TITLE: Probability of obtaining a heads on the coin before a 1 or 2 on the die?
QUESTION [7 upvotes]: I came across this question recently and can't seem to find the correct approach.
Any help would be appreciated!

An experiment consists of first tossing an unbiased coin and then rolling a fair die.
If we perform this experiment successively, what is the probability of obtaining a heads on the coin before a $1$ or $2$ on the die?
$\mathbb P(\textrm{Heads})=\frac12$
$\mathbb P(1,2)=\frac13$
If $A_i$ represents the event that a $1$ or a $2$ is rolled on the $i^{th}$ toss, then I have to find the following:
$$\bigcup^{\infty}_{i=1}\mathbb P(A_i).$$

But I am  not sure how to find this and also incorporate the probability of landing on heads before this...
Am I approaching this correctly or should I be assigning random variables and working from there?

REPLY [3 votes]: What you are describing is a series.
You could think of this as a game between Alice and Bob, where Alice flips the coin (wins with a head) and Bob rolls the die (wins with 1 or 2). Essentially you are asking what is the probability that Alice wins before Bob
$$P(A<B).$$
Well, she could win before Bob in the 

First round $(A_1)$ with chance $P(A_1) = 1/2$. 
Second round $(A_2)$, which means Alice lost, Bob lost and then Alice flipped a winning Head. This occurs with chance
$$P(A_2) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}.$$
Third round $(A_3)$, which means Alice lost, Bob lost, Alice lost, Bob lost, and then finally Alice flips a winning head. This occurs with chance
$$P(A_3) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}.$$
Etc.

Since the events are disjoint, we can add up the probabilites. This gives us (the series you want),
$$P(A<B) = \sum_{k = 1}^\infty P(A_i) = \sum_{k = 1}^\infty \left(\frac{1}{2}\right)^{k-1}\left(\frac{2}{3}\right)^{k-1}\frac{1}{2} = \frac{3}{4}.$$
The reason why I frame this in terms of a game is because alongside evaluating a series, we can also use craps principle; regarding a particular round,
\begin{align*}
P(A<B) &= \frac{P(\text{Alice wins})}{1-P(\text{Draw})} = \frac{1/2}{1-(1/2)(2/3)} = \frac{3}{4}\\
&= \frac{P(\text{Alice wins})}{P(\text{Alice wins})+P(\text{Bob wins})} = \frac{1/2}{1/2+(1/2)(1/3)} = \frac{3}{4}.
\end{align*}