Randy takes a walk using the following rules. First, he tosses a fair coin. If it falls heads, he walks 10 feet north. If it falls tails, he walks 10 feet south. He repeats this process every 10 feet and thus executes a "random walk." What is the probability that after 80 feet of walking, he will end up exactly 20 feet from the starting point?

Question

>Randy takes a walk using the following rules. First, he tosses a fair coin. If it falls heads, he walks 10 feet north. If it falls tails, he walks 10 feet south. He repeats this process every 10 feet and thus executes a "random walk." What is the probability that after 80 feet of walking, he will end up exactly 20 feet from the starting point?

anonymous · Answer

To end up exactly 20 feet away from starting point, he must go 50ft north and 30 ft south. OR 50ft south and 30ft north. 

So he must toss heads and tails in the number of 5 and 3. 

If he toss the coin 8 times, the combination of heads and tails he can get are:
(H for heads, T for tails)
8H 0T
7H 1T
6H 2T
5H 3T
4H 4T
3H 5T
2H 6T
1H 7T
0H 8T
(note that it does not matter the order in which he gets them)

Total: 9 combinations.

So the probability is 2/9 = 22.2%