Question: Show that R(1)=1 Consider generalizing in the following way: N customers line up for their sandwiches, and you are the last customerin line; the first customer grabs a sandwich at random, and every customer afterwards attempts to take their ownsandwich, or will grab one at random. Define R(N) to be the probability that you get your own sandwich in thissituation.

We will assume that there is equal probability for other customers to grab a sandwich and not to grab a sandwich. And there are enough sandwich for all the customers.As the chances of any other customer to take and not to take a sandwich is same, the probability of a customer (not me) to take a sandwich is : 1/2 = 0.5 And as the choice of one customer to grab a sandwich is not dependent on the choice of any other customer, the events that two or more that two customers grabbing/not grabbing sandwiches are disjoint. Hence if the event that the kth customer grabs a sandwich is Ak then : p(Ai ∩ Aj) = p(Ai) p(Aj) where p(A) = probability of event A Now as the probability of any customer (except me) to grab a sandwich is 0.5 , we have : p(Ak) = 0.5 for all : k = 1,2,3,...,(N-1) Next as I will surely grab a sandwich , in my case the probability is : p(AN) = 1 Thus we get : R(N) = p(A1∩ A2∩...∩AN-1∩AN) ➞ R(N) = p(A1) p(A2) ••• p(AN-1) p(AN) ➞ R(N) = (0.5) (0.5) ••• (0.5) • 1 ➞ R(N) = (0.5)N-1 ➞ R(1) = (0.5)1-1 = (0.5)0 = 1 ∴ R(1) = 1