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

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 $$4/9$$.

• Solution 1 definitely makes sense to me now. Thank you! Solution 2 is a bit more fuzzy to me for some reason, despite answering the same question with the same result. And what if we were trying to answer the coin flip problem? This would be solved differently, correct? Instead, we would say P(Heads on 1st & 3rd of 10 flips) = 1/(10!/(2!8!)/2^10 ? Aug 30, 2020 at 22:47
• The coin problem is different because there is no constraint on number of heads and tails like 5 Heads and 5 Tails. The answer would be simply 1/2, since we want HH, TT out of HH, TT, TH, HT. Aug 31, 2020 at 10:33