A die is rolled. A coin is flipped four times. Find the probability that the number on the die is twice the number of heads from the coin flips. I absolutely have no idea how to begin this problem. I am not looking for an exact answer as much as I am a proper solution so I can learn how to handle similar problems.
 A: You have 1 head with probability ${4\choose 1}\cdot \frac1{2^4}=\frac14$ and a matching die result of two with $\frac16$, thus this gives $\frac1{24}$.
You have 2 heads with probability ${4\choose 2}\cdot \frac1{2^4}=\frac38$ and a matching die result of four with $\frac16$, thus this gives $\frac1{16}$.
You have 3 heads with probability ${4\choose 3}\cdot \frac1{2^4}=\frac14$ and a matching die result of six with $\frac16$, thus this gives $\frac1{24}$.
Woth 0 or 4 heads, the die cannot macth twice this number. Thus the total probability is $\frac1{24}+\frac1{16}+\frac1{24}=\frac{7}{48}$.
A: We use the strategy of breaking the problem  up into simpler cases. Our desired event can happen in $3$ different ways: (i) $2$ on die, $1$ head; (ii) $4$ on the die, $2$ heads; (iii) $6$ on the die, $3$ heads.
Let's find the probability of (i). The probability of $2$ on the die is $\dfrac{1}{6}$. The die and coin don't talk to each other, so given we got $2$ on the die, the probability of $1$ head on the coin is, as it always is,  $\dfrac{\binom{4}{1}}{2^4}$. The technical term is that the event $2$ on the die and the event $1$ head are independent. 
So the probability of (i) is 
$$\frac{1}{6}\cdot\frac{\binom{4}{1}}{2^4}.$$
Do a similar calculation for (ii) and (iii), and add up.
