A and B toss a dice with 1 to n faces in an alternative way, the game is over when a face shows up with point less than the previous toss and that person loses. What is the probability of the ﬁrst person losing the game and the expected number of tosses?

I can solve this question for 6 face dice: If the first die is 1, there are 5 greater numbers on the second: 1/6 \* 5/6 = 5/36. If the first die is 2, there are 4: 1/6 \* 4/6 = 4/36 and so on. T/hus prob of first person losing the game is:

5/36 + 4/ 36 + 3/36 + 2/36 + 1/36 = 15/36 = 5/12

don't know how to generalize it to n faces.