Pick a Card Part 2

Continuation of our puzzle from yesterday.

Let's solve this problem using brute force. We do this by looking at all of the possibilities. Let's label the cards as follows:
R1 - the first red card
R2 - the second red card
B1 - the first black card
B2 - the second black card

Next, list all of the possibilities of the cards from left to right:

1. R1 R2 B1 B2
2. R1 R2 B2 B1
3. R2 R1 B1 B2
4. R2 R1 B2 B1
5. R1 B1 R2 B2
6. R1 B1 B2 R2
7. B1 R1 R2 B2
8. B1 R1 B2 R2
9. B1 B2 R1 R2
10. B1 B2 R2 R1
11. B2 B1 R1 R2
12. B2 B1 R2 R1
13. B2 R2 R1 B1
14. B2 R2 B1 R1
15. R2 B2 R1 B1
16. R2 B2 B1 R1

We have exactly 16 different ways these cards can be dealt. Check it for yourself.

Now since these cards are dealt face down, we don't know the card values. Let's say we pick the left 2 cards. How many of the 16 possibilities gives us either 2 red cards or 2 black cards? Just check the list above and count them. We get 8.

Thus our probability of getting a pair of cards of the same color is 8/24 or 1/3!

Is that our final answer? Yes it is! :) Not 2/3, nor 1/2 as we might have expected intuitively.

Next time we will see if we can devise some shortcuts and see if we can extend this to more general problems.