If six coins are flipped, there are a total of 64 (or 2^6) possible outcomes. Let’s count the number of “successes” — outcomes in which no two adjacent people will stand. We’ll have successes when 0, 1, 2, or 3 people flip heads. If 4, 5, or 6 flip heads it’s guaranteed that two adjacent people will stand.

Let’s count the successes for each number of heads:

0 heads flipped – There is **1** way for this to happen (TTTTTT) and it is a success.

1 heads flipped – There are **6 **ways for this to happen (HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH) and all of them are successes.

2 heads flipped – If there are 6 people and 2 flip heads, then there are 6 people who could flip the first head and 5 people who could flip the second head, giving 6*5 or 30 possible pairings. This must be divided by 2 because the order doesn’t matter (Bob and Mary is the same as Mary and Bob). So, there are 15 ways for 2 heads to be flipped. It’ll be a success unless the people are adjacent. There are 6 ways that they could be adjacent (HHTTTT, THHTTT, TTHHTT, TTTHHT, TTTTHH, HTTTTH), leaving us with **9** successes.

3 heads flipped – For this to be a success, every other person must flip heads. There are only **2** ways for this to happen (HTHTHT, THTHTH).

Let’s add them up! 1 + 6 + 9 + 2 = 18. So the probability is 18/64, about 28%, that no two adjacent people will stand. Woo!