# 4-alternative forced choice problem

#### ClementFerrand

##### New Member
Hi there,

I have the following problem:

Participants in my experiment were presented with 4 objects and 4 categories and asked to map each of the items to its correct category. It was a forced choice, so one and only one item could be mapped to each of the four category bins. I'd now like to compute the probability that a participant got 0, 1, 2, 3, or 4 correct mappings if he randomly assigned items to categories (chance).

Getting the chance probability for 4 correct mappings seem easy, it should be 1/4*1/3*1/2*1=1/24. 3 correct mappings are not possible, because once you got 3 correct, there's only one remaining option for the last mapping.

However, I'm struggling with computing the p's for 0 correct mappings. I though it would be 3/4*2/3*1/2*1=6/24, but the problem is more complex. The first mapping constrains the second mapping, so there are different probabilities dependent on each choice. Is there an easy way to compute this or do I have to go through all possible branches of the decision tree?

Thanks,
Klemens

#### Dason

It's impossible to get only 3 correct if I understand you correctly (think about why that must be true). I can't think of a better way than just brute forcing every possibility though. Fortunately with 4 outcomes there are only 4! = 24 different ways to do the mapping.

9 of the mappings get 0 correct
8 of the mappings get 1 correct
6 of the mappings get 2 correct
1 of the mappings get 4 correct

So P(0 correct) = 9/24 and so on.

If you've ever used R here is some code to automate that for you
Code:
library(combinat)
table(sapply(permn(1:4), function(x){sum(x == 1:4)}))

#### ClementFerrand

##### New Member
Great, thanks. Was aware of the 3 correct issue. Indeed I'm a part-time R user, so thanks a lot for suggesting the package!