Probability that the first $2$ outcomes are one, given that outcome three is the last outcome to occur

Question

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes $1$, $2$, or $3$. Given that outcome $3$ is the last of the three outcomes to occur, find the conditional probability that

1. the first two trials both result in an outcome of $1$
   

my attempt: let

1. {one $1st$} = event that outcome of first trial is one

2. {one $2nd$} = event that outcome of second trial is one

3. {third last} = event that outcome three occurs after outcomes one and two have occurred.

$P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) = \dfrac{P(\text{one 1st}\cap \text{one 2nd}\cap \text{third last})}{P(\text{third last})} = \dfrac{P(\text{third last}) \cdot P(\text{one 1st}|\text{third last}) \cdot P(\text{one 2nd}|\text{one 1st}\cap \text{third last})}{P(\text{third last})}$ $= P(\text{one 1st}|\text{third last}) \cdot P(\text{one 2nd}|\text{one 1st}\cap \text{third last})$

now, since each trial is equally likely to be either $1$, $2$, or $3$ and we are given that the $1^{st}$ trial is not $3$ hence, $P(\text{one 1st}|\text{third last})=0.5$

similarly, $P(\text{one 2nd}|\text{one 1st}\cap \text{third last}) = 0.5$ since all trials are independent, each trial is equally likely to be either $1$, $2$, or $3$ and the second trial's result cannot be $3$(since outcome $3$ occurs after outcomes $1$ and $2$ have both occurred)

hence, $P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) =0.25$, but the given answer is $\dfrac{1}{6}.$

what did I do wrong?

edit: the given answer(which I understand) is
$P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) = \dfrac{P(\text{one 1st}\cap \text{one 2nd}\cap \text{third last})}{P(\text{third last})} =\dfrac{P(\text{one 1st})\cdot P(\text{one 2nd}|\text{one 1st})\cdot P(\text{third last}|\text{one 2nd}\cap \text{one 1st})}{P(\text{third last})} = \dfrac{\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2}}{\frac{1}{3}} = \dfrac{1}{6}$

Answer 1

There are some assumption that is wrong.

For instance, the sequence $(1,1,3)$ is not a legitimate "third last" event but is counted as legitimate in your calculation.

The first "one first|third last" is correctly calculated to be $1\over 2$. However the "one second| one first and third last" is not $1\over 2$ because a $2$ has to occur somewhere after the $1$ and before $3$ so given the first is $1$ and the last is $3$, there is more chance that a $2$ occurs at the second trial.