Throw a dice until get 6.What's the probability that total number of throw is odd?
My approach: throw 1 time : $1/6$
throw 3 times : $(5/6)^2*1/6$
throw 5 times : $(5/6)^4 * 1/6$
...and sum them up.It is a geometric sequence with first term $a_1 = 1/6$ and ratio = $ 25/36$
So the answer is $(1/6)/(1-25/36) = 6/11$
Am I on the correct path? My friend says I should use conditional probability,but I don't quite get it.
Thanks in advance!