# Stochastic

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Unformatted text preview: ple, we consider: Example 1.44. Roulette. If we bet \$1 on red on a roulette wheel with 18 red, 18 black, and 2 green (0 and 00) holes, we win \$1 with probability 18/38 = 0.4737 and lose \$1 with probability 20/38. Suppose we bring \$50 to the casino with the hope of reaching \$100 before going bankrupt. What is the probability we will succeed? Here ✓ = q/p = 20/18, so (1.22) implies P50 (V100 < V0 ) = Using (20/18) 50 20 50 18 20 100 18 1 1 = 194, we have P50 (V100 < V0 ) = 194 1 1 = = 0.005128 (194)2 1 194 + 1 Now let’s turn things around and look at the game from the viewpoint of the casino, i.e., p = 20/38. Suppose that the casino starts with the rather modest capital of x = 100. (1.22) implies that the probability they will reach N before going bankrupt is (9/10)100 1 (9/10)N 1 If we let N ! 1, (9/10)N ! 0 so the answer converges to 1 (9/10)100 = 1 2.656 ⇥ 10 5 If we increase the capital to \$200 then the failure probability is squared, since to become bankrupt we must ﬁrst lose \$100 and then lose our second \$100. In this case the failure probability is incredibly small: (2.656 ⇥ 10 5 )2 = 7.055 ⇥ 10 10 . From the last analysis we see that if p > 1/2, q/p < 1 and letting N ! 1 in (1.22)...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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