Consecutive Tails

Before counting anything, let us pin down the structure of the problem.

Ten flips of a fair coin give $2^{10}=1024$ possible sequences, and every one of them is equally likely, each with probability $\frac{1}{1024}$. That is the key simplification: because the sequences are equiprobable, we never have to weigh one against another. The answer is just

$$P=\frac{\#\{\text{sequences where tails come only in pairs}\}}{1024}$$

So this is purely a counting problem. All the work is in the numerator.

Step 1: What does a valid sequence look like?

Read the condition carefully. Tails may only appear in consecutive pairs, so a valid sequence is nothing more than a string built out of two kinds of pieces:

• the block $[TT]$ — a pair of tails
• the single letter $H$

Glue any number of these together in any order and you get a valid sequence. Note that this correctly allows $TTTT$, which is just two $[TT]$ blocks sitting next to each other, and it correctly forbids a lone $T$, since there is no piece that can produce one.

Say we use $i$ blocks of $[TT]$. Those eat up $2i$ of our ten flips, so the rest must be heads:

$$\#\text{heads}=10-2i$$

and the total number of pieces we are arranging is

$$i+(10-2i)=10-i$$

Since $2i\le 10$, the possible values are $i=0,1,2,3,4,5$. So the plan is clear: count the arrangements for each $i$, then add them up.

Step 2: Counting the arrangements for a fixed $i$

We are arranging $10-i$ pieces in a row: $i$ copies of $[TT]$ and $10-2i$ copies of $H$.

Start by pretending every piece is distinguishable — label the heads $H_1,H_2,\dots ,H_{10-2i}$ and the blocks $B_1,B_2,\dots ,B_i$. Then there are simply

$$(10-i)!$$

orderings, since we have $10-i$ distinct things to permute.

But of course they are not distinguishable, and this badly overcounts. Swap $H_3$ with $H_7$ and the string of H's and T's you actually see does not change at all. Same if you swap $B_1$ with $B_2$. So every genuine sequence has been counted once for every way of relabelling the heads among themselves and the blocks among themselves — that is,

$$(10-2i)!\cdot i!$$

times over. To cancel this redundancy, we divide it out:

$$\#\text{arrangements}=\frac{(10-i)!}{i!(10-2i)!}=\left(\genfrac{}{}{0ex}{}{10-i}{i}\right)$$

Which reads very naturally in the end: out of $10-i$ positions in the row of pieces, choose the $i$ that hold the $[TT]$ blocks.

If you are being extra careful, you might also want to distinguish the two tails inside each block, multiplying the top by $2^i$. But those two tails are identical as well, so you would divide the very same $2^i$ straight back out. It cancels, and we can ignore it entirely.

Step 3: Summing over $i$

Now just add up the six cases:

$$\#\text{valid sequences}=\sum _{i=0}^5\left(\genfrac{}{}{0ex}{}{10-i}{i}\right)$$

Term by term:

• $i=0$: $\left(\genfrac{}{}{0ex}{}{10}{0}\right)=1$  (all heads)
• $i=1$: $\left(\genfrac{}{}{0ex}{}{9}{1}\right)=9$
• $i=2$: $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$
• $i=3$: $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$
• $i=4$: $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$
• $i=5$: $\left(\genfrac{}{}{0ex}{}{5}{5}\right)=1$  (all tails, $TTTTTTTTTT$)

$$1+9+28+35+15+1=89$$

Step 4: The probability

$$P=\frac{89}{1024}\approx 0.0869$$

Step 5: Sanity check

Let us verify $89$ a completely different way, by building the count up from short sequences. Let $A_n$ be the number of valid sequences of length $n$.

Look at how a valid sequence of length $n$ starts. There are only two possibilities, since it cannot begin with a lone tail:

• it starts with $H$, and the remaining $n-1$ flips form any valid sequence — $A_{n-1}$ ways
• it starts with $TT$, and the remaining $n-2$ flips form any valid sequence — $A_{n-2}$ ways

These cases are disjoint and cover everything, so

$$A_n=A_{n-1}+A_{n-2}$$

with $A_0=1$ (the empty sequence) and $A_1=1$ (just $H$). Rolling it forward:

$$1,1,2,3,5,8,13,21,34,55,89$$

So $A_{10}=89$, agreeing with the sum in Step 3. (Those are the Fibonacci numbers, which is a nice thing to fall out of a coin problem — the recursion above is exactly the Fibonacci rule.)

It is also easy to spot-check the small cases by hand. For $n=4$ the formula gives $\left(\genfrac{}{}{0ex}{}{4}{0}\right)+\left(\genfrac{}{}{0ex}{}{3}{1}\right)+\left(\genfrac{}{}{0ex}{}{2}{2}\right)=1+3+1=5$, and indeed the only valid sequences are $HHHH$, $TTHH$, $HTTH$, $HHTT$ and $TTTT$.

Thus, the probability that all tails occur in consecutive pairs is $\frac{89}{1024}$