Probability with card games

Question

I was doing my exercises and I had this one question.

If I drew all cards from a deck of 36 cards ( 4 different colors and 9 numbers per color) how big is the probability that the second card is a Ace.

I started finding my element set which is 36! And there are 4 ace in the whole deck.

Then I thought, well for the first pull i'll have 4/36 And at the second 3/35 but I think I should add the other numbers (where is not important what number I get) too cos I have 36! Permutations. But I don't understand how should I do that....

Do I do it like (4×3×34!)/36! Or am I completely wrong?

I'm sorry if my english is not good enough.

Answer 1

Then I thought, well for the first pull i'll have 4/36 And at the second 3/35 but I think I should add the other numbers (where is not important what number I get) too cos I have 36! Permutations.

This is an interesting thought and you're on the right track but you have a few mistakes: first you are forcing both draws to be aces instead of just the second draw, second it's not $36!$ total ways, since we are only drawing the first two cards. You should stop here and try to figure out where you went wrong and fix it before reading below.

We have two options: First we draw a non-ace and then an ace, or we draw an ace then an ace. Either one results in an ace on the second draw, but they have different amounts of aces on the second draw. We will call these events $A$ and $B$, and then by inclusion exclusion we have:

$$ \mathcal{P}(A \text{ or } B) = \mathcal{P}(A) + \mathcal{P}(B) - \mathcal{P}( A \text{ and } B)$$

Where $\mathcal{P}$ denotes the probability of some event. Then, we want each probability, for the event $A$ we have $32$ cards that are non aces, and $4$ that are aces, out of $36 \cdot 35$ total combinations since order matters here. Next, for $B$ we have $4$ that are aces for the first draw and $3$ that are aces for the second draw, again out of $36 \cdot 35$ total combinations. And of course, since our first draws our different (ace and non-ace) we can never have both event $A$ and $B$ in the same event, so that event has probability $0$. Thus, we conclude to:

$$ \mathcal{P}(A \text{ or } B) = \frac{32 \cdot 4}{35 \cdot 36} + \frac{4 \cdot 3}{36 \cdot 35} - 0 = \frac{1}{9}$$

This is kind of well known but unintuitive result that the second draw is equally as likely as the first.

Answer 2

The way I thought about it is the following: for the first place you have 35 choices (leaving out 1 ace) for the second place you have 1 choice ( the specific ace you left out) for the rest 34 place it can be anything, so 34! Now you have 4 different aces, so number of arrangements in which you have ace in second place is = 4 * 35 * 1 * 34!

Total number of arrangements = 36!

So the probability of the second card being ace = 1/9

Answer 3

When you are in stuck with a problem, you can try to add some intermediate questions :

1. What is the probability that the 2nd card is ace of spades.
2. Deduce the probability that the 2nd card is an ace.

You can even continue to re-write the questions :

1. What is the probability that the ace of spades is at spot n°2
2. Deduce the probability that the 2nd card is an ace.