Odds of drawing exact cards

[Laires]

Hello,

I have been playing a card game called Hearthstone (the only reason I'm mentioning this is - if you know the game and it's draw mechanics, you will better understand what I'm talking about). I found a deck in this game that performs really really well if I manage to draw 2 exact cards early in the game and I would love to calculate the chance of drawing one of those game winning cards in early stage of the game, but I was never really good at math, so after hour I gave up and I leave this to the experts.

It's quite simple to explain at least:
--- deck consists of 30 cards in total, 2 of them are the game winning cards you want in your hand by Turn 4.
--- At first you draw 3 random cards from the deck. If any of them isn't one of those 2 cards, you toss them back in the deck and draw 3 different cards.
--- The game goes on for exactly 4 Turns in which you draw 4 more cards in total, which all can also be the cards you tossed back in the deck.
--- What are the odds of drawing each on Turn 4? What are the odds of drawing both on Turn 4?

It's enough complicated for my little brain to get lost in it. If anyone can calculate this for me and show the calculation step by step that would be great!

 

[Steven G]

Can we see your work so we can help you figure this out?
I am confused about something. You said At first you draw 3 random cards from the deck...draw 3 different cards. That is a total of six cards that you have taken. Then you go onto say The game goes on for exactly 4 Turns in which you draw 4 more cards in total. You need to say how many cards you pick in turn 3 and 4. Is it 0 and then 4 OR 1 and then 3 OR....???
You also said At first you draw 3 random cards from the deck. If any of them isn't one of those 2 cards, you toss them back in the deck and draw 3 different cards. So what happens if you do draw one of these 2 cards?
You say can also be the cards you tossed back in the deck. Sorry but I do not know what it means to toss the cards back into the deck.
Are you playing this game alone or with other people? You never mentioned this!
You say It's quite simple to explain at least but you did not do a very good job explaining it.
If you want help, then explain the game and show us your work.

 

[Laires]

This is why I said it's easier for people who know the game. I found a very straight visualization on YouTube:

BTW: you dont need to know the actual cards or anything else you see, only the draw of the cards matter in this. Also the deck is only yours, nobody else interfere with it, your opponent has his own deck which is irrelevant to this.

First you draw 3 cards, then you have a choice to throw each card back into the deck and draw completely different card.
BUT if you drew one of the 2 cards we are looking for, you can keep it and replace only the other 2 cards. Or you can have both of the cards you are looking for and keep them both.

Each Turn you draw a single card, starting at Turn 1 going till Turn 4, so you draw extra 4 cards.
With 3 starter cards and those extra 4 cards that's total of 7 cards by Turn 4, on the Turn you need the game winning card. Nothing after Turn 4 matters, only what are the odds of getting each of the 2 cards in your hand by Turn 4.

My initial approach was very simple. You can go through up to 10 different cards by the time you need the winning card, so that's 10/30 = 1/3 chance to draw each of them, right? NO, because you have to include a chance to draw the cards you shuffled back into your deck, which could make it from 7/30 to 10/30 depending on your luck. That's where I figured this would need bigger brain that knows how to calculate odds and combine them with other odds to make a one odd mess from it. Thanks for trying to keep up, I really appreciate it!