Let's play heads or tails in the following way: We toss the coin four times and then as many times as there were heads in the first four tosses. What is the probability that we will get at least 5 heads in all our tosses?
Solution. The result of each coin toss is either heads or tails, so the total number of possibilities for 4 tosses is: $2^{4}$. To get at least 5 heads, we need to get at least 3 heads in the first round.

The first favorable case is when we get heads 3 times out of 4 tosses. This can happen in 4 different ways (depending on which toss results in tails).

Then we can toss three more times, and we need to get at least 2 heads. In this case, the total number of possibilities is $2^{3}$ and the number of favorable outcomes is 4. The probability of getting at least 5 heads this way is:

$$
\frac{4}{2^{4}} \cdot \frac{4}{2^{3}}=\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8}
$$

The second favorable case is when we get 4 heads out of 4 tosses, which can only happen in one way. Then we can toss four more times, and we only need one more head, which can happen in $2^{4}-1$ ways (we need to exclude the one case where we get all tails.) The probability of getting at least 5 heads in this scenario is:

$$
\frac{1}{2^{4}} \cdot \frac{2^{4}-1}{2^{4}}
$$

Summing up the two cases, the desired probability is:

$$
\frac{1}{8}+\frac{2^{4}-1}{2^{8}}=\frac{1}{8}+\frac{15}{256}=\frac{47}{256}
$$