Consider a game where you flip a coin until you get 2 heads in a row.

Let X be a random variable that counts the #of tails.

1. What is the distribution of X?
2. If you win X dollars, how much should you pay for this to be a fair
game?(Expected value = 0)

The answer to question 2 is fairly simple.  Let ET be the expectation
of the game (Assume we are paid $1 for each tail until two heads in a
row occur.)  Let EH be the expectation of the game if the first coin
toss is a head.  Obviously,

ET = 1/2*("Expectation given that the first coin toss is a head") 
   + 1/2*("Expectation given that first coin toss is a tail").

Notice that if the first coin is a tail then the total expectation for
the game is 1 + ET.  So

(A) ET = 1/2*EH + 1/2*(1+ET).

If the first coin toss is a head, then two things can happen.  1)
Another head is tossed and the game ends with zero payoff.  2) A tail
is tossed after the original head toss and the expectation for the
player is 1+EH.  So

(B) EH = 1/2*0 + 1/2*(1+ET).

Substituting the expression for EH in (B) into equation (A) yields

ET = 1/2*1/2*(1+ET) + 1/2*(1+ET).

Solving for ET yields

  ET = 3/4 + 3/4*ET
4*ET = 3 + 3*ET
  ET = 3.

So the expectation for the game is 3.  

Question 1 is harder.  I expect the answer involves the Fibbinocci
sequence.  See "An Introduction to Probability Theory and its
Applications" by Feller 1968 Vol 1, Chapter 8, section 7, pages
322-326 of the Third Edition.

For another source of information goto www.twoplustwo.com probability
forum.  There were at least two threads discussing sucessive coin
tosses.  See
"Great problem solved - original material" posted on 07/12/03 06:34
PM.

The distribution of X can be found by analyzing the following equation

(A) p(i) = 1/2*pt(i) + 1/4*pht(i) + 1/4*phh(i)

where p(i) is the probability that X will equal i, pt(i) is the
probability that X will be i given that the first coin toss was tails,
pht(i) is the probability that X will be i if the first two coin
tosses are heads then tails, and phh(i) is the probability that X is i
given that the first two coin tosses are two heads.  Equation (A) is
true because all sequences of coin tosses begin with T, HT, or HH with
probabilities 1/2, 1/4, and 1/4 respectively.

We can express the right hand side of (A) in terms of p to get a
recurrence relation.  First

(B) pt(i) = p(i-1)

because i is the number of tails.

(C) phh(i) = 0 if i > 0  because the game ends on the second toss.

(D) pht(i) = p(i-1) also.  

Substituting (B), (C), and (D) into (A) yields:


    p(i) = 1/2*p(i-1) + 1/4*p(i-1) + 0  
(E) p(i) = 3/4*p(i-1)                  

for i>0.  The only way that X can be zero is two heads in a row, so 

p(0) = 1/4.

If we repeatedly use (E) to generate the other values for p we get

p(i) = 1/4*(3/4)^i  =  P(X=i)

That is the distribution of X and the answer to for question 1.