Recitation 2 Substituting in the boundary conditions Ro =0 and Rw= l gives two linear equa tions 1=a+b The solution to this system is a =0, b=1/. Therefore, the solution to the recur- fence Rn=n/w (c) So you know the probability that Stencil falls off the right side. Can you quickly deduce the probability that he will . falls off the left side?. lives on forever? g at pi hich is(w-n)/ This is bad news. The probability that Stencil eventually falls off one cliff or the other Is Theres no hope! The probability that he hops around on the island forever is zero (d) Now let's go back to the original probler m, where Stencil is 1 inch from the left edge of an infinite plateau. What is the probability that he lives on forever? Solution. The probability that he eventually falls into the sea is 1 lim Our little friend is doomed!Recitation 23 3 Substituting in the boundary conditions R0 = 0 and Rw = 1 gives two linear equa￾tions: 0 = a 1 = a + bw The solution to this system is a = 0, b = 1/w. Therefore, the solution to the recur￾rence is: Rn = n/w (c) So you know the probability that Stencil falls off the right side. Can you quickly deduce the probability that he will . . .falls off the left side? . . . lives on forever? Solution. We exploit the symmetry of the problem: the probability that he falls off the left side starting at position n is the same as the probability that he falls of the right side starting at position w − n, which is (w − n)/n. This is bad news. The probability that Stencil eventually falls off one cliff or the other is: n + w − n = 1 w w There’s no hope! The probability that he hops around on the island forever is zero. (d) Now let’s go back to the original problem, where Stencil is 1 inch from the left edge of an infinite plateau. What is the probability that he lives on forever? Solution. The probability that he eventually falls into the sea is: w − 1 lim = 1 w→∞ w Our little friend is doomed!