What is the probability that Fra wins? Fra and Sam want to play a game. They have two classic coins Head-Tail.
They flip the coins at the same time. 
If the result is $HH$, Fra wins. If the result is $HT$ (or $TH$), they flip again and result is again $HT$( or $TH$) Sam wins. In the other cases they continue.
So for example if happens $HT$ and after $HH$, Fra wins. The important for Fra is that $HH$ is an outcome.
The question is: what is the probability that Fra wins?
My work:
There is the outcome $TT$ that is the canceler of the game in the sense that is like they start again from the beginning. So for finishing the game the possible outcomes are:
$HHXX,HTHH,HTHT,HTTH,THHH,THHT,THTH$ where $XX \in \{HH, HT, TH, TT\}$ so Fra wins in $6$ cases on $10$. So the probability is $\frac{3}{5}.$
What do you think about it? Thanks and sorry for my bad English.
 A: Your result is correct, it can be obtained in a cleaner way as follows. Let $p$ be the probability of Fra to win. Fra wins in the following cases:
1) First throw is HH (prob. $1/4$).
2) First throw is TT (prob. $1/4$) and then Fra wins (prob. $p$).
3) First throw is TH or HT  (prob. $1/2$) and second throw is HH (prob. $1/4$).
4) First throw is TH or HT  (prob. $1/2$), second throw is TT (prob. $1/4$) and then Fra wins (prob. $p$).
Hence:
$$
p={1\over4}+{1\over4}p+{1\over8}+{1\over8}p,
\quad\text{that is:}\quad
p={3\over5}.
$$
A: I did it in a  slightly different way and got a result that seems to follow yours.
I first drew a tree (sadly, I don't know any way to draw it with mathjax), by considering TT as a draw (as you told, the game virtually start over from the beginning). 
I found this probabilities:
$$\begin{array}{lcr}
\text{Throw} & \text{Probability} & \text{Winner} \\
\hline
HH & \frac{1}{4} & Fra \\ 
HTHH & \frac{1}{16} & Fra \\ 
HTHT & \frac{1}{16} & Sam \\
HTTH & \frac{1}{16} & Sam \\
HTTT & \frac{1}{16} & Draw \\
THHH & \frac{1}{16} & Fra \\
THHT & \frac{1}{16} & Sam \\
THTH & \frac{1}{16} & Sam \\
THTT & \frac{1}{16} & Draw \\
TT & \frac{1}{4} & Draw
\end{array}
$$
So, 
$$p(Fra) = \frac{1}{4} + 2\times \frac{1}{16} = \frac{3}{8} $$
$$p(Sam) = 4\times \frac{1}{16} = \frac{1}{4} $$
$$p(Draw) = \frac{1}{4} + 2\times \frac{1}{16} = \frac{3}{8} $$
Now, if we decide not taking care of the throws leading to draws, we get:
$$p(Fra)_{without\_draws} = \frac{p(Fra)}{(1-p(Draw))} = \frac{\frac{3}{8}}{\frac{5}{8}} = \frac {3}{5}$$
A: It is a Markov chain with $4$ states :
$X$ is the start of the game, $HH$ is Fra winning the game, $HT/TH$ is when Sam scores one point, $L$ is when Fra loses the game.
Starting at $X$, we have a probability of $\frac{1}{4}$ that we reach state $HH$, a probability of $\frac{1}{4}$ that we start again ( outcome is $TT$) and a probability of $\frac{1}{2}$ that Sam scores one point. That is 
$$p_X=\frac{1}{4}p_{HH}+\frac{1}{2}p_{HT}+\frac{1}{4}p_{X}$$
At states $HT$, we have a probability of $\frac{1}{4}$ that we reach state $HH$, a probability of $\frac{1}{4}$ that we start again ( outcome is $TT$) and a probability of $\frac{1}{2}$ that Sam scores another point.
$$p_{HT}=\frac{1}{4}p_{HH}+\frac{1}{2}p_{L}+\frac{1}{4}p_{X}$$
where $p_{XYZ}$ is the probability that FRa wins when he is at state $XYZ$. Obiviously, we have $p_{HH}=1$ and $p_L=0$. Solving the two equations, we have $$p_X=\frac{3}{5}$$
A: ns=10000
For t=1 to ns
[alfa]
x=rnd(1)
if x<0.25 then f=f+1:goto[beta]
if x>0.75 then goto[alfa]
y=rnd(1)
if y<0.25 then f=f+1:goto[beta]
if y>0.75 then goto[alfa]
[beta]
next t
print f/ns
Simulation with Just Basic; ns is number of simulations, try other values. 
