac{1}{7776}\approx 0.08\mathrm{\%}$.

I recently learned the binomial probability formula: $P=nCr\times {p}^{r}\times (1-p{)}^{n-r}$.

- $nCr=\frac{n!}{r!(n-r)!}$ << The different possible combinations

- r is the number of dice you need to match

- n is the number of dice you're rolling

I use $\frac{1}{6}$ in the formula because that's dice probability

Because I am rolling 5 dice and need 5 matches, I set $r=5$ and $n=5$ and came up with this equation: $P=\frac{5!}{(5-5)!5!}\times (\frac{1}{6}{)}^{5}\times (1-\frac{1}{6}{)}^{5-5}\approx 0.012\mathrm{\%}$

But when I set $r=4$, $n=4$ with the formula below I do get the desired result. $P=\frac{4!}{(4-4)!4!}\times (\frac{1}{6}{)}^{4}\times (1-\frac{1}{6}{)}^{4-4}\approx 0.08\mathrm{\%}$

Why is this? Shouldn't r and n be 5 because I'm rolling 5 dice and need 5 matches?