Finding expectation through conditioning
A coin, having probability p of coming up heads is successively flipped until at least one head and one tail have been flipped.
(a)Find the expected number of flips needed.
X=1, if the first flip results in heads
X=2, if the first flip results in tails
N: the total number of flips needed
![E[N]=E[N|X=1,h](P(X=1))+E[N|X=1,t](P(X=1))](http://latex.codecogs.com/png.latex?E[N]=E[N|X=1,h](P(X=1))+E[N|X=1,t](P(X=1)))
)+E[N|X=0,t](P(X=0))](http://latex.codecogs.com/png.latex?+E[N|X=0,h](P(X=0))+E[N|X=0,t](P(X=0)))
![E[N|X=1,h]\frac{1}{p}+E[N]](http://latex.codecogs.com/png.latex?E[N|X=1,h]\frac{1}{p}+E[N])
![E[N|X=1,t]\frac{1}{p}+1](http://latex.codecogs.com/png.latex?E[N|X=1,t]\frac{1}{p}+1)
![E[N|X=0,h]\frac{1}{1-p}+1](http://latex.codecogs.com/png.latex?E[N|X=0,h]\frac{1}{1-p}+1)
![E[N|X=0,t]\frac{1}{1-p}+E[N]](http://latex.codecogs.com/png.latex?E[N|X=0,t]\frac{1}{1-p}+E[N])
![E[N]=(\frac{1}{p}+E[N])(p)+(\frac{1}{p}+1)(p)](http://latex.codecogs.com/png.latex?E[N]=(\frac{1}{p}+E[N])(p)+(\frac{1}{p}+1)(p))
![+(\frac{1}{1-p}+1)(1-p)+(\frac{1}{1-p}+E[N])(1-p)](http://latex.codecogs.com/png.latex?+(\frac{1}{1-p}+1)(1-p)+(\frac{1}{1-p}+E[N])(1-p))
![1+pE[N]+1+p+\frac{1}{1-p}-\frac{p}{1-p}+1-p+\frac{1}{1-p}-\frac{p}{1-p}+E[N]-pE[N]](http://latex.codecogs.com/png.latex?1+pE[N]+1+p+\frac{1}{1-p}-\frac{p}{1-p}+1-p+\frac{1}{1-p}-\frac{p}{1-p}+E[N]-pE[N])
![2+\frac{2}{1-p}-\frac{2p}{1-p}+E[N]=E[N]](http://latex.codecogs.com/png.latex?2+\frac{2}{1-p}-\frac{2p}{1-p}+E[N]=E[N])
At this point I'm obviously not able to solve for E[N] because they cancel each other out. What am I doing wrong? I've seriously been looking at this problem for four hours now.
There's another way that I just thought of doing the problem..
![E[N]=E[N|X=1]P(X=1)+E[N|X=0]P(X=0)](http://latex.codecogs.com/png.latex?E[N]=E[N|X=1]P(X=1)+E[N|X=0]P(X=0))
![E[N|X=1]=1+\frac{1}{1-p}](http://latex.codecogs.com/png.latex?E[N|X=1]=1+\frac{1}{1-p})
![E[N|X=0]=1+\frac{1}{p}](http://latex.codecogs.com/png.latex?E[N|X=0]=1+\frac{1}{p})
![E[N]=(1+\frac{1}{1-p})(p)+(1+\frac{1}{p})(1-p)](http://latex.codecogs.com/png.latex?E[N]=(1+\frac{1}{1-p})(p)+(1+\frac{1}{p})(1-p))
})
But I don't have the solution to the problem, so I can't be sure it's correct. Although I'm not very confident it is anyway..
Re: Finding expectation through conditioning
So if your first flip is a H, then you're waiting for a T. And if your first flip is a T, then you're waiting for a H. In either case the number of additional flips needed follows a geometric distribution (but with different values for "p"). Your second approach seems correct to me.
Re: Finding expectation through conditioning
Quote:
Originally Posted by
Random Variable 
So if your first flip is a H, then you're waiting for a T. And if your first flip is a T, then you're waiting for a H. In either case the number of additional flips needed follows a geometric distribution (but with different values for "p"). Your second approach seems correct to me.
Great, that's the same line of thought I was using for the problem.
