finding the conditional probability that outcome of the second trial is $1$, without any explicit calculations 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 of the first trial is $1$, and outcome $3$ is the last of the three outcomes to occur, find the conditional probability that outcome of the second trial is $1$, without any explicit calculations.
let

*

*$A_1$ = outcome of first trial is one


*$A_2$ = outcome of second trial is one


*$\text{third last}$ = outcome three occurs after outcomes one and two have occurred.


*${A}$ denote the unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes $1$, $2$, or $3$.
what needs to be found is $P(A_2|A_1 \cap \text{third last})$, without any explicit calculations
My reasoning is as follows:

*

*given that outcome $3$ is the last of the three outcomes to occur, we are now reduced to working with a subset $A$ where only outcomes $1$ and $2$ can occur.

*this subset of $A$ may itself be an unending sequence of independent trials.

*since outcomes $1$ and $2$ were equally likely in $A$, they should be equally likely in this subset of $A$ where only outcomes $1$ and $2$ can occur.

*now, the probability that outcome of the second trail is one is $0.5$ (since there are only 2 outcomes, each of which are equally likely)

*also, information about the outcome of the first trial won't change this probability, since the first and second trials are still independent.

hence, $P(A_2|A_1 \cap \text{third last}) = \dfrac{1}{2}$, but the given answer is $\dfrac{1}{3}$.
edit:

*

*$1,2,3,2,1,2,...$ is an example where the outcome of the first trial is one

*$2,1,2,2,1,3,1,...$ is an example where outcome $3$ is the last of the three outcomes to occur

*$1,1,2,2,1,2,3,2,1.....$ is an example where the outcome of the first trial is one and outcome $3$ is the last of the three outcomes to occur

edit2: to summarize all answers and comments

*

*let $X_i=j$ denote that outcome $i$ occurred for the first time at the $j^{th}$ trial. Then,what I wanted to find was $P(X_2 \neq 2\mid X_1=1, X_3>X_2)$, but what I ended up calculating is $ P(X_2 \neq 2\mid X_1=1,X_3\neq 2)$ -- JMoravitz


*$P(A_2|A_1 \cap \text{third last}) = \dfrac{1}{2}$ is is not true because there must be a $2$ between $1$ and $3$. This means there is no symmetry between outcomes $1$ and $2$ anymore for the second trial. -- cr001


*Knowing $3$ is the last of the numbers to appear does at each moment, that $2$ has not yet appeared, slightly skew the numbers in favor of $2$ next appearing as though it were "fated to appear". -- JMoravitz
 A: $P(A_2|A_1 \cap \text{third last}) = \dfrac{1}{2}$ is not true because there must be a $2$ between $1$ and $3$. This means there is no symmetry between outcomes $1$ and $2$ anymore for the second trial.
Consider the extreme sub-case where the experiment ends in exactly $3$ trials. In this sub-scenario, the second trial has a probability of $1$ to result in an outcome of $2$, not $1\over 2$.
Of course, the above sub-case is not the whole sample space and the calculation of this probability is not simple. To do this you need to sum up an infinite amount of things, scaled by each case's probability.
The easy way is to go for Bayes Theorem to solve the original problem.
In the Bayes Theorem way, we have
$$P(A_2 \cap A_1 | \text{third last})\cdot P(\text{third last})$$ $$=P(\text{third last}|A_2 \cap A_1 )\cdot P(A_2 \cap A_1 )$$
Notice, $$P(\text{third last})={1\over 3}$$
$$P(\text{third last}|A_2 \cap A_1 )={1\over 2}$$
$$P(A_2 \cap A_1 )={1\over 9}$$
Notice, the second equality comes from the fact that outcome $1$ cannot be last and there is a complete symmetry between outcomes $2$ and $3$.
