A and B play rolling an $N$-faced die having numbers $1,2,3..,N$. A takes the first turn. After that both players take alternative turns. Whoever gets a number which is lesser or equal to the number obtained by the opponent in the previous turn loses the game. Let $P(k)$ be the probability of losing after rolling the die if opponent obtained $k$ in the previous turn. Write a recursive expression for $P(k)$. Also find the probability that A loses the game.
If my opponent obtains some number $1 \le k \le N$ in the previous turn, then $p(k)$, the probability that I lose, should be $k/N$ since favourable outcomes of a die roll are $[1,k]$ and there are obviously $N$ possible outcomes. So I wish to obtain a recursive expression for this.
Now I think $p(k+1)>p(k)$ so I think the recursive expression should be $p(k+1)=p(k)+1/N$, $1/N$ added for one more extra term $k+1$.
Where do these ideas need correction?
Thanks Hagen and Peter for the hint , i ll attempt a recursive solution P(N)=prob to lose if the opponent gets N is the previous turn Clearly P(N) = 1 P(loose) = loose in this chance + P(u dont loose).P(opponent doesnot loose).P(then u loose)+...
P(N-1)=N-1/N + (1/N).(1-P(N)).P(lose this time)=N-1/N . . P(N-2)=N-2/N + (1/N)(1-P(N-1)) + 0 please help me generalize this