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There are $n$ distinct/distinguishable beads arranged in a circle. Let's color each bead either black or white with probability $1/2$ independently. What is the probability that there are exactly $k$ pairs of neighboring beads whose colors are different?

For example, the bead circle $BWWB$ has two such pairs while the circle $BWWBW$ has four such pairs.

Thank you for your help!

Alex
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1 Answers1

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Hint 1: Since the probability model is uniform over all possible colorings, it suffices to count the number of colorings satisfying your condition, and then divide by $2^n$.

Hint 2: Note that $k$ must be even for the condition to make sense. [If you go around the circle, you switch colors $k$ times and then end up at the color you began with.] This ended up being irrelevant.

We now assume $k$ is even.

Hint 3: To construct a coloring satisfying your condition, one can first mark where the $k$ "color changes" occur. After setting this, there are two possible ways to fill in the colors.

There are $n$ "gaps" between neighboring beads, so there are $\binom{n}{k}$ ways to choose where these color changes are. The probability is thus $$\frac{2 \binom{n}{k}}{2^n} = 2^{-n+1} \binom{n}{k}.$$ As a sanity check note that this result implies $$2^{-n+1}\sum_{\substack{0 \le k \le n\\\text{$k$ even}}}\binom{n}{k} = 2^{-n+1} \frac{2^n}{2} = 1.$$

Thanks to lulu for catching my mistakes.

angryavian
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  • The method seems sensible, but is the answer correct? If $n=2$, say, you either have no changes or two, so non-zero probabilities only for $k=0,2$ and probability $\frac 12$ for either. But you give a non-zero probability for $k=1$, no? Also...do your values sum to $1$? – lulu Jun 07 '16 at 00:49
  • On further review...sticking to the $n=2$ case, you have two "gaps" but filling them is not an independent choice. If you fill one, you automatically fill the other. I think your crossed out parity observation has meaning. – lulu Jun 07 '16 at 00:52
  • @lulu Thanks for catching my embarrassing mistakes; will try to fix! – angryavian Jun 07 '16 at 00:59
  • No problem. I think all you have to do is to un-cross out your comment, and note that your formula only applies to the case of even $k$. – lulu Jun 07 '16 at 01:00
  • Edited version looks good (+1). – lulu Jun 07 '16 at 01:11