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I wanna to calculate a probabilities of buying a box of lucky cards. However I am messed up with the equations.

The question: Let there is a box with 30 cards. and 6 of them are lucky cards.

Calculate the probabilities of buying cards one by one until 6 cards are all brought.

At which number would the probability that the chance is higher than 80%?

2 Answers 2

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Presumably each purchase is independent and can buy each card with equal probability $\frac{1}{30}$.

Suppose $P(n,k)$ is the probability that after $n$ purchases you are missing $k$ lucky cards. Then $P(n+1,k)=\frac{30-k}{30}P(n,k)+\frac{k+1}{30}P(n,k+1)$ starting at $P(0,6)=1$ and $P(0,k)=0$ for $k \not = 6$. This can easily be calculated, for example using a spreadsheet.

You are interested in the smallest $n$ such that $P(n,0) \gt 0.8$. It turns out that $P(97,0) \approx 0.79402$ while $P(98,0) \approx 0.80031$.

As a check on the calculations, the expected number of cards needed to be bought to get all the lucky cards is both $\displaystyle\sum_{n=0}^\infty (1-P(n,0)) = 73.5$ and $30\left(\frac16 + \frac15 + \frac14 + \frac13 + \frac12 + \frac11\right)=73.5$.

Added

If purchases were without replacement (meaning that at $30$ purchases you have all the cards, lucky and unlucky), the recurrence becomes

$P(n+1,k)=\frac{30-n-k}{30-n}P(n,k)+\frac{k+1}{30-n}P(n,k+1).$

You now find $P(29,0) = 0.8$. This is not a surprise as the probability the final card purchased is lucky is $\frac{6}{30}=0.2$.

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    Many thanks for solving my mess :)2012-12-18
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If you buy cards one by one, the box of 30 doesn't matter. How many total cards are there? You have the Coupon collector's problem but need to specify the conditions better before there can be a specific solution.

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    @Kitw: In the classic Coupon collector's problem, each coupon is randomly selected from the universe, so yes, it replaces every time. You can use the classic coupon collector's problem for 6 cards and multiply by 5, because $\frac 45$ of the time you don't get a lucky card and can ignore it. You want the number of draws to complete your set of 6.2012-12-18