Don't monte carlo this

Let $X$ be the Probability that heads first appears on odd numbered toss. Let $Y$ be the probability that heads appears first on even numbered toss.

We have $Y$ = $1 - X$

A) We throw heads on our first toss with probability $\dfrac{1}{2}$ .

B) We throw tails on our first toss with probability $\dfrac{1}{2}$ and now after that event, we wish to throw a heads first on any even numbered toss which is probability $Y$ .

So $X$ = $\dfrac{1}{2}$ + $\dfrac{1}{2}$ $Y$

Substitute $Y$ = $1 -X$ and solve...

$X$ = $\dfrac{2}{3}$

The probablitity that heads first appears in the $n^{th}$ toss is $\frac{1}{2^n}$ . For odd $n$ we get the geometric series $\frac{1}{2}+\frac{1}{8}+\frac{1}{32}...+\frac{1}{2}(\frac{1}{4})^m+...=\frac{1}{2}(\frac{1}{1- { }1/4})=\frac{2}{3}.$

A fair coin tossed until a head appears on an odd numbered toss can happen in the following ways: H,TTH,TTTH,TTTTTTH, ... Therefore, the desired probability is P=1/2+(1/2)(1/2)(1/2)+(1/2)(1/2)(1/2)(1/2)(1/2)+...=(1/2)/(1-(1/4)= 2/3

(I apologize for not knowing how to type exponents)

Of the possible answers listed, only one was more than 1/2. <cheating>

A is an event that H first appears on an odd filp.

Total probability:

$P(A) = P(A|H) \cdot 1/2 + P(A|TT) \cdot 1/4 + P(A|TH) \cdot 1/4 =$ $= 1 \cdot 1/2 + P(A) \cdot 1/4 + 0 \cdot 1/4$

$P(A) = 2/3$

There are any number of ways to conceptualize this. I think one of the easier ways uses a simple conditioning trick. Let $O$ be the event that the first occurrence of a heads is observed on an odd flip, and $N$ be the flip number for the first head.

By conditioning we can simply see that: $\mathbf{P}(O) = \mathbf{P}(O|H=1)\mathbf{P}(H=1)+\mathbf{P}(O|H=2)\mathbf{P}(H=2)+\mathbf{P}(O|H>2)\mathbf{P}(H>2)$ But it should be noted that if we know that H>2, then the system has probabilistically restarted. Hence $\mathbf{P}(O|H>2) = \mathbf{P}(O)$ . Additionally, the only way for H>2 is to get two tails in a row, hence $\mathbf{P}(H>2) = 1/4$ . Therefore the answer reduces to:

$\mathbf{P}(O) =\frac{1}{2}+\mathbf{P}(O) \frac{1}{4}$ . Solving we get the answer of 2/3.

draw the probability three to sufficient amount of branches, notice that the probabilty is the infinite sum of : $0.5 + 0.25 * 0.5 + (0.25)^2 * 0.5 + ...$ by using infinite geometric series, we get $\boxed{2/3}$

1 - 1/4 - 1/16 - 1/(4^3) - ... {{total probability of 1 and subtracting the probability of landing heads on 2nd, 4th, 6th, etc. turn}}

1 - ((1/4)/(1-(1/4)) {{1 - the geometric sum}}

1 - 1/3 {{since (1/4)/(3/4) = 1/3}}

2/3 {{simplifying}}

Using Bayes Theorem, $P(Odd Turn|Head)=P(head|odd turn)*p(odd turn)/P(head)$