# Flipping a coin (similar to the ballot problem)

1. Sep 17, 2008

### Random Variable

1. The problem statement, all variables and given/known data

You flip a coin n times. The probabilty of getting a head on any flip is p. What is the probability that the number of heads flipped is always greater than the number of tails flipped?

3. The attempt at a solution

For example,

if n=1, the only possibility is H
if n=2, the only possibility is HH
if n=3, the two possibilities are HHH or HHT
if n=4, the three possibilites are HHHH, HHHT, or HHTH
if n=5, the six possibilites are HHHHH, HHHHT, HHHTT, HHHTH, HHTHH, and HHTHT

Of course you could keep doing this until you notice a pattern, take a guess at the formula, and then try to prove that forumula by induction. (I tried that approach but didn't get anywhere). Conditioning on the first flip clearly won't simplify the problem, and conditioning on the (n-1) flip or the nth flip doesn't seem to simplify matters either. I'm stumped.

2. Sep 17, 2008

### SiYuan

There's more possibility to n=3 and n=4 isnt it?

HHH, HHT, HTH, THH?

3. Sep 17, 2008

### Random Variable

No. After every flip the total number of heads flipped up to that point must be greater than the number of tails flipped up to that point.

4. Sep 17, 2008

### Dick

I'll tell you where I am so far. Let H(n,k) be the number of good sequences of length n (where heads lead all the time) and k is number of heads-number of tails. You can just define H(n,k)=0 if k>n or k<=0. Ok, so far? Write a recursion relation for H(n+1,k) in terms of H(n,...). That's easy. Now take the whole list [H(n,n),H(n,n-1)...H(n,1)] and figure out how to use it to construct the list [H(n+1,n+1),H(n+1,n),...,H(n+1,1)]. It's pretty easy. You just take the original list, chop off a bit, add some zeros in two different ways and add them. That let's you write a really simple computer program to compute the H(n+1) list from the H(n) list. That's great. Now that you have a simple way to construct the recursion you sit back an wait for your intuition to kick and tell you how to sum the whole thing up. Unfortunately, I'm still waiting. Any good ideas?

5. Sep 19, 2008

### stewartcs

I would suggest writing the question in mathematical terms to see what you are missing.

HINT: What type of probability distribution is this?

CS