I've been playing a ton of Marvel Snap lately. It's a card game with simple rules. You have 12 cards in your deck. You start with 3 cards and draw 1 card every turn for the 6 turns you play. This means that you draw 9 cards by the end of the game.
What I'm having trouble with is the chance of drawing combinations of cards by different turns. For example, let's say I want to draw Spider-Man by turn 4 and Miles Morales by turn 5. Say there's a combo play that requires me to play Spider-Man by turn 4 and Miles Morales by turn 5, so I have to draw them by those turns. I know that I have a 58.33% chance (7/12) of drawing Spider-Man by turn 4, and a 66.67% (8/12) chance of drawing Miles Morales by turn 5.
How do I find the chance of both events happening? I feel like simply multiplying the two probabilities together ignores edge cases, but is that how I solve this?
For ease of syntax, label the cards A,B,C,...,J,K,L.
You want the probability of drawing A by turn 4, and (simultaneously) B by turn 5.
The probability is $~\dfrac{7}{12}~$ that A is one of the first $~7~$ cards.
There are a variety of ways of formally completing the problem. I will use two methods.
You are interested in the probability of also drawing a B by turn 5, given that you drew an A by turn 4.
$\underline{\text{Method 1}}$
Warning
This method, while valid, does not generalize well.
The only way to fail in drawing a B by turn 5, is if the B is one of the last $~4~$ cards, out of the $~11~$ cards remaining, given that the A was one of the first $~7~$ cards.
So, the probability of drawing the A by turn 4 and the B by turn 5 is
$$\frac{7}{12} - \left[ ~\frac{7}{12} \times \frac{4}{11} ~\right] $$
$$= \frac{7}{12} \times \left[ ~1 - \frac{4}{11} ~\right] $$
$$= \frac{7}{12} \times \frac{7}{11} = \frac{49}{132}.$$
$\underline{\text{Method 2}}$
Let $~E_1~$ denote the event that you draw the A on or before the 1st 4 turns.
Let $~E_2~$ denote the event that you draw the B on or before the 1st 5 turns.
You want to compute
$$p(E_1) \times p(E_2 ~| ~E_1). \tag1 $$
In (1) above, the second factor represents the probability of event $E_2$ occurring given that event $E_1$ has occurred.
As discussed, $~p(E_1) = \dfrac{7}{12}.~$
To compute $~p(E_2 ~| ~E_1), ~$ assume that event $~E_1~$ has occurred and then ask yourself how that assumption has affected whether event $~E_2~$ will also occur.
In consideration of event $~E_2~$ by itself, without any regard for whether event $~E_1~$ has occurred, you would be presuming that you have a $~12~$ card deck, and you want to know the probability that the B is one of the first $~8~$ cards.
Under the assumption that event $~E_1~$ is one of the first $~7~$ cards, the circumstances in the previous paragraph have changed in two ways:
With the A occurring somewhere in the first $~7~$ card positions, there are only $~(8-1) = 7~$ remaining positions that the B can satisfactorily occur.
With the A occurring somewhere in the first $~7~$ card positions, in your attempt to receive the card B soon enough, you are (in effect) battling an $~11~$ card deck, since the A has been eliminated from consideration.
Therefore,
$$p(E_2 ~| ~E_1) = \frac{7}{11} \implies p(E_1) \times p(E_2 ~| ~E_1) = \frac{7}{12} \times \frac{7}{11}.$$
Addendum
This section is added purely as a sanity check. I strongly suspect that the problem composer intended that Method 2 above be used.
There are two ways of succeeding in getting the A and B in a timely manner.
You get the A in the first 7 cards, and then get the B specifically on card 8.
The probability of this happening is
$\displaystyle \frac{7}{12} \times \frac{1}{11}.$
You get both the A and the B in the first $~7~$ cards.
The probability of this happening is
$\displaystyle \frac{\binom{7}{2}}{\binom{12}{2}} = \frac{7 \times 6}{12 \times 11}.$
Summing the two mutually exclusive cases give you
$$\frac{[~7 \times 1 ~] + [~7 \times 6 ~]}{12 \times 11}.$$