<s>[INST] More than 4 at Even throw of Fair Dice A dice is being thrown till we get number greater than 4 at Even Throw. What is the Probability of this Event?
I have two approaches:
Method $1$. 
The probability of number greater than $4$ is $\frac{1}{3}$ and Probability of number less than or equal to $4$ is $\frac{2}{3}$
let $T_1$, $T_2$, $T_3$ $\cdots$ be Throw numbers. Then the Required Probability is
$$P(A)=P(T_1T_2)+P(T_1T_2T_3T_4)+P(T_1T_2T_3T_4T_5T_6)+\cdots$$
But $P(T_1T_2)=\frac{2}{3}.\frac{1}{3}$
$P(T_1T_2T_3T_4)=\frac{2}{3}.\frac{2}{3}.\frac{2}{3}.\frac{1}{3}$ and so on finally we get an infinite GP as
$$P(A)=\frac{1}{3}\left(\frac{2}{3}+\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^5+\cdots\right)=\frac{2}{5}$$.
In Method $1$, I considered that at odd throws we get number less than or Equal to $4$ and at Even  throw a number greater than $4$.
Method $2$.  In This i will consider that at odd throws we can get any number, but still  i need Probability of number greater than $4$ at Even throw.
Now here $P(T_1T_2)=1. \frac{1}{3}=\frac{1}{3}$
$P(T_1T_2T_3T_4)=1.\frac{2}{3}.1.\frac{1}{3}=\frac{1}{3}.\frac{2}{3}$
$P(T_1T_2T_3T_4T_5T_6)=1.\frac{2}{3}.1.\frac{2}{3}.1.\frac{1}{3}=\frac{1}{3}.\left(\frac{2}{3}\right)^2$  Hence
$$P(A)=\frac{1}{3}\left(1+\frac{2}{3}+\left(\frac{2}{3}\right)^2+\cdots\right)$$
hence $$P(A)=1$$...
Can any one Resolve this Ambiguity
 [/INST] If you insist that number greater than 4 comes on an even throw, you can ignore all the odd throws. They are not events, because they cannot determine success or failure. The simple answer is you keep trying with some chance each time.  Eventually you will succeed.  The probability is $1$.  You need to understand that=that is the point of the problem.  Your error in method 1 is that all the $T_n$ with $n$ odd are $1$ because that doesn't matter.
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