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.
Best Answer
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: $$ p={1\over4}+{1\over4}p+{1\over8}+{1\over8}p, \quad\text{that is:}\quad p={3\over5}. $$