Probability of obtaining a heads on the coin before a 1 or 2 on the die? 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?
 A: This is the probability of no-heads and probability of no-1,2 before than a head appear in $n$ play, where a play is tossing in order first a coin and after a dice.
So a play before the last play (when a head happens) is no-head AND no-1,2. Because the two events are independent one of each other (coin and dice) then we have that the probability for some $n$ that a head happen before a 1 or 2 in the dice is
$$\left(\frac12\cdot\frac46\right)^{n-1}\cdot\frac12$$
because the probability that the dice show something different than one or two is $\frac46$, and the probability than the coin show tail or a head is $\frac12$. Then we have $n-1$ plays where we cant have a head or a 1 or 2, and in the last play we can have in the coin a head (in the dice doesnt matter what we get after we toss the coin).
Then the probability that this happen in any $n$ number of plays is the probability that this happen in one play OR two plays OR three plays OR..., i.e.
$$\sum_{n\ge 1}\left(\frac13\right)^{n-1}\cdot\frac12=\frac12\sum_{n\ge 0}\left(\frac13\right)^n=\frac12\cdot\frac1{1-\frac13}=\frac34$$
A: Use recursion; Let $p$ be probability of your event.
Then, we have
$$p= \frac{1}{2} + \frac{1}{2}\times\frac{2}{3}\times p,$$
where first term is probability of having head in first toss and second term results from tail in coin toss and 3-6 in first roll and having head before 1-2 in the next tries.
Thus, $p=0.75$.
A: 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*}
A: A simple way to find the probability is to condition on the result of the first round. It is clear there is some probability $p$ of obtaining a head before (though not necessarily immediately before) a $1$ or $2$. Call that the probability of winning.
We win if (i) we get a head on the first round or (ii) we get a tail, don't roll a $1$ or $2$, but ultimately win.
The probability of (i) is $\frac{1}{2}$.
For (ii), note that the probability of tail and then something other than $1$ or $2$ is $\frac{1}{2}\cdot \frac{4}{6}$. Given this has happened, the probability of ultimately winning is $p$. Thus
$$p=\frac{1}{2}+p\cdot \frac{1}{2}\cdot \frac{4}{6}.$$
Solve this linear equation for $p$. We get $p=\frac{3}{4}$.
