Combinatorics
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
If your answer is in the form of for coprime , input it as .
The answer is 303.
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1 solution
Here we can use the fibbonacci sequence to generate allowable coin-flip possibilities. For the moment, it will be helpful to imagine the coin flips on a line, rather than on a circle.
Let f(n) denote the number of ways you can throw the first n coins without disrupting our alternating seating arrangement, given a certain starting flip. We can either throw a tails as out last throw (which would have as many possibilities as the previous throw had) or we can throw a heads, which implies the previous throw was a tails, which is as many ways as f(n-2). Thus, f(n) = f(n-1)+f(n-2).
Now we have two cases. If we flip a tails at first, we then have 2 possibilities for the first flip, and thus in 8 flips starting with a tails, we have 34 possibilities.
If we flip a heads, we must end in tails. Thus, we need only look for the 7th term in the head-starting series. There is 1 way to start with heads, then 1 possibility after that, getting 13 possibilities if we start with heads.
Since there are 8 positions, we have 2^8=256 possible ways to flip coins. Thus, the probability of no two adjacent people standing up is 47/256.