Card game odds/probability help

[Cardgameplayer]

Hello all, first of all I want to say that I really don't know much about math at all. However, I do love me some card games and was wondering if I could get help on something.

1. I have a deck of 60 cards
2. Opening hand consists of 7 cards
3. I want to find at least 1 copy of X
4. X =10

This is simple enough that even I was able to figure out how to calculate it. However, card games can be a bit complicated at times. Here is where I am completely lost...

As I said before, Inside that deck of 60 there is 10 copies of X.
Alongside that we have 4 copies of Y
We also have 4 copies of Z
The remaining 42 cards are irrelevant

Let's talk Y and Z

Y has the ability to be discarded to draw 2 more cards
Z has the ability to be discarded to draw 7 more cards but can only be used a total of one time. Meaning if you were to draw another copy, you would NOT be able to use it.

So my question to you all becomes this

Given the abilities of Y and Z, what are the odds of finding at least 1 copy of X in a 60 card randomized deck? Opening hand is 7

 

[blamocur]

What are the meanings of X, Y and Z? For example when you say X=10 and "there is 10 copies of X" do you mean the same thing?
What do you mean by "Y has the ability to be discarded to draw 2 more cards"? If I get four Y's can I discard them all and get 8 new cards? And why would I want to do this? Is my goal to get as many X's as possible or just to get one X?

 

[Cardgameplayer]

What are the meanings of X, Y and Z? For example when you say X=10 and "there is 10 copies of X" do you mean the same thing?
What do you mean by "Y has the ability to be discarded to draw 2 more cards"? If I get four Y's can I discard them all and get 8 new cards? And why would I want to do this? Is my goal to get as many X's as possible or just to get one X?

Sorry for not being clear in my post, let me answer your questions and hopefully it helps

X, Y, and Z are just different cards

X = 10 and "there is 10 copies of X" are the same thing

You are correct, if you draw four Y's you can discard them all to get 8 new cards. You can even start with one and draw into the others. However, with Z you can only do it once even if you draw multiple copies

The reason for drawing a bunch of cards is because the goal is to find at least 1 X or in other words greater than or equal to 1

I want to know what the probability is given all that information. I think it might be called true probability or estimated probability? I'm not really sure what the correct term is

I hope that cleared some things up

 

[blamocur]

This is much clearer -- thank you.

Can you figure out the probability of drawing at least one 10 on the first try, i.e. before replacing Ys and Z ?

 

[Cardgameplayer]

This is much clearer -- thank you.

Can you figure out the probability of drawing at least one 10 on the first try, i.e. before replacing Ys and Z ?

I know of something called hypergeometric distribution but being completely new to math I wasn't sure how to approach it. I did things a little different

I looked at the odds of not drawing a certain card so -

50/60 for the 1st draw
49/59 for the 2nd draw
48/58 for the 3rd draw
47/57 for the 4th draw
46/56 for the 5th draw
45/55 for the 6th draw
44/54 for the 7th draw

Now we do some math (50*49*48*47*46*45*44) / (60*59*58*57*56*55*54) = .258 or we can round it up to 26%
The odds of not drawing the certain card I want is about 26% so lets subtract that from 100% to see the odds of getting the card I want

100 - 26 = 74

So about 74% of the time I will find the card I want in my first 7 draws

 

[Cubist]

So about 74% of the time I will find the card I want in my first 7 draws

I agree with your calculation and result. Well done!

The next step is to work out what happens with the Y and Z cards that can be used to (draw extra cards and) increase the probability of getting an X. I can think of the following approaches...

1) Write a Monte Carlo simulation to estimate the result (can you write code?)
2) Write a program which goes through all possibilities of Y and Z cards drawn (and use recursion for any extra cards drawn). This could give you an exact result.
3) Estimate using maths (this will be a poor estimate compared to option 1). Work out an (average) expected number of drawn cards. Then use this to calculate a new probability of obtaining an X card.

What do you think, would you be able to make a start on any of these ideas?

I'm also interested if @blamocur has any further ideas (or any other helper) before proceeding.