# Conditional probability of multiple coin flips

I flip $$9$$ fair coins. There are more heads than tails. What is the probability all flips resulted in heads?

Since coins are fair, probability of more heads is the same as more tails, therefore $$p = 0.5$$. Let's call the event where all $$9$$ flips are heads H. Does that mean the conditional probability I mentioned is $$2P(H)$$?

• Hint: condition on the number of heads and tails (the event 'more heads than tails' branches into 5 cases: 5 heads 4 tails, 6 heads 3 tails, ..., 9 heads 0 tails.) Commented Nov 15, 2022 at 2:50
• Yes you are correct Commented Nov 15, 2022 at 2:51

Let's say the event of all heads is $$A$$. You want $$\mathsf P(A\mid H)$$.
You have argued that $$\mathsf P(H)=1/2$$, because $$\sum_{t=5}^9\binom 9t=\sum_{h=5}^9\binom 9h$$ . The count of outcomes for five or more tails equals the count of outcomes for five or more heads. Indeed, however this also means the count of equiprobable outcomes in event $$H$$ is: $$\sum_{h=5}^9\binom 9h = 2^8$$
Now there is only one outcome where all flips show heads, and this is a subset of event $$H$$. Thus: \begin{align}\mathsf P(A\mid H) &=\dfrac{\mathsf P(A\cap H)}{\mathsf P(H)}\\&=\dfrac{\mathsf P(A)}{\mathsf P(H)}\\&=\dfrac{1/2^9}{1/2}\\&=\dfrac{1}{2^8}\end{align}
Note This is $$\mathsf P(A\mid H)=2\mathsf P(A)$$ rather than $$2\mathsf P(H)$$.