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:

1234   2134   3124   4123
1243   2143   3142   4132
1324   2314   3214   4213
1342   2341   3241   4231
1423   2413   3412   4312
1432   2431   3421   4321
Take a minute to understand the order in which I listed those, because that’s important. I listed them from least to greatest, taking care to list all the ones that started with “12,” then all the ones that started with “13,” etc. I broke them into columns every time I had a new digit in the first position. I can check my work to make sure I’ve listed them all by making sure all the columns are the same length.
Now that I’ve listed every single possibility, I only need to count the ones that fit the question. How many have the same two people adjacent to each other? Let’s say Scooter is 1 and The Big Man is 2:
1234 2134 3124 4123
1243 2143 3142   4132
1324   2314 3214 4213
1342   2341   3241   4231
1423   2413 3412 4312
1432   2431 3421 4321
Count ’em up! Scooter and The Big Man are next to each other 12 of the 24 times! That’s a probability of 12/24, which simplifies to 1/2, or 0.5.
Remember that the numbers will stay small on the SAT. If you get fast at listing possibilities, this is probably the biggest matrix of this kind you’ll ever have to make. If you have to make one with 5 possibilities (which I’ve seen 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.
Ok, but what about the math? What makes this question so difficult to solve from a math perspective (and it is difficult — hundreds of you viewed this post for an average time of 3:23 without answering) is that you have a combination of ORs and ANDs:
• 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.
Whoa. Break that down into a big ol’ probability expression (more on how to do this in my massive probability post):
So there you go. Of course, the math works out nicely as well, once you know how to set it up. I stress again: 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.