# Odds of 2 people answering exactly the same on a 1000 question multiple choice test

What are the odds of 2 people answering the same on a 1,000 question multiple choice test with approximately 50,000 people taking the test ?

The test was about random things. Life, the world, likes, dislikes, etc. They had four options to choose from

I have a friend that claims that her and someone else took a 1,000 question test and her and another person got a perfect match. They became friends and he verifies the story also. The next closest was a 85% match, with most people scoring having a match around the 70-75% range.

So, we all want to know what are the odds of that happening.

Thank you.

• How many choices per question? – Pedro Tamaroff Jun 2 '13 at 21:54
• 4 choices per question. – John Jun 2 '13 at 21:54
• Yes, so 25% chance on each question. – John Jun 2 '13 at 21:56
• Well, one has a much better than $\frsc{1}{4}$ probability on a particular question if one knows the answer. – André Nicolas Jun 2 '13 at 21:58
• The people taking the test were trying to guess correctly. This isn't the same as $50,000$ guessing randomly. – John Douma Jun 2 '13 at 21:58

Since each question has four options and there are $1,000$ questions, there are a total of $4^{1000}$ different ways to complete the test. We can select $50,000$ distinct submissions from this collection in $(4^{1000})_{50,000}$ ways (the falling factorial). The total number of possible submissions is $(4^{1000})^{50,000}$. Finally, the probability that no two tests had exactly the same answers is $$\frac{(4^{1000})_{50,000}}{(4^{1000})^{50,000}},$$ which is quite close to $1$.
Returning to the original question, the probability that at least two answered the same is $$1 - \frac{(4^{1000})_{50,000}}{(4^{1000})^{50,000}},$$ which is quite close to $0$.
• You could maybe get some crude estimate as follows: Set $n= 4^{1000}, \, m = 50'000$. Then $$\frac{\prod_{k=0}^{m-1} (n-k)}{n^m} = \prod_{k=0}^{m-1} \left(1-\frac kn\right) = \exp\left(\sum_{k=0}^{m-1} \log \left(1-\frac kn\right)\right) \overset{ m\ll n}\approx \exp\left(-\sum_{k=0}^{m-1} \frac{k}{n}\right)$$ Now $\sum_{k=0}^{m-1} k = \frac{m(m-1)}{2}$ yields that the probability of no two identical results is about $\exp\left(-\frac{m^2}{n}\right)$. So the probability of at least two identical tests is $$1- \exp\left(-\frac{m^2}{n}\right) \approx \frac{m^2}{n}$$ – Sam Jul 13 '13 at 20:58
• To get the order of magnitude of this we take logs, which comes out at about $\log_{10}(\mathrm{probability}) \approx -592$, i.e. $\mathrm{probability} \approx 10^{-592}$. Or in other words, the probability of this happening by pure chance is ridiculously small. But of course 1000 questions about personal preferences most definitely will be heavily correlated, so who knows... Also 1000 questions seems like a lot of questions. At 5 seconds per question, filling in the questionnaire would take $5000\, \mathrm{sec} \approx 1\, \mathrm{ hr} \, 23\, \mathrm{min}$. – Sam Jul 13 '13 at 21:04