Probability of a die ends up showing 1 So the question asks:  A fair six-sided die is rolled repeatedly until one of the numbers 1, 2 or 3 shows up. Find the probability that the experiment ends with the die showing the number 1.
So so far I have: 
Suppose n th time the first 1 shows up, then 
P(experiment ends with 1 ) = $(\frac{3}{6})^{n-1}  \frac{1}{6} $ =  $\frac{1}{6}(\frac{1}{2})^{n-1}   $
But in this way I cannot get rid of the n, so I tried 
Suppose event A is the die shows 4, 5, or 6; and event B is the die shows 1.
And P(A) = $\frac{1}{2}$ ; P(B) = $\frac{1}{6}$
So I want to know the probability of event A happens exactly before event B happens, 
So I guess the probability 
P(A before B) = $\frac{1}{2}$ /( $\frac{1}{2}$+$\frac{1}{6}$) = $\frac{3}{4}$
However, the answer given is $\frac{1}{3}$. So where am I doing wrong and what is right way to do this problem? 
 A: You could uses series, but there are a couple of shortcuts. 


*

*You could argue that by symmetry, the probability is $1/3$

*Let $A$ be the event that the game ends on $1$. Then you could use this trick. The way the game is structured, then you could win with $1$ on the first trial with probability $1/6$, or you draw with chance $1/2$ and restart the game:
$$p_A = \frac{1}{6}+\frac{1}{2}p_A,$$
which gives $p_A = 1/3$.

*Finally, by craps principle
$$P(A) = \frac{P(A)}{P(\text{End game})} = \frac{P(A)}{1-P(\text{Draw})} = \frac{1/6}{1/2} = \frac{1}{3}.$$
A: 
So I want to know the probability of event A happens exactly before event B happens, 

No, you want the probability that event $B$ (rolling a $1$) happens before the event of rolling one of $\{2,3\}$.   Call it event $C$.
Ignore all rolls where event $A$ happens; the game will continue and each subsequent roll is independent of the last.   Eventually the game ends on the first roll when $B$ or $C$ happens; (who knows when, but it is almost certain to happen after some finite tries).   So, given that one will happen on the last roll (whenever it is), what is the probability that the one you want happens?
$$\mathsf P(B\mid B\cup C) = \frac{1/6}{3/6}$$
A: Doing this using series, as you tried.  
The event "experiment ends with 1" is the disjoint union of the events:
"experiment ends with 1 after rolling the die once (n=1)",
"experiment ends with 1 after rolling the die twice (n=2)",
"experiment ends with 1 after rolling the die thrice (n=3)",
etc. 
So what you have is kind of correct $(\frac{3}{6})^{n-1}  \frac{1}{6}$, 
but you need to sum over $n\ge1$.
That is, $(\frac{3}{6})^{n-1}  \frac{1}{6}$ is not P(experiment ends with 1 ). It is P(experiment ends with 1 at the n-th roll). 
So, the answer will be $\sum\limits_{n=1}^\infty\Bigl((\frac{3}{6})^{n-1}  \frac{1}{6}\Bigr) = \frac16 \sum\limits_{n=1}^\infty(\frac{1}{2})^{n-1} = \frac16\cdot2=\frac13$. 
