# SAT & ACT Brain Teaser

Let’s face it: preparing for the SAT and ACT is not the most exciting activity for students. Keeping students engaged and motivated is half of the battle for teachers and tutors. In order to keep students interested, I like to incorporate logic puzzles into classes and tutoring sessions. Often times these are of a different ilk and certainly more challenging than a typical SAT or ACT math question, but they value a similar type of thought process.

The makers of the SAT and ACT, as well as college admissions officers, are not so much interested in a student’s rote memorization of rules and formulas as they are in a student’s ability to think critically.

Logic puzzles are a great way for students to engage in this type of thought. They also act as a respite from your run-of-the-mill math questions. I find the best puzzles are simple but elegant: a straightforward solution, but something that requires a series of logical steps. Also, many times they have a very unexpected answer, which is why students will usually find them interesting and are more likely to remember the process necessary to solve them.

While most students do not end up getting the correct solution (I try to always keep them challenging), I am more concerned with how they approach the problem. One thing I always remind them to do is to try to take an intimidating question and alter it slightly or break it apart to make it manageable. A lot of questions are based on patterns. If a puzzle involves large numbers, try replacing them with smaller numbers and see if you can develop that pattern.

With that said, let me present to you one of my favorites. Note that this is taken from a fairly well known puzzle, with the situation altered slightly. See if you can get it without any outside help! It may seem like it could involve high-level mathematics, but only logic and a basic understanding of probability is necessary.

#### Here is the puzzle:

*Fifty students show up to their high school to take the ACT. They all have assigned seats in a room with exactly fifty seats. The students walk in one at a time and take their seats. The first student forgets her assigned seat number and decides to sit in a random seat. Every student thereafter sits in his or her assigned seat if it is unoccupied. Otherwise he or she will sit in a random seat.*

*What is the probability the fiftieth and final student sits in his assigned seat?*

Comment below if you think you have it!

Hint: Try the question for two, three, and four students and see if you notice a pattern.

Answer: ½ or 50%. Here's a video explanation or scroll down for a written explanation.

Solution: At first, people may think there is some sort of complicated probability formula necessary to calculate this. In fact, simple logical reasoning is all that is necessary. There are a few ways you can think about it, but this is one that I find easiest to digest:

Let’s say you are student 50. There are 2 outcomes that matter: either your seat is empty by the time you sit, or someone is sitting there. At some point between students 1 – 49, the outcome will be determined. It will be determined the moment a student either sits in **your** seat OR the first student’s seat. If at any point a student sits in your assigned seat, the probability of you getting your seat would be 0. If a student sits in the first student’s seat at any point (including the 1/50 chance the first student sits there herself), then everyone thereafter, including you, would sit in the correct seat, and the probability would be 1. It should make sense that both of these outcomes are equally likely, which is why the probability is simply ½.

If that explanation does not make sense, think about the case with two students.

There are only two possibilities here:

- The first student can sit in the correct seat, which means you will get the correct seat
- The first student sits in your seat and you end up in the wrong seat.

Because the first student is sitting in a random seat, each of those scenarios has a 50% chance of happening.

Similarly, you can work it out for three students. You will see it is still 50%. The total number of students does not change the probability. You can certainly come up with a more mathematical solution for the case with 50 students, but the above solution is a more intuitive way to think about it.