TTTTTT (1)

TTTTTH (6)

HTHTTT (4+3+2+1)

HTHTHT (2+2)

= 21 ways

2^6 combinations = 64

21/64

I can't think of a faster way off the top of my head

Hard question but awesome for really testing knowledge of probability!

Here, we flip a coin 6 times. Think about all the different types of flips you can have if you were to do that.We want to avoid flipping two consecutive heads in a row. That means \(HHTTTT\), or any variant shouldn't be counted. But we could, for example, have \(HTHTTT\), or \(THTHTH\).

Notice, if we flip more than three heads in our six flips, there will 100% be two heads being flipped consecutively. This means we have to flip more than two tails in order to count the flips we want.

There are six flips:

_ _ _ _ _ _

Each of those can have two outcomes, heads or tails. So there are 2*2*2*2*2*2 = 64 different possibilities of flips. This will be our denominator.


I would use Carcass' diagram to see the different possibilites. Here's how you'd arrive to them:


Case 1: Flip all tails

This one is trivial. If I flip all tails, no heads are consecutive.

That's 1 possibility out of the 64 possible flips



Case 2: Flip 5 tails

There are six flips. Five of them need to be tails. That would be 6 Choose 5, or in other words:

_ _ _ _ _ _

Fill in five T's (for tails) in those six positions. How many ways can we place the H (for heads) in the remaining spot? 6 ways.

That's 6 possibilities out of the 64 possible flips



Case 3: Flip 4 tails

Now it gets a bit tricky.

There are again six positions. Four of them need to be tails. That's 6 Choose 4, which is 15 different possibilites. However, we need to be careful, as we are counting consecutive heads in those 15 possibilities. For example, one of those possibilities is: \(HHTTTT\).

How many consecutive head flips are there in those 15 possibilites? If we could find them, we could subtract them out from the 15 possibilities.

Here's a neat trick for dealing with two objects that need to be adjacent while counting:

_ _ _ _ _ _

x _ _ _ _

Let that x represent the two consecutive heads. We've collapsed two spaces into one space, which will make the counting a bit easier. Since they can't be seperated we can count them as just one object. So, how many ways can we move that x in those positions? Five ways.

Out of the 15 ways to distribute four tails, we need to take out 5 possibilities that have consecutive heads. That's 15-5 = 10 possibilities.

That's 10 possibilities out of the 64 possible flips



Case 4: Flip 3 tails

This one is also tricky.

For this one, notice that out of the six positions:

_ _ _ _ _ _

We get a sense that there will need to be some alternating flips, so heads then tails, then heads again, etc. So let's throw in the tails with some positions between them:

_ T _ T _ T _

Notice I've added an additional space. This is because the H could appear on either side of each T. Once we select our three positions for H, we can remove the additional position we added to go back to our original six flips.

So there are 4 positions, and we need to fill three heads. That's 4 Choose 3, which is 4.

That's 4 possibilities out of the 64 possible flips


Summing up

Adding all four cases we get: 1+6+10+4 = 21 possibilites out of 64.

That is: \(\frac{21}{64}\), which is choice A.