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I was wondering is anyone could help me check if my math is correct here. I'm trying to calculate the probability of my drawing a certain combination of cards from a deck. Question 1: I have a deck of 60 cards, within that deck are 4 copies of card "x". I'd like to figure out the probability of drawing 2 of card "x", being able to change the total number of cards drawn until I find a where the probability becomes greater than 50%. I think my equation is suppose to be (4C2)(56C"y")/(60C"z") where y represent the total number of cards drawn minus the 2 I want, and z represents the total number of cards drawn. Is this correct? Every time I calculate the answer is muh lower than I expect so I'm not sure if I'm fusing something up.

Question 2. I'd like to be able to do the same thing as above but change the need to draw multiple instances of multiple cards. Is there a way to do this as well?

Thanks

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  • $\begingroup$ Do you want exactly two or at least two? The probability of drawing exactly two will peak below 50% and eventually start to drop as you draw more cards. $\endgroup$ – amd Mar 30 '17 at 19:35
  • $\begingroup$ I was looking for at least 2, which may be why my calculations get wonkey as I increase the number of cards drawn. $\endgroup$ – Carlos Luna Mar 30 '17 at 21:50
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If you’re looking at the probability of drawing exactly two of these cards, that will reach some peak and then start dropping, reaching zero on the 59th draw. This makes intuitive sense: the more cards you draw, the more likely it is that you’ll get more than two of these cards. In this particular case, the probability peaks well below 50%.

To instead compute the probability of drawing at least two, you can use the same multinomial distribution. It’s easiest to compute the probability of drawing less than two of the special card and subtracting that from $1$. If we let $X_n$ be the number of special cards drawn after drawing $n$ total, then $$1-\Pr(X_n\lt2)=1-\Pr(X_n=0)-\Pr(X_n=1)=1-{\binom40\binom{56}{n-0}\over\binom{60}n}-{{\binom41\binom{56}{n-1}\over\binom{60}n}}$$ which can be simplified somewhat. It looks like this first passes 50% at $n=24$. This feels like it’s in the right ballpark. Since you’re trying to draw at least half of the special cards, it should take drawing, on average, less than but near half of the deck to do so.

I’m not entirely sure what it is you’re asking in your second question, but this analysis can certainly be expanded to include multiple special card types. The resulting expressions for the probabilities will become much more complex rather quickly.

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