# 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?

• What are your thoughts on how to proceed? – Alex Feb 6 at 16:09
• Welcome to MathSE. What have you tried? – jvdhooft Feb 6 at 16:10
• I think conditional probability might come in to place. Also, the probability that flip 3 time and get k "6" is (5^(3-k))/6^3. But how conditional probability apply next? – Peter Gamer Feb 6 at 16:15

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.

• Do you think binomially distribution is nessary for this problem? – Peter Gamer Feb 6 at 17:25
• You could invent binomial distribution anew for this simple problem. – Christian Blatter Feb 6 at 19:18

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.

$$P(X=k) = {1\choose 1}(\frac{1}{2})^2.{3\choose 1}(\frac{1}{6})^1.(\frac{5}{6})^2+{1\choose 0}(\frac{1}{2})^2.{3\choose 2}(\frac{1}{6})^2.(\frac{5}{6})^1$$