Probability to draw a card AND not draw both copies of another card.

Question

I'm working on a project involving the card game Hearthstone. Right now I'm looking at the chance to draw AT LEAST one copy of a certain card while NOT drawing both copies of another card after a certain number of draws.

So, 30 card deck. There are two cards of interest, which I will call S. I want to draw at least one of these in the first 9. This can be written as follows:

P(Draw at least one copy of S) = 1 - P(draw no copies of S).

So this would be 1 - $\binom{28}{9}$/$\binom{30}{9}$ giving a probability of .51724.

For the second event, I want to draw at most 1 copy of the card, call it U. Remember, there are 2 copies of U in the deck So:

P(Draw 0 or 1 U) = 1 - P(draw 2 copies of U) = 1 - $\frac{\binom{2}{2}*\binom{28}{7}}{\binom{30}{9}}$ = .91724.

Now if these were independent events, we could simply multiply the probability of both to find the probability of both occurring. I wrote an R function that calculates the probability mathematically and get 0.4804598 to draw at least 1 copy of S and 0 or 1 copies of U in 9 draws.

Multiplying the probability of each event gives .91724*.51724 = .4744339, which does not equal .4804598. So this tells me that the two events are dependent.

Can someone verify this or offer another way to calculate this by hand, instead of finding the probability of each event then multiplying?

Best

Answer 1

To be clear on what we're talking about, is this the situation?

There are 30 cards in the deck, of which 2 are 'S' and two are 'U'.   Let $S$ be the count for 'S'-cards, and $U$ the count for 'U' cards, within nine draws without replacement.

These will clearly be dependent random variables, as you cannot draw an 'S' and a 'U' in any particular draw.   As can be witnessed by quickly evaluating the conditional probability for drawing no 'S' cards given both 'U' cards were drawn: $$\mathsf P(S=0\mid U=2)=\left.{\tbinom{26}{7}}\middle/{\tbinom{28}{7}}\right.~~\neq~~ \left.{\tbinom{28}9}\middle/{\tbinom{30}9}\right.= \mathsf P(S=0)$$

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Then the probability for drawing at least $1$ 'S' cards is indeed: $$\begin{split}\mathsf P(S\geq 1) ~&=~ ~1-\mathsf P(S=0)\\~&=~1-\dfrac{\binom{2}{0}\binom{28}{9}}{\binom{30}{9}}\end{split}$$

Likewise the the probability for having at most 1 'U' card, is: $$\begin{split}\mathsf P(U\leq 1)&=1-\mathsf P(U=2)\\&=1 - \dfrac{\tbinom{2}2\tbinom{28}7}{\tbinom{30}9}\end{split}$$

Now the probability for the join of these evens, by the Principle of Inclusion and Exclusion: $$\begin{split}\mathsf P(S\geq 1\cap U\leq 1) & = 1-\mathsf P(S=0\cup U=2)\\ & = 1 -\mathsf P(S=0)-\mathsf P(U=2)+\mathsf P(S=0\cap U=2)\\&~~\vdots\end{split}$$