Prize this week: you get to decide, once and for all, whether Marshmallow Peeps or Cadbury Creme Eggs are the better candy. No longer will disagreements be chalked up to “difference of opinion.” Once you issue your decree, it will be a matter of record. Minor dissent will be tantamount to outright prevarication. Think you can handle all that power?

Right, the question. Here we go.

Four adjacent offices are to be assigned to four employees at random. What is the probability that Scooter and The Big Man (two of the employees) will be placed next to each other?

UPDATE: Props to Elias, who nailed it on Facebook. He’s going with whichever candy has the least packaging (a man after my own sustainable heart). I’m thinking that’s the Creme Egg.

Solution below the cut…

The biggest mistake kids make with probability questions is to overcomplicate them. If you’ve done a unit on probability in school, you know things can become very complicated very quickly, and that it’s pretty easy to make a mistake under pressure. On the SAT, of course, that’s doubly true. However, on the SAT you’ll always be dealing with small numbers (remember: it’s a *calculator optional* test), so you can sidestep all the necessary calculations if you just get fast at LISTING COMBINATIONS!

Let’s just number the employees 1-4, and list all the possible ways they could be distributed to the four offices:

*once*), it’s still doable. It’s a much faster and more sure-fire way to get a question like this correct than trying to work through all the crazy probability calculations.

*is*difficult — hundreds of you viewed this post for an average time of 3:23 without answering) is that you have a combination of

*OR*s and

*AND*s:

- Scooter could get the first slot AND The Big Man could get the second slot,
- OR Scooter could get the second slot AND The Big Man could get the first slot OR the third slot,
- OR Scooter could get the third slot AND The Big Man could get the second slot OR the fourth slot,
- OR Scooter could get the fourth slot AND The Big Man could get the third slot.

*for this kind of question it is easier and less time consuming to do it the “long” way, unless you are an absolute prodigy at probabilities. When I take the test, I who PWNs the SAT and gets 2400s, I list all the possibilities and count. But hey man, it’s your life.*

## Comments (6)

This is how I did it, but I had to write out the matrix to be sure I was correct. Now that I’ve though about it I’m pretty sure this would work generically.

There are 4! combinations (24 possible layouts)

There are three cases where the two offices can be next to each other (on the left end, in the middle, or on the right end). [This will always work out to number of employees minus 1]

In each of these three case there are two configurations ( Scooter to the left or the right of The Big Man) that will work, so six configurations in total.

In each of these six cases there are two configurations for the remaining two employees (employee #3 to the left or right of employee #4). So 12 configuration in total or 50%. [This will always work out to the factorial of the remaining employees]

Generic formula for two employees being next to each other regardless of how many total employees there are:

1 – [(# of employees)! – 2*(# of employees – 1) * (# of employees-2)!] / (#of employees)!

1 – [4! – 2*(4-1) *2!] / 4! …

1 – [24 – 12] / 24 …

0.5

So if there were 5 employees instead of 4 you would have…

1 – [5! – 2*(5-1) * 3!] / 5! …

1 – [120 – 48] / 120 …

0.4

If there we 10 employees…

1 – [10! – 2*(10-1) * 8!] / 10! …

1 – [3,628,800 – 725,760] / 3,628,800 …

0.2

Wow, what a completely useless algorithm. Good brain exercise though.

Much easier to do the math your way, which simplifies for just about any number of employees. For 10 employees it is just…

2*(1/10 * 1/9) + 8*(1/10 * 2/9) = 0.2

Thanks

Wow JD, that’s great! My solution completely ignores the configurations of the two employees we don’t care about, which is why it ends up producing a fraction with a denominator of 12 instead of 24. As you pointed out, it’s still scalable, but your more rigorous solution might be a better way to go in a general sense.

Still, the message I’m hoping kids will get from this is that while it can be a fun brain exercise to calculate the probability, it’s prudent for SAT purposes just to practice creating the possibility matrix and counting the favorable outcomes. In most cases, it takes less time AND results in greater certainty. That’s a win-win, boys and girls.

Wow JD, that’s great! My solution completely ignores the configurations of the two employees we don’t care about, which is why it ends up producing a fraction with a denominator of 12 instead of 24. As you pointed out, it’s still scalable, but your more rigorous solution might be a better way to go in a general sense.

Still, the message I’m hoping kids will get from this is that while it can be a fun brain exercise to calculate the probability, it’s prudent for SAT purposes just to practice creating the possibility matrix and counting the favorable outcomes. In most cases, it takes less time AND results in greater certainty. That’s a win-win, boys and girls.

There is a much more elegant and teachable trick that can be used to answer almost any kind of adjacency question. If two specific objects out of N have to be adjacent, you can just imagine them glued together, giving (N-1)! permutations. But we have to double that because once we’ve arranged them while glued together, we could let them switch sides once we’ve fixed their glued position. So the number of favorable outcomes, where the two objects are adjacent, is always 2*(N-1)!. The number of total outcomes is just the total number of permutations, N!. So the answer is always 2(N-1)!/N! = 2/N. Here, N=4 so you get 1/2 as you did. JD’s approach is the right one but he failed to see the simplicity of the general answer, which is just 2/N.

As a follow-up exercise, imagine the problem for 3 objects that have to be adjacent out of N. Glue them together and you get (N-2)! permutations. (Gluing three things together forces the number of objects to decrease by 2 this time.) But you can permute the three internally once you’ve fixed their adjacency, so there are 3!(N-2)! favorable outcomes out of N! total outcomes and thus the probability is 6/[(N-1)*N]. Even more generally, for k adjacent objects out of N, the probability will be k!*(N-(k-1))!/N!

A good point, nicely explained.