Calculating Significantly Above Chance for 3-AFC Test I am trying to figure out which participants who took a 18-item multiple choice test scored significantly above chance. Each item has 3 choices, so the probability of getting each question correct is 0.33. To know how many items correct a person would have to get in order to be significantly above chance, does it make sense to use a binomial probability calculator like this: https://stattrek.com/online-calculator/binomial.aspx and try different numbers of successes until the probability is &lt; p=0.05 (assuming this is my significance criteria)?

For 18 items where the probability correct is 0.33, the probability of getting 9 correct is 0.061 and the probability of getting 10 correct is p=0.027, so does it make sense to use "10 items correct" as my criteria for "performed significantly above chance?"
The hypothesis would normally be expressed as a one-sided test as p(X≥10). The online calculator you have selected is fine and the significant result could be expressed in an article as follows: (p=0.33, q=0.66, K=10, n=18, p-value = 0.043).  

Another way to think of this is that 9 or fewer correct answers is more likely to be random chance.

At the threshold of 10, it is unlikely that the result could be explained by random chance.