What is the probability of rolling 2 before a second odd? Consider a game of dice:

*

*You win if you roll $2$.

*You lose if you roll two odds ( need not be consecutive ).

*If you roll a $4$ or $6$, you keep playing as you have neither lost nor won.

Eg: $1$, $3$ is a loss. $1$, $6$, $4$, $3$ is a loss.

What's the probability of winning?
The answer is $7/16$.
My approach:
Suppose you roll $2$ on the first go. You win and probability of rolling $2$ is $1/6$.
Suppose you roll evens other than $2$ before you roll $2$. The probability of rolling an even other than $2$ is $1/3$. So for the probability of this event, we have,
$$\small
(1/3)(1/6) + (1/3)(1/3)(1/6) + (1/3)(1/3)(1/3)(1/6) + ... = (1/6)[(1/3)/(1-1/3)] = 1/12
$$
Now suppose we roll evens and one odd before rolling $2$. Probability of rolling an odd is $1/2$. So the probability of this event is $(1/2)(1/6)\left[1 + (1/3) + (1/3)(1/3) + \cdots + \cdots\right] =  1/8$
So required probability is $1/6 + 1/12 + 1/8 = 3/8$.
Obviously, my approach is wrong but I don't understand why. Please help me figure this out. Thanks :)
 A: so how would approach it is first calculating the probability if winning where you only have 1 life, meaning rolling 2 before any odd you have 1/6 chance to roll 2 and 1/2 chance to roll odd meaning you are 3 times as likely to loose than to win' from here its quit wasy to show that the non 2 evens don't matter and the probability of winning is 1/4 which is third of the probability to loose 3/4. but we want the probability of getting 2 odds before the 2 so we can look at it like we played the  game twice and lost each time which is
$ \frac{3}{4}* \frac{3}{4}=\frac{9}{16} $  to loose, so $1-\frac{9}{16}=\frac{7}{16} $ to win
A: You can neglect all throws showing $4$ or $6$. Therefore you throw $2$ with probability ${1\over4}$ and odd with probability ${3\over4}$. You win when you (a) throw $2$ on the first move or (b) odd on the first move and $2$ on the second move. The probability that one of these happens is
$${1\over4}+{3\over4}\cdot{1\over4}={7\over16}\ .$$
All other game stories lead to a loss.
A: Your approach is right, but you haven´t calculated correctly the probability of rolling evens and one odd before rolling 2, since you´re supposing you always get the odd in the first roll.
If you get a 2 in the second roll you had (odd, 2) with probability of $\frac{1}{2}*\frac{1}{6}$
Getting 2 in the third roll might be either (odd, even, 2) or (even,odd, two), with probability of $\frac{1}{2}*\frac{1}{3}*\frac{1}{6} + \frac{1}{3}*\frac{1}{2}*\frac{1}{6} = 2*(\frac{1}{2}*\frac{1}{3}*\frac{1}{6})$
In the fourth roll you have (odd,even,even,2) , (even,odd,even,2) or (even,even,odd,2) , so the probability is $3*(\frac{1}{2}*\frac{1}{3}*\frac{1}{3}*\frac{1}{6})$
So the real probability of rolling evens and one odd before rolling 2 is
$\frac{1}{2}*\frac{1}{6}*[1+2*\frac{1}{3}+3*\frac{1}{3}*\frac{1}{3}+...]$
Let´s call $S=1+2*\frac{1}{3}+3*\frac{1}{3}*\frac{1}{3}+...$ and $R=1+\frac{1}{3}+\frac{1}{3}*\frac{1}{3}+...=\frac{3}{2}$
Now you can realize that $S-R=\frac{1}{3}*S$ so $S=\frac{3}{2}*R=\frac{9}{4}$
Then, the probability of rolling evens and one odd before rolling 2 is
$\frac{1}{2}*\frac{1}{6}*\frac{9}{4}=\frac{3}{16}$
Finally, we have $\frac{1}{6}+\frac{1}{12}+\frac{3}{16}=\frac{7}{16}$
