# Coin flip to determine the amount of die rolled

A coin is flipped to determine whether two or three die are rolled, heads for two and tails for three. What's the chance of two die being rolled when the point count is eight? Honestly I have really hard time trying to understand this intuitively. I thought it would simply be the probability of flipping heads and then rolling an eight

$$\dfrac{1}{2}*\dfrac{5}{36} = \dfrac{5}{72}$$

I doubt this is the case and even if it was, I have hard time understanding why. Shouldn't the probability of rolling eight with three die affect the answer?

If I understand correctly, you want the conditional probability that the coin came up heads, given that the sum was eight, or in other words

$$P\big[ \text{heads} \big| \text{sum of } 8 \big]$$

In other words, we want to restrict attention to situations where you rolled an 8. In some fraction of these you flipped heads and then rolled a total of eight, whereas in the rest you flipped tails and then rolled a total of eight. We want to know what fraction of the total probability comes from the former situation; that is

$$P\big[ \text{heads} \big| \text{sum of } 8 \big] = \frac{P\big[ \text{heads, sum of } 8 \big]}{P\big[ \text{sum of } 8 \big]}$$

Above, you noted that

$$P\big[ \text{heads, sum of } 8 \big] = P\big[ \text{heads} \big] P\big[ \text{sum of } 8 \big| \text{heads} \big] = \frac{1}{2}\left(\frac{5}{36}\right) = \frac{5}{72}$$

You can work out

$$P\big[ \text{tails, sum of } 8 \big]$$

in a similar way, and use this to determine

$$P\big[ \text{sum of } 8 \big].$$