- Devin Rossiter

# Make Richer Questions by Broadening Those Multiple Choices

The notification "ding" of my work e-mail meant it was time to check in with ** Brilliant**, the web-based math task curation service that allows rich problems to be discussed, revisited, and analyzed. The problems shared by the community at Brilliant are rarely immediately apparent, intellectually appealing, and entertaining in their complexity. The forum discussions regarding solutions provide a porthole view into multiple ways of thinking that challenge the user to consider alternate approaches in a "peek-behind-the-curtain" sense.

And, as a means of entry and accessibility, all problems shared are presented in multiple-choice format.

Some are traditional in their choices: 4 or 5 options as possible answers, some designed off of common misconceptions. These are the kind of problems all too familiar, and fairly recently discouraged as not truly reflective of student learning. The shift in state- and national-level assessments consisting ENTIRELY of this question types is encouraging, and though they still have a place, they do not dominate instruction and assessment as they have in recent generations.

Alternately, many of the problems posed by Brilliant choose an approach to multiple-choice that invites and engages richer thought and processing: open-ended choices.

Consider this problem from Brilliant that appeared in my inbox today:

If it seems "easy", it's because there are only three choices. Rationally, this means that a simple guess gives you a 1/3 chance of being correct.

Look at the solutions, and you may notice something that invites some consideration.

**What can we make of the fact that only one of the choices calls out an exact value as a solution?**

When I see a problem like this, I am invited into the problem more easily since the only thing required to be successful is a defensible reason why the chance of the top and bottom cards being the same color is even, less than even, or greater than even.

Let's analyze the skills and knowledge needed to be successful with this problem:

Playing card familiarity - helpful, but the key information is provided: a complete deck contains 52 cards, 26 of the cards are red (hearths and diamonds), and the rest black (spades and clubs).

Percentage equivalence - a handy tool for connecting values into a standard measure. Familiarity with converting probabilities in any format (decimal, fraction, percent) allows for flexibility with different kinds of values.

Compound probability - an essential here, but one that can be built through reasoning and discourse. And that's the key here: we are not asked to identify the precise probability, but the *range* where the probability would fall on a likelihood scale.

There is the probability model that can be built, where the bottom card has a 25/51 chance (since the top car cannot also be on the bottom). But in this problem, that depth of mathematical reasoning is not absolutely necessary to be successful.

One could reason that since there are as many red as black, once I commit a red to put on the top of the deck there are instantly more black cards than red remaining, which gives us more of a chance that both cards will NOT share the same color.

Another approach may involve modeling involving a tree, visualizing and testing all possible outcomes based on a selected top card. Broken down by suit, one might connect that one color will have one fewer option than the other.

The argument here favoring multiple choice problems with non-precise options is that it encourages not only multiple approaches and strategies, but facilitates discourse by depending less on calculation and more on problem-solving and reasoning as tools for success. The potential for anxiety is lowered due to the intentionally general nature of the options and not on a specific value as an end-goal.

Can we apply this approach to traditional multiple choice problems? Let's consider one where calculation is generally required, remaining on the topic of percentages:

**24,998 ÷ 500**

**A) Less than 500**

**B) 500**

**C) Greater than 500**

The focus of this calculation is no longer centered on a value that may or may not have any meaning to students who are disinterested with number sense; by asking students to approach near a centered range, they can work around simpler values and logic. For example, if 500 x 500 is equal to 25,000, then a lower value divided by 500 must itself be lower than 500 (whatever it ends up being is irrelevant in this case).

Another example might include ranges:

**How would you describe the sum of -15 and +10?**

**A) Less than -15**

**B) Between -15 and -5**

**C) Between -5 and +5**

**D) Between +5 and +15**

**E) Greater than +15**

Going with ranges allows for a greater number of options, as well as encourage elimination as a strategy for reasoning (frequently and pejoratively regarded as a "test-taking strategy" instead of a valid tool for problem-solving), while still offering the potential for multiple approaches.

Where the structure of multiple-choice problems have generally caused a great deal of math trauma over the years, the problem type does have leverage as a tool for deeper conversations when the options are of a less specific variety. Though we understand the need for precision in math, it is often at the expense of general reasoning opportunities. By making your options broader, students are invited back into the conversation by simply thinking enough to get to the ballpark.

**-DR**