Suppose 6 people are queueing for a bus. If all orderings are equally likely, what is probability that Jack is in front of Jill?

Question

Suppose 6 people are queueing for a bus. If all orderings are equally likely, what is probability that Jack is in front of Jill? I know that the total number of permutations is 6! which is 720. I am not sure about how to permutate Jack in front of Jill

Answer 1

If "in front" doesn't mean "directly in front": By symmetry, there are as many orderings with Jack in front of Jill as those with Jill in front of Jack. In other words, the probability that Jack is in front of Jill is just $1/2$. (The number of orderings in which Jack is in front of Jill is $\frac{1}{2}\cdot 720=360$.)

If "in front" means "directly in front": For any position Jill is in, out of the $5$ remaining positions, Jack has to go in the position in front of her. Except if Jill is at the front already, in which case it is impossible. This happens with probability $1/6$. So the probability that Jack is directly in front of Jill is $\frac56\times\frac15=\frac16$. (The number of orderings in which Jack is directly in front of Jill is $\frac16\cdot 720=120.$)

Answer 2

I'm not sure if you mean when Jack is directly in front of Jill or just in front of her in general.

If it is the latter, then @bof has posted a perfect solution. The probability that Jack is in front of Jill is the same as the probability that Jill is in front of Jack (due to the symmetry; you could just swap the names) and since the two probabilities add up to $1$ (either Jack is in front of Jill or Jill in front of Jack), the probability is $\frac{1}{2}$.

If your question meant the former, as you have rightly identified, there are 720 possible orderings. So to find the probability, you just need to find how many possible orderings there are of 6 people where Jack is directly in front of Jill. Let us call the six people $A,B,C,D,Jill, Jack$. Since we want $Jack$ in front of $Jill$, let us treat $[Jill \rightarrow Jack]$ as one entity called $X$. Now we just need to find how many possible orderings there are of $A,B,C,D,X$. This is just $5! = 120 $ and so there are 120 ways to order the 6 people with Jack directly in front of Jill. Thus the probability of this is $\frac{120}{720} = \frac{1}{6}$

Answer 3

All of your answers are correct. I realized that the question was ambiguous in the way it was asked.

If Jack is directly in front of Jill, yes, they would be treated as a single group such that instead of 6 groups there would be 5, and the number of permutations=5!=120

If Jack and Jill can be placed anywhere as long as Jack is in front of Jill, there would be 15 ways to arrange them. The answer would then be 15*4!=360 (taking into account the permutation of the other people in the queue).

Thank you for pointing this out!