Two people: A and B play a board game with a fair six sided die. They step with their figures as many as they rolled with the die.
(a) What is the probability that A will step on the $n$th place?
(b) They play a five rounds game. They have already rolled two times and as a result A is $m=2$ steps ahed of B. What is the probability that they will stand on the same place at the end of the game? (A wins if his figure ahed of B's figure and vice versa)
(c) With the same initial situation discussed in (b) what is the probability that B is going to win? What is the probability that B is going to win if $m=3$?
(d) Playing with two dice, how can we change the rules to make a martingale process?
So far I've just started the (a)... the outcome of the rolled numbers can be: 1,2,3,4,5,6 with the same $1/6$ probability. I can get to the $n$th place with 1,2,3,...,n steps. To get to the first place, I can do it in only one case (if i roll a $1$ with the die) with probability $1/6$, to get to the second place I can roll two 1s OR I can roll one 2 with probability $1+(1/6)^2$. To get to the 3rd place I can roll a 3 OR I can roll a 2 and a 1 OR I can roll three 1s. The probability here is $1/6+2*(1/6)^2+(1/6)^3$ ... the coefficients follow a pattern, they are the binomial coefficients...these events are disjoint so I can sum them up... Hence, the probability getting to the $n$th place is $\sum_{k=0}^{n-1}\binom{n}{k}(\frac{1}{6})^{k+1}$ ... I think its working till the $6$th place (its not a surprise), but how can I continue? I start to realize it has some connection with Markov chains, and transition matrices... I don't even dare to start (b),(c),(d).