Probability of die and coin , probability law? conditional probability? You roll a fair die 3 times, say k rolls are 6's.You then flip a fair coin till you got k head. what is the probability that you flip two time and get k head?
 A: The number $k$ of sixes you throw in three trials is binomially distributed; the resulting probabilities are
$$(p_0,p_1,p_2,p_3)=\left({125\over216},{75\over216},{15\over216},{1\over216}\right)\ .$$
Since we do $2$ throws of the coin afterwards we have $1\leq k\leq2$. When $k=1$ the probability that we need exactly $2$ throws for the first $H$ is ${1\over4}$, and if $k=2$ the probability that we arrive at $HH$ in two throws is also ${1\over4}$.
Now put it all together.
A: From what I understand of your scantily worded problem.
$P(Y=k)$ where $Y$ is the number of $k$ rolls of $6$.
$P(X|Y=k)$ where $X$ is the number of tosses to get $k$ heads.
$Y$ is distributed as Binomial $(3,\frac{1}{6})$
$X|Y$ is distributed as negative Binomial $(k,r,\frac{1}{2})$. where k is the number of heads and r is the number of tails)
$P(X=k) = P(X=1/Y=1).P(Y=1) + P(X=2/Y=2).P(Y=2)$
You flip two times to get k heads.
$P(X=k) = {1\choose 1}(\frac{1}{2})^2.{3\choose 1}(\frac{1}{6})^1.(\frac{5}{6})^2+{1\choose 0}(\frac{1}{2})^2.{3\choose 2}(\frac{1}{6})^2.(\frac{5}{6})^1$
