a question for the experts!

#1
Could anyone work this out?

At a recruitment event, 34 candidates are interviewed by 4 groups of interviewers (each group consisting of 2 interviewers);

What is the probability of the following happening:

1. Only candidates (A and B) have the same group of interviewers (C and D);
and
2. A and B are given the same mark by C and D- 34/50;
and
3. The interview marking form used by C and D for A and B have been filled out incorrectly. All other forms by other pairs of interviewers have been filled out correctly for the other 32 candidates;
and
4. Interviewer C was involved in 8 other interviews and correctly filled out the interview marking form.

Can any boffins on here tell me how they would work out the chances of all this happening?

Many thanks!
 

hlsmith

Not a robit
#4
Well this doesn't seem like a controlled experiment performed in a vacuum, so it may get wishy-washy.

So 34 candidates, that is straightforward. But later you are calling them A,..,D?
How many interviewers are there in total?
Was interviewer assignment suppose to be random?
Was interviewer pairing suppose to be random?
I don't think we can answer much about the "marks" since that is likely based on some subjective/objective mechanism (may be able to look into interrater reliability), the other questions may be able to be likened to ball draws from an urn, but I wonder if there are just too many intangibles. Say some interviewers go longer or self-select to interviewing with other interviewers or the physical location in the building results in certain pairings. This is a real convoluted series of questions and context. What do you hope to get out of it.