Probability of Coin Game Suppose we play the following game. We take turns tossing a fair coin, and whoever is the first to reach $2$ heads (not necessarily in a row) wins. You go first. Draws are not allowed, so for example, if you flip heads, then I flip heads, and then you flip heads again, you win (I don't get a chance to toss the coin again).
What is the probability that you will win the game? Enter your answer as a fraction, such as $\frac23$. 
I'm not sure how to approach this. I understand the probability of getting a head on the first two tries is $\frac14$. I also understand that the first person who goes should have better odds.  Just not sure how to approach this with infinite turns until you get two heads.
 A: At any point in the game when it's your go, you are at one of 4 situations:
A) You have thrown 1 head and your opponent has thrown 1 head.
B) You have thrown 1 head and your opponent has thrown 0 heads. 
C) You have thrown 0 heads and your opponent has thrown 1 head.
D) You have thrown 0 head and your opponent has thrown 0 head
Then we have to think about how we get from one situation to another:
We write for simplicity 'P(X)' to mean 'the probability of winning the game given you are at situation X'. 
P(A) = 0.5 + 0.25P(A) => P(A) = 2/3
P(B) = 0.5 + 0.25P(B) + 0.25P(A) => P(B) = 8/9
P(C) = 0.25P(A) + 0.25P(C) => P(C) = 2/9
P(D) = 0.25P(A) + 0.25P(B) + 0.25P(C) + 0.25P(D) => P(D) = 16/27
A: The game has $10$ states that can be addressed in the form $(F,a,b)$,where $F\in\{A,B\}$ denotes the immediately foregoing player, $a$ denotes the number of heads obtained so far by $A$, and $b$ denotes the number of heads so far obtained by $B$. When $a=2$ or $b=2$ the game is over, and the value $\leq1$ of the other variable is irrelevant.
Denote by $p(F,a,b)$ the probability that $A$ ("I" in the question) wins when the game is in state $(P,a,b)$. Then $p(A,2,b)=1$ and $p(B,a,2)=0$, and our aim is to compute $p(B,0,0)$.
This setup furnishes $8$ more equations of the form
$$p(A,1,0)={1\over2}\bigl(p(B,1,0)+p(B,1,1)\bigr)$$
and similar. Solving the resulting system of equations gives
$${\mathbb P}\bigl({\rm I\ win}\bigr)=p(B,0,0)={16\over27}\ .$$
A: U - you , O - Opponent
1st Try: No body wins
2nd Try: You toss a head, you win with a probability of 1/4. (Because you could have thrown (HT,TH,TT)
2rd Try:  Your opponent throws the unwinning 2nd try with P(O') = 3/4
After 2nd try, P(U) = 1/4
3rd Try: 
Number of winning counts out of unwinning 2nd Try = P(U) = 2/6 & P(O') = 3/4
At this moment, P(U) = $(1/4)+(3/4)^2*(2/6)$ meaning both of you failed to win in the second try and you win in the third try
3rd Try: Your opponent throws the unwinning No of heads = {HTT,THT,TTH,TTT} of {HTT,HTH,THT,THH,TTH,TTT} giving a probability of 4/6 Now the P(O')
4th Try: P(O').P(U)
Number of winning counts out of unwinning 3rd try = P(U) = 3/8
At this moment, P(U) = $(1/4)+(3/4)^2*(2/6)+(3/4)^2*(4/6)^2*(3/8)$ meaning both of you failed to win in the second try  and the third try and you win in the fourth try
Thus you get the series
$$P(U) = (1/4)+(3/4)^2*(2/6)+(3/4)^2*(4/6)^2*(3/8)+(3/4)^2*(4/6)^2*(5/8)^2*4/10+...$$
This reduces to  $$\sum_2^{\infty}\frac{n(n-1)}{2^{(2n-1)}}$$
Which when you evaluate gives you the answer 16/27 and concurs with answer of the other responder.  You can evaluate the series in wolfram alpha.  This is a simple way to make students think about games. Answer: goo.gl/eUj8O5
Thanks
Satish
A: Let us suppose that the person who starts tossing the coin wins the game in $(2k+1)^{th}$ toss. So each player has tossed $k$ times before the winning toss.
Winner has tossed $1$ Head and $k-1$ Tails in his k tosses. 
Loser may have tossed all Tails(Case 1) or $1$ Head + $k-1$ Tails.(Case 2)
Case 1. 
Probability of Winning = $k*(\frac{1}{2})^{2*k+1}$ 
Because there are $k$ ways to achieve one head and and $k-1$ tails(for winner) and only one way to achieve all tails(for loser) in $k$ tosses
Case 2.
Probability  of Winning  = $\frac{k*k}{2}*(\frac{1}{2})^{2*k+1}$
Similar to above case now the number of ways(arrangements of T,H etc is $\frac{k*k}{2}$ 
Total probability of winning in $(2k+1)^{th}$ toss is Case 1 + Case 2.
$P_{2*k+1}$ = $\frac{k*(k+1)}{2}*(\frac{1}{2})^{2*k+1}$
Now answer to your question is summation of $P_{2*k+1}$ from $k=1$ to $\infty$
Let $S$ = $\frac{1*2}{2^3}$ + $\frac{2*3}{2^5}$ + $\frac{3*4}{2^7}$ ..
This series is a simple Arithematico-Geometric Series which adds up to $\frac{16}{27}$ 
