Infinitely Long Multiple Choice Tests and Your Old Pal e!

Let’s make up a very ridiculous multiple choice exam and try to figure out how likely it is that you’ll get every question wrong on it. Instead of just thinking about one exam, we’re going to think about a sequence of exams (already sounds like a nightmare, but let’s keep going!), where the n^{th} exam has n questions with n answer choices. So the tests get progressively longer, and each question has more answer choices each time.

If every answer is choice is equally likely and there are n answer choices, then the probability of getting one question right is \frac{1}{n}, and the probability of getting that question wrong is \frac{n-1}{n}. We can also write that as 1-\frac{1}{n}. So, the chances of getting every question wrong is (1-\frac{1}{n})^n. Since we care about what happens as the test gets longer and the number of answer choices increases, we want to evaluate \lim_{n\to \infty}(1-\frac{1}{n})^n.

Instead of evaluating this limit, we’re going to evaluate the \lim_{n\to\infty}(1-\frac{x}{n})^n and then set x equal to 1.

We want to evaluate \lim_{n\to\infty}(1-\frac{x}{n})^n, and the first step we’re going to is take the natural log of that expression and exponentiate with e. We can do this since those operations undo each other. So, we want to evaluate e^{ln(\lim_{n\to\infty}(1-\frac{x}{n})^n)}. Let’s bring the n down from the exponent out in front of the logarithm and bring the limit inside of the logarithm to get e^{n\cdot ln \lim_{n\to\infty}(1-\frac{x}{n})}.

Now we’ll just start writing the exponent because that’s the part we’re going to manipulate. If we try to evaluate ln \lim_{n\to\infty}n\cdot(1-\frac{x}{n}) directly it won’t really work. So we’re going to rewrite this as \lim_{n\to\infty}\frac{ln(1-\frac{x}{n})}{1/n}. If we try to let n approach infinity, we’ll get an expression with indeterminate form, so we’re going to use L’Hôpital’s rule. That means we’ll take the derivative of the top and the bottom of the fraction separately, and THEN see if we can evaluate the limit. Since we’ll eventually be setting x equal to 1, we’ll treat the n in this expression as the variable and the $x$ as the constant.

Let’s do it. For the numerator, \frac{d}{dn}\left(ln(1-\frac{x}{n})\right)=\frac{1}{1-\frac{x}{n}}\cdot\frac{x}{n^2}. And for the denominator, \frac{d}{dn}\left(\frac{1}{n}\right)=-\frac{1}{n^2}. Now we can get the limit of the original expression by taking \frac{\text{numerator}}{\text{denominator}}.

When we do that, and cross out terms that cancel in the numerator and denominator, we get -\frac{x}{1-\frac{x}{n}}.

So, if we evaluate the original limit we were trying to evaluate, we get \lim_{n\to\infty}e^{-\frac{x}{1-\frac{x}{n}}}. As n approaches infinity, the second term in the denominator goes to zero. so that whole fraction approaches -x.

That means that the limit of the original expression is e^{-x}. When we set x equal to 1, this expression evaluates to e^{-1}, or \frac{1}{e}\approx 0.37. That’s the chance of getting every question wrong on a tests of increasing length and number of answer choices.

That means there’s roughly a 63\% chance that you’ll get at least one question right! That’s pretty high, considering these this would be for tests with a lot of answer choices.

One way to understand why this is so high is because even though there are more answer choices as you go along in the sequence of tests, there are also more questions, which increases your chance of getting one of them right!

I just think it’s cool that the number e pops up in a probability question like this! I hope you do too!


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