Chance of marrying a girl

Question

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.

1. What is the possibility that I can marry my girlfriend?

2. 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?

Answer 1

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 $1$.

Answer 2

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.

Answer 3

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