I just recently had an interview for a new job, so hurrah for me. The thing about this job is, they’re interviewing 18 people for 5 positions. They’re probably not picking randomly, but if they were, what’s the chance I’ll get the job? I’d ask, because my first instinct…18 jobs, 5 people, means I have a 5 in 18 chance of getting the job, or about 27.8%.
My problem is, another method of figuring out my chances comes to mind, if we treat each job distinctly. So, I have a 1 in 18 chance of getting the first job, and then a 1 in 17 chance of getting the second job, a 1 in 16 chance of the third, a one in 15 chance of the 4th, and a 1 in 14th chance of the third, or…
Probabilities of sequential events don’t add that way. Think of dice rolls to make it easy. A 6 sided and a 10 sided you need a 1 on either to get the job. So you have a 5/6 chance to not get the first and a 9/10 to not get the second. Multiply those to see the probability of neither. That would be (59)/(610) of failure. Success would be 1-(59)/(610).
Apply that method to your second example for the correct number for a sequential job pick. 1-(1716151413)/(1817161514)
Another way to do it would be to note that there are (18x17x16x15x14)/(5x4x3x2x1)=8568 possible outcomes for selecting 5 people from a pool of 18, and then figure that (17x16x15x14x13)/(5x4x3x2x1)=6118 of them don’t include you. So your chance of being picked is 2450/8568 which simplifies to 5/18.
As someone who has interviewed hundreds of applicants over the years I can tell you that if you did a good job in the interview you probably have about 5 chances in 9. It is my experience that about half of all the interviewees may as well walk in and punch a panel member in the face because they will totally ruin their prospects anyway.
a friend of mine went for a job and stumbled through the interview. When it finished, he got up, opened a door and went through. It was a cupboard door. :smack:
I went for a specialist job that was ideal for me (involving chess.) There were two other applicants. One had just come along because he was applying for everyything (he didn’t know anything about chess) and the other was a fellow chess player. He immediately remarked “Glee - if I’d known you were interested, I wouldn’t have bothered trying!”
If you get the first job, there is a zero percent chance of getting the 2nd job. Therefore the odds of getting the 2nd job is (1/17) times (17/18), or 1 chance in 18 of getting the 2nd job.
Another example: my husband went in for a job interview in a tie, sport jacket and dockers. The guy who left the interview just before him was in jeans and a dirty T-shirt. DH got the job.
With exceptions, like jobs that require a law degree etc., there are always going to be a few people trying not to get the job. Unemployment makes people fill out three applications (it may vary by state: it’s three in Indiana) a week. If you get called for an interview and don’t go, you lose your benefits.
If you do get the first job opening, you’re not eligible for the 2nd job opening. So the probability of getting the 2nd job is <probability of not getting the 1st job opening> x <probability of getting the 2nd job opening, given that you didn’t get the 1st>. Which is (17/18) x (1/17) = 1/18.
For the 3rd position, it’s <probability of not getting 1st opening> x <probability of not getting 2nd opening, given that you didn’t get the 1st> x <probability of getting the 3rd opening, given that you didn’t get the 1st or 2nd> = (17/18) x (16/17) x (1/16) = 1/18.
Imagine the filling of the five jobs as five sequential events. To be eligible for one of the later slots, you must have been passed over for all of the earlier slots.
Your odds of filling the first job: 1/18
Now there are 17 applicants remaining. Your odds of filling the second job are the product of your odds of being passed over for the first job (17/18) and your odds of being the one applicant of 17 chosen for the second job. in other words, 17/18 * 1/17. The 17s cancel, and you are left with 1/18.
Now there are 16 applicants remaining. Your odds of filling the third job are the product of your odds of being passed over for the first two jobs (16/18) and your odds of being chosen for the third job. In other words, 16/18 * 1/16. The 16s cancel, and you are left with 1/18.
You can see the pattern. To be a candidate for the fifth and final slot, you must be one of the fourteen candidates passed over for the first four jobs (14/18), and you must be chosen for that fifth and final slot (1/14). Again, it works out to 1/18.
So your equation should be:
(1/18) + (17/18)(1/17) + (16/18)(1/16) + (15/18)(1/15) + (14/18)(1/14) = 27.8%, same as your first analysis.
edit: ninja’d by scr4; what were the odds of that?
This is the right instinct, at least if we imagine that the employers choose five “winners” at random from the 18 applicants.
In this case, you can model the employers’ procedure as follows: Imagine that the employers put every applicant’s name on its own slip of paper. They put all 18 slips into a bucket and mix them up. Then a blindfolded person pulls out slips one-at-a-time (without putting slips back in the bucket). As each name is pulled, it is added to a list. In the end, they have a list of all 18 names in random order. Finally, they give jobs to the first 5 people on the list, and they reject everyone else.
Intuitively, your name has an equal chance of being in any particular position in the list. That is, your chance of being the first name is 1/18. Your chance of being the second name is 1/18. Your chance of being the third name is 1/18. And so on. There is nothing to bias your chances toward being in any particular position on the list over any of the others, so you have an equal chance of being in any one of them.
This means that your chance of being among the first five names is 5/18. This is therefore the probability that you will be hired.
