Chance of marrying a girl My girlfriend's father has a magic - fair - coin, he agrees to let me marry his daughter if I play his game: I have to toss the coin couple times until I see the head comes up. Then if the number of times I have tossed is divisible by three, I cannot marry his daughter, otherwise I can marry his daughter.


*

*What is the possibility that I can marry my girlfriend?

*Suppose $0 < \alpha < 1$; can you design a game like this, such that the probability of winning is $\alpha$? Your game still requires the player to toss the coin but you are allow to change when is pass/no pass case. ($\alpha\in\mathbb{Q})$
This is a really good and interesting question but I cannot solve it. Can you please help?
 A: The length of the longest initial run of tails is a random variable that can take on values $0,1,2,\ldots$ with probability $1/2, 1/4, 1/8,\ldots$.  The probability of winning is generally
$$
P_{\text{win}}(\mathbf{a}) = \sum_{i=0}^{\infty}\frac{a_i}{2^i},
$$
where $a_i=1$ if a run of $i$ tails followed by a head is a win and $a_i=0$ if it's a loss.  You can see that this is just the number whose binary representation is
$$
0.a_0a_1a_2\ldots$$
For instance, the original question (total flips divisible by three is a loss) has winning probability
$$
0.110110\overline{110} =\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\ldots=\frac{3/4}{1-1/8}=\frac{6}{7}.
$$
In general, if you want to winning probability to be any $0<\alpha<1$, declare that a run of $i$ tails followed by a head is a win if and only if the $(i+1)$-th binary digit of $\alpha$ is a $1$.
A: Hints: For 1, define $a$ as the probability you win if the number of tails is $0 \pmod 3$, $b$ as the probability you win if the number of tails is $1 \pmod 3$,  $c$ as the probability you win if the number of tails is $2 \pmod 3$  Now $a=\frac 12b$ because if you have already tossed a multiple of $3$ times, you can only hope for tails and then a win.  You should be able to write two other equations, and have a set of three.  Solve.
For 2, think of the flips as forming a  binary expansion, heads=1, tails=0.  When you have enough information to conclude whether the number being formed is greater or less than $\alpha$ you announce the result, otherwise you keep flipping.  Flesh this out. 
A: I can answer the first part.
Probability that you get a head in the 3rd flip given that the first two are tails is = (1/2)^3
Probability that you get a head in the 6th flip given that the first five are tails is =(1/2)^6
Probability that you cannot marry his daughter = (1/2)^3 + (1/2)^6 + ...
= (1/2)^3 / (1-(1/2)^3) = 1/7
Probability that you can marry his daughter = 1- 1/7 = 6/7
