determining the amount of total questions needed in a game given the probabilty 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%?
 A: 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 $4.7\%$
A: 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. 
A: 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!
A: Hint:
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
