Odds of having exactly 12 correct answers out of 13

[Addez123]

TL;DR

On a lottery ticket you can choose 1,x,2 as answer on 13 questions.
Whats the odds that you get exactly 12 correct answers?

The odds are:
$$\frac {1}{3}^{12} * \frac {2}{3} * 13P1$$
And it's the 13P1 that I don't understand.

It's basically saying, among 13 lines, how many ways can 1 line be choosen to be incorrect?
Shouldn't 13P12 be equally correct?
That would be like saying: Among 13 lines, how many ways can 12 lines be choosen correctly?

My question is, why 13P1 and not 13P12?
Maybe it should be 13C1?
But that doesn't make sense since combinations doesn't regard order. Having first answer wrong isn't the same as having second answer wrong. So I should be using permutations, right?

 

[etotheipi]

It's a binomial problem, so it should be $n C r$. If you have 13 trials. (labelled 1,2,3,...,13) and get $r$ successes, you have $13 C r$ combinations for possible sets of trials on which you succeed. It's a combination because we already fixed the order of the trials, so for anyone combination there is only one permutation that makes sense.

That is to say, for example, that the events that the successful trials are $\{1,4,7\}$ and $\{4,1,7\}$ are the same event, so we need only count it once! If all permutations of a set are equivalent in the context of our problem, then it is the number of combinations that is meaningful.

Luckily for you (or unluckily?), $13C1 = 13P1$.