## Hints for Homework 1

#### 1.42 and 1.44

*(See the update/correction on the homework sheet regarding 1.44.)*

These two problems are probably the hardest on this assignment; most of the rest of the problems are computational. I don't want to give these problems away entirely, but you can expend your effort in one of two ways (1) hard calculations, or (2) a little hard thinking which would allow you to solve the problem with a few lines of nice equations.

On the assignment I wrote that 1.42 is similar to an example in class -- namely, computing the expected number of flips of a fair coin needed for *one* head to appear. However, I didn't mean to imply that you have to solve it the same way. While it's possible to use a similar approach, it can lead to some messy calculations. For one thing, the sample space is nowhere near as nice as our in-class example; there's only one way to get your first H on the fifth flip (**TTTTH**), but there are many ways to get your first HH ending on the fifth flip (**TTTHH**, **THTHH**, **HTTHH**). With some careful work and analysis of the sample space and associated probabilities, you could write down an infinite sum and probably (no pun intended) work out the answer. I'd recommend a different approach.

You're interested in E, the expected number of flips until you get HH. Notice that flipping a T is essentially the same as starting over. So if we define E_{T} to be the expected number of flips to get an HH after a T, then

E = E_{T}
You could also define E_{H} to be the expected number of flips until you get HH after flipping an H. This is definitely a different number than E_{T}. If you analyze the possibilities, you should be able to write down some equations involving E_{T}, E_{H} and E. Solve the system and you'll have your answer!

If you can solve 1.42 with P(H)=p and P(T)=q, then it's ok with me if your answer for 1.43 is essentially plugging the new probabilities into your formula from 1.42. However, if you have a lot of trouble with this problem, I'd just work with p=q=1/2 until you straighten things out.

Math 5251: Rogness