First consider the total number of paths, including those crossing the x axis, which is 50 choose 101 or 101!/(50!*51!). I will this (101:50).

Next consider the number of paths which never touch the x axis, except at the beginning. Another way to express this is the number of paths from (1,1) to (101,1) minus the number of those ways that every touch the x axis. Remember the first vote must be for Richard and the path must go to (1,1).

The number of paths from (1,1) to (101,1) that ever touch the x axis is the same number of paths that go from (1,-1) to (101,1). This is the hardest step of this problem to explain. The reason is that any path from (1,-1) to (101,1) has a 1:1 counterpart of paths from (1,1) to (101,1) that touch the x axis at least once. To find the counterpart for any given path just reflect the path from (1,-1) to the point of first contact with the x axis over the x axis and keep the rest of the path the same.

So the number of paths from (1,1) to (101,1) that never touch the x axis is (100:50)-(100:49), where (100:50) is the number of paths from (1,1) to (101,1) and (100:49) is the number of paths from (1,-1) to (101,1). Remember that (100:50) is the number of ways to have a 50/50 tie and (100:49) is the number of ways to have a specific person win by two by a vote of 51 to 49.

(100:50)-(100:49) =
100!/(50!*50!) - (100!/(49!*51!) =
(100!*51)/(51!*50!) - (100!*50)/(50!*51!) =
100!/(50!*51!) =
101!/(50!*51!*101) =
(101:50)/101 .

Remember that the total number of possible paths from (0,0) to (101,1) was (101:50) from the second paragraph. The final answer is the number of valid paths divided by the number of total paths =

(101:50)/101 / (101:50) = 1/101.

In general the probability that a path goes from (0,0) to (x,y) without ever touching the x axis after (0,0) is y/x.