Consider a simple case: persons A and B take turns flipping coins, with A flipping first, followed by B. The first person to throw a head wins. What is the probability of A winning?
Here's what I'm confused about. I know a method to solve for the probability is to use first step analysis and applying the law of total probability (LTP). However, I don't understand why. From what I understand, LTP needs these conditions to work:
There needs to be a partition of a sample space S, say of events B1, B2, and so on. The particular event we are interested in, A, must belong to S. However, the method used in first step analysis involves setting A=the event that person A wins, while B1=the event that the coin lands on Heads on the first flip, and B2=the event that the coin lands on Tails on the first flip.
In order to use LTP, don't we have to show that A belongs to the sample space S? And just what is S in this case?? I have absolutely no idea.
Thanks for your answers in advance!
Define $X$ as the number of the flip on which for the first time a head shows up.
Actually we find easily that $P(X=n)=2^{-n}$ for every positive integer $n$.
Essential is here that $$P(X=n+1\mid X\neq1)=P(X=n)$$ for positive integer $n$.
This leads to the observation that: $$P(X\text{ is odd}\mid X_1=1)=1-P(X\text{ is odd})\tag1$$
The event that player A wins can recognized as $A:=\{X\text{ is odd}\}$
We find by means of the law of total probability :$$P(X\text{ is odd})=P(X\text{ is odd}\mid X=1)P(X=1)+P(X\text{ is odd}\mid X\neq1)P(X\neq1)$$
Here $P(X=1)=P(X\neq1)=0.5$ and on base of $(1)$ we find:$$P(X\text{ is odd})=1\cdot0.5+(1-P(X\text{ is odd})\cdot0.5\tag2$$ From $2$ it follows easily that $P(A)=P(X\text{ is odd})=\frac23$.
Underlying sample space is the set $\Omega:=\{0,1\}^{\mathbb N}$ containing as elements all functions $\mathbb N\to\{0,1\}$. Here $0$ corresponds with tail and $1$ corresponds with head. On this sample space function $X:\Omega\to\mathbb R$ is prescribed by: $$\omega\mapsto\min(\{n\in\mathbb N\mid\omega(n)=1\})$$