Expected number of coin tosses before obtaining at least 1 head

Question

We have covered conditional probabilities and law of total expectation so far but I have a very poor understanding of those concepts. I have absolutely no clue on how to proceed with this question:

A fair die is thrown and a coin is tossed the number of times as the score shown on the die. If any heads are shown in the tosses of the coin, we stop, otherwise, we continue the experiment of throwing the die and coin toss until at least one head is shown. Find the expected number of coin tosses before we stop.

Answer 1

Let $q(k)$ be the probability that a head fails to appears when the die shows $k.\;\,$Then $$ q(k)=\left({\small{\frac{1}{2}}}\right)^k $$ Let $p(k)$ be the probability that a head appears when the die shows $k.\;\,$Then $$ p(k)=1 -q(k) = 1 - \left({\small{\frac{1}{2}}}\right)^k $$ Let $x$ be the expected number of tosses until the game ends.

Either a head appears in the first round, or if not, we are effectively starting a new game, with a prior toss count equal to the value of the roll in the first round.

It follows that $$ x = \sum_{k=1}^6 \left({\small{\frac{1}{6}}}\right) \Bigl(p(k)(k) + q(k)(k+x)\Bigr) $$ which simplifies to $$ x = \frac{7}{2}+\frac{2}{128}x $$ hence $$x = \frac{448}{107}\approx 4.186915888$$

Answer 2

Let the expected number of tosses from the present time to stopping be $E$. If a 1 is rolled, which happens with probability $\frac16$, we flip the coin once. With probability $\frac12$ there are no heads and we spend an expected $E$ more flips from there.

The same situation happens with a roll of 2; because two flips are done, the $\frac12$ becomes $\frac14$. Similarly for a roll of 3, when it becomes $\frac18$, and so on.

All this can be encoded in a relation for $E$: $$E=\frac16\left(1+\frac12E\right)+\frac16\left(2+\frac14E\right)+\dots+ \frac16\left(6+\frac1{64}E\right)$$ $$6E=1+\frac12E+2+\frac14E+\dots+6+\frac1{64}E=21+\frac{63}{64}E$$ $$E=\frac{21\cdot64}{64\cdot6-63}=\frac{448}{107}=4.1869\dots$$