Probability Question for a Game Design: There are 22 cards; 4 are identical...

Hi everyone, this is my first time posting, so I apologize if I do anything wrong. I've looked through the forums but couldn't find a similar question.

In the game, there are 22 cards.
4 of them are identical and create a special effect, and the other 18 are blanks (for the purposes of this question).
Cards are drawn one at a time and not replaced.

How do I calculate the probability that 3 of the 4 special cards will be drawn within the first N draws. (I'd like to know the probability for N being 4, 5, 6, etc.)

I know how to calculate a version of the game where you must draw all 4 of the special cards. But what's throwing me off is only needing to draw 3 of the 4 special cards.

Thank you!

-- Danny

There are 22 card, 4 of them "special". The probability one of the "special" cards is drawn first is 4/22. Given that, there are 21 cards left, 3 of them "special". The probability the second card is "special" is 3/21. Given that two "special" cards are drawn on the first two draws, there are 20 cards left, 2 of them "special". The probability the third card drawn is "special" is 2/20. The probability of drawing three "special" cards in the first three draws is (4/22)(3/21)(2/20)= (1/5)(1/7)(1/11). Given that three "special" card are drawn in the first three draws there are 19 cards left, 1 of them "special". The probability the fourth card drawn is not special is 18/19. The probability the first four cards drawn are "special", "special", "special", "not special" in that order are (4/22)(3/21)(2/20)(18/19). It is not hard to see that the probability that three of four cards drawn are "special" and one not, in any specific order is also (4/22)(3/21)(2/20)(18/19) (in the fractions, the denominators are the same, the numerators are just in a different order). There are \(\displaystyle \begin{pmatrix} 4 \\ 3 \end{pmatrix}= \frac{4!}{3!}= 4\) such orders so the probability exactly 3 of four cards are "special" is 4(4/22)(3/21)(2/20)(18/19).

The same kind of calculation shows that the probability that exactly 3 of n cards drawn are "special", for n greater than 3 is \(\displaystyle \begin{pmatrix}n \\ 3\end{pmatrix}\left(\frac{4}{22}\frac{3}{21}\frac{2}{20}\frac{18}{19}\frac{17}{18}\cdot\cdot\cdot \frac{21- n}{22- n}\right)\).

(You asked for the probability of drawing 3 "special" cards. The probability of drawing "at least 3" is slightly different. You would need to add the probability of drawing all 4 which, for drawing n cards, is \(\displaystyle \begin{pmatrix}n \\ 4\end{pmatrix}\left(\frac{4}{22}\frac{3}{21}\frac{2}{20}\frac{1}{19}\right)\).)

Thank you for your quick reply. There's a lot of stuff in here that's new to me, which is really helpful.