Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm creating a game and can't seem to quite figure this out - driving me crazy.

  • There are 8 questions in my game
  • You can play the game an unlimited amount of times
  • the test bank doesn't change. so when a new game starts it draws from the same test bank (the questions are not replaced by new ones). within a game, however, there are 8 unique questions.

how many total questions do I need in the test bank, where if i play the game twice, the probability of getting 4 questions repeated is less than 5%?

share|cite|improve this question
clarify something. How many questions per game? – Sabyasachi Mar 12 '14 at 17:58
"getting 4 questions repeated" exactly 4 or at least 4? – Hoda Mar 12 '14 at 18:09
wow very sorry. i meant at least 4 questions. Thanks a lot guys. – user2989523 Mar 12 '14 at 18:12

4 Answers 4

Let $n$ be the number of questions. We will calculate the chance that exactly four questions are repeated out of eight. There are ${n-8 \choose 4}{8 \choose 4}$ ways to choose four matching and four non-matching, and $n \choose 8$ ways to choose the questions overall. So we want $$\frac {{n-8 \choose 4}{8 \choose 4}}{n \choose 8}=0.05$$ I find $36$ gives a probability of $4.7\%$

share|cite|improve this answer
OP meant at least 4. I think replacing $\binom{n-8}{4}$ with $\binom{n-4}{4}$ would answer the new question. – Hoda Mar 12 '14 at 18:13
Yes, OP wanted at least 4. But if the chance of 4 is small, the chance of 5 is smaller yet, so it will not be much of a change. It should be clear how to modify this to add up 5,6,7,8. – Ross Millikan Mar 12 '14 at 18:16

In a game context, we should probably interpret the question as asking that the probability of $4$ or more duplicates is less than $5\%$. And in a game context, for successive games, we would presumably not choose questions at random.

But the problem seems to ask us to assume that for each new game, the $8$ questions are chosen at random from the question bank, with all choices equally likely.

Suppose the test bank has $n$ questions, and that the first game has been played. We find the probability that the next selection of $8$ overlaps by $4$ or more with questions from the first game.

There are $\binom{n}{8}$ equally likely ways to choose the second set of $8$ questions. We count the number of choices with overlap $4$.

Which $4$? These can be chosen in $\binom{8}{4}$ ways, and for each way the $4$ non-overlapping questions can be chosen in $\binom{n-8}{4}$ ways.

If (as we should) we want to count also overlaps $5$, $6$, $7$, $8$, the number of ways is $$\binom{8}{4}\binom{n-8}{4}+\binom{8}{5}\binom{n-8}{3}+\binom{8}{6}\binom{n-8}{2}+\binom{8}{7}\binom{n-8}{1}+\binom{8}{8}\binom{n-8}{0}.$$ Divide by $\binom{n}{8}$ for the probability.

We want to make this probability $\lt 0.05$. The easiest way to get the appropriate $n$ is to use software. One can for example ask Wolfram Alpha to compute the probability for various values of $n$. Quite soon we can locate the appropriate $n$. One can also do it "by hand" by making appropriate estimates.

share|cite|improve this answer

If there are $n$ question is the bank and we have already picked our 8 for the first time, then the $n$ questions are split into two categories: the 8 chosen the first time and the remaining $n-8$ not chosen the first time. Thus the question can be restated as: what is the smallest $n$ for which $$ \frac{\binom{8}{4}\binom{n-8}{4}}{\binom{n}{8}}+\frac{\binom{8}{5}\binom{n-8}{3}}{\binom{n}{8}}+\frac{\binom{8}{6}\binom{n-8}{2}}{\binom{n}{8}}+\frac{\binom{8}{7}\binom{n-8}{1}}{\binom{n}{8}}+\frac{\binom{8}{8}\binom{n-8}{0}}{\binom{n}{8}}<0.05 $$

Clearly we have $n\ge 12$. Time to bust out your favorite number crunching software from here!

share|cite|improve this answer


In the second draw, the four repeated questions should be drawn from the first 8. This you can draw it in ${8\choose4}$. Assume there are n questions in the test bank. The other 4 questions in the second draw should come from n-8. The number of ways that could happen is${{n-8}\choose4}$. Extending the same to i = 5,6,7,8 and the total number of ways you can pick 8 answers from the test bank = ${n\choose8}$

Thus the probability that you will have atleast 4 repeated questions = $\sum_{i=4}^8[\dfrac{{8\choose i}{{n-8}\choose{8-i}}}{{n\choose8}}] \approx 0.05$

Find n from that. n seems to be equal to 37

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.