Calculating probabilities for drawing specific cards from multiple decks?

I've been trying to get an achievement in a game called 'Sentinels of the Multiverse' (for context, its a superhero-themed card game). The achievement requires you to get insanely lucky with your draws (the achievement has a luck theme). I want to know the exact probabilities of you winning, but I don't know how to calculate it.

Here's how the situation works. There are four decks. Three of them belong to players (they are the heores). They start the game with 4 cards in hand. Each deck has 40 cards in it. The fourth deck is called the villain deck. It starts with 30 cards in it. The top card of the deck is played every turn (no one actually 'plays' the villain deck).

To win the game, very specific things need to happen.

Deck 1 has to have one of 4 specific cards in his beginning hand. Deck 2 and 3 need to have one of 2 specific cards in their beginning hands.

What are the probabilities of this happening?

And further more, the villain deck has 2 cards that pretty much spell doom if he doesn't draw them early. Essentially, the deck has some really powerful mobs in it and this card puts them into play from the discard pile. There's a total of 5 of these 'underboss' cards. Essentially, the only way to survive this card is if the villain deck draws them BEFORE you're forced to put underbosses in the discard pile. Either he draws them early on, or you have no hope.

First, what is the probability of the players having the cards they need on any specific turn? Note that their decks have little card draw, so they will 99.9% of the time be drawing one card a turn. The villain deck can draw more than one card a turn, also it fishes out an underboss from the deck each turn. That doesn't matter. I want to know the probability of it drawing those two specific cards early on. Let's say what is the probability of all this happening for the first 5 to 10 card draws.

And yes, the requirements for this achievement are ridiculous.

• Player 1 has at least one of his correct cards in his opening hand with probability $1-\frac{\binom{26}{4}}{\binom{30}{4}}$. Player 2 has at least one of his correct cards in his opening hand with probability $1-\frac{\binom{28}{4}}{\binom{30}{4}}$. Similarly for player 3. Multiplying these probabilities gives the probability of them occurring simultaneously. You may further modify these with numbers other than $4$ if you are interested in finding these cards within a certain number of turns, effectively by "giving them a bigger hand." – JMoravitz Sep 14 '18 at 0:10