Dependent Bead Draw Copied below is a question that I recently tried to solve from a website:
"There are 5 black beads and 5 white beads in your pocket. You pull the beads out of your pocket one at a time, at random and without replacement. What is the probability that the third bead drawn from your pocket is the same color as the first?"
Here is how I think the problem should be solved.
There are 10! possible ways to draw beads from the bag. Next, there are a = 10!/(2!8!) combinations of 2 bead draws from the bag. Finally, there are b = 5!/(2!3!) combinations of drawing 2 of the 5 colored beads.
So,  P(1st & 3rd Same) = a/b*2 = 4/9. I multiplied by 2 because there are two sets of colors in the bag. The website says that 4/9 was the correct answer.
The reason that I am posting this question is because I'm not sure if I entirely understand why I got the answer correct. Was I not actually just solving for the first and second draws? I would really appreciate it if someone could post a better proof of the answer than what I came up with. Another similar question might be, "If we flip a coin 10 times, what's the probability that we get a heads on the 1st and 3rd flips?". I would solve this problem similarly to the way I solved the first, but again, something feels wrong about the way I solved it and the answer that I got.
Thanks,
Jacob
 A: Solution 1 - Brute Force:
Pick an arbitrary bead. It doesn't matter which colour it is, we have four situations for the next two draws: SS, DS, SD, DD, where S means same colour, D means different colour. Here, we're interested in finding the probability of SS or DS:
$$P(\text{Third is the same colour})=P(SS)+P(DS)=\frac{4}{9}\frac{3}{8}+\frac{5}{9}\frac{4}{8}=\frac{32}{72}=\frac{4}{9}$$
If you were to calculate for the fourth or fifth draw, you'd end up with $4/9$ again. That's why when you solved for the second draw, the answer is the same. There is no difference between $k$-th or $l$-th draw in this question. They're all equally likely to be the same with the first draw.
Solution 2: After drawing one bead, the other draws are either the same colour or a different one. So, all the next draws can be represented by one of the permutations of 4S's and 5D's: SSSSDDDDD, where S is the same colour and D is the different colour. For any of the subsequent draws, say the third, having an S is with probability $4/9$.
