what is the probability xth of card not being destroyed? Initially i have N cards with value(1,2,..N) , each time i will destroy cards based on number of card left.
If N is odd , will leave smallest value card and destroy the half of the remaining cards.
If N is even will destroy half of the card.
what is the probability that Xth card is not being destroyed till last round?
Note: Each card has equal probability of not being destroyed.

It very much depends on $N$ and not in a simple way.

So, lets take thirteen cards as an example.  In four steps :


*

*You keep 1 and destroy six, leaving seven cards.

*You keep 1 and destroy three, leaving four cards.

*You destroy two cards, leaving two. 

*You destroy one card.
The probability for retaining 1 is $1/4$, and that for retaining any other particular card is: $1/16$.

Take fourteen cards.


*

*You destroy seven, leaving seven cards.


*

*The lowest card left is: $k\in\{1,2,3,4,5,6,7,8\}$ with probability $\binom{14-k}{8-k}/\binom{14}{7}$


*You keep the lowest(which?) and destroy three cards, leaving four cards.

*You destroy two cards, leaving two cards. 

*You destroy one card.
So the probability of a card being retained is ...?
