Calculating probabilities in a children's card game

Question

So I want to know how to calculate the probabilities of opening a specfic hand in a 44 card deck in which you draw 5 cards in you 1st turn. Lets say I have 3 copies of Stratos, 3 copies of Faris, 3 copies of A Hero Lives, 2 copies of Mali and 33 blank cards, I want to know the chances of opening Stratos+Faris, Stratos+A Hero Lives, Stratos+Mali, Faris+A Hero Lives, Faris+Mali OR Mali+A Hero Lives

Answer 1

$\frac{245520\text{ desired combinations}}{1086008\text{ total combinations}} \approx 22.6\text{%}$

Why?

1. There are $44 \times 43 \times 42 \times 41 \times 40$ ways to deal 5 cards from a set of 44, when order matters (permutations).
2. There are $5 \times 4 \times 3 \times 2 \times 1$ ways to order 5 cards.
3. $1,2 \implies$ There are $\frac{44 \times 43 \times 42 \times 41 \times 40}{5 \times 4 \times 3 \times 2 \times 1} = 1086008$ ways to deal 5 cards from a set of 44, when order does not matter (combinations).
4. There are $3$ ways to deal 1 card from a set of 3.
5. $4 \implies$ There are $3 \times 3$ ways to deal 1 card from 3 sets of 3.
6. $5 \implies$ There are $(3 \times 3) \times (2 \times 3)$ ways to deal 1 card from 3 sets of 3 and 1 card from 2 sets of 3, when order matters (permutations).
7. There are $2 \times 1$ ways to order 2 cards.
8. $6,7 \implies$ There are $\frac{(3 \times 3) \times (2 \times 3)}{2 \times 1}$ ways to deal 1 card from 3 sets of 3 and 1 card from 2 sets of 3, when order does not matter (combinations).
9. There are $2$ ways to deal 1 card from a set of 2.
10. $4,9 \implies$ There are $(3 \times 3) \times (1 \times 2)$ ways to deal 1 card from 3 sets of 3 and 1 card from 1 set of 2.
11. $6,10 \implies$ There are $\frac{(3 \times 3) \times (2 \times 3)}{2 \times 1} + (3 \times 3) \times (1 \times 2) = (3 \times 3) \times (3 + 2)$ ways to deal 1 card from 3 sets of 3 and 1 card from either a set of 3 or a set of 2.
12. There are $33 \times 32 \times 31$ ways to deal 3 cards from a set of 33, when order matters (permutations).
13. There are $3 \times 2 \times 1$ ways to order 3 cards.
14. $12,13 \implies$ There are $\frac{33 \times 32 \times 31}{3 \times 2 \times 1}$ ways to deal 3 cards from a set of 33, when order does not matter (combinations).
15. $11,14 \implies$ There are $\frac{33 \times 32\times 31 \times (3 \times 3) \times (3 + 2)}{3 \times 2 \times 1} = 245520$ ways to deal 3 cards from a set of 33, 1 card from a set of 3, and 1 card from either a set of 3 or a set of 2, when order does not matter (combinations).
16. $3,15 \implies$ There is a $\frac{245520}{1086008} \approx 22.6\text{%}$ chance that the desired combinations will be drawn.

Answer 2

The existing answer by Jeffrey Kumley assumes that the hand should include only those two special cards, and the remaining cards must be blank. Another possible interpretation of the question is (as discussed in the comments) that the hand should include at least those two special cards, and the remaining cards can be any cards. Here’s a solution for this latter interpretation.

So we want to calculate the probability to draw at least two types out of the four types Stratos, Faris, A Hero Lives and Mali. Often it’s easier to calculate the probability of the complementary event, in this case, drawing either none or only one of those types.

The number of hands that include at most a particular one of those four types that has $k$ copies is $\binom{44-11+k}5$. We need to add this for the four types, but then we’ve counted the hands with none of the four types four times, so we need to subtract them $3$ times. There are $\binom{44-11}5$ such hands. Thus there are

$$ \binom{44-11+2}5+3\binom{44-11+3}5-3\binom{44-11}5=648024 $$

hands that contain at most one of those four types, so the probability to draw at least two of the types is

$$ 1-\frac{648024}{\binom{44}5}=\frac{3891}{12341}\approx32\%\;. $$