In a trading card game I am playing, the following rules apply:
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The deck consists of N cards total.
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Among these N cards are X identical copies of a particular card I wish to calculate the probability of drawing.
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I initially draw 7 cards. If I wish, I may shuffle these 7 cards back into the deck and draw 6 new cards. I can repeat this any number of times, each time drawing 1 card less.
Given N and X, how can I calculate the probability of drawing this particular card as a function of the number of times I redraw?
To be clear, I wish to know the probability that one of the hands I drew would have contained this card, as I continue to redraw. So the probability should go up with each successive draw. I do not mean the probability that the hand I decide to keep has it, in which case each successive smaller hand would have a lower probability of containing the card.
Best Answer
As Doug tried to begin., we may assume that every card in the deck is labeled.
There are $\binom{n}{7}$ ways to pick a hand of seven cards. There are $\binom{n-x}{7}$ ways to pick seven cards where none of which are the desired card-type.
Since every hand is equally likely, we apply the definition of probability in an equiprobable sample space, $Pr(E)=\frac{|E|}{|S|}$ to find the probability of not drawing a desired card in the first hand as $\dfrac{\binom{n-x}{7}}{\binom{n}{7}}$
Applying the multiplication principle of probability, if we were to continue to not draw our desired card, we would have a probability of:
$\dfrac{\binom{n-x}{7}}{\binom{n}{7}}\dfrac{\binom{n-x}{6}}{\binom{n}{6}}\dfrac{\binom{n-x}{5}}{\binom{n}{5}}\dfrac{\binom{n-x}{4}}{\binom{n}{4}}\dfrac{\binom{n-x}{3}}{\binom{n}{3}}\dfrac{\binom{n-x}{2}}{\binom{n}{2}}\dfrac{\binom{n-x}{1}}{\binom{n}{1}}$
The probability of having drawn at least one of the desired card types is then the opposite of the probability we found and will be one minus the probability found above.
as an aside, the notation $\binom{n}{r}$ is the binomial coefficient as mentioned previously in the comments