Let pN(n) be the probability that an N-step walk ends at position n. Some examples for...

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Let pN(n) be the probability that an N-step walk ends at position n. Some examples for small n: • Po(0) = 1 • p₁(−1) = p₁(1) = // • P2(−2) = P2₂(2) = 1; P2(0) = // • P3(−3) = µ3(3) = {}; P3(−1) = P3(1) = ²³ Generally, PN(n) depends on the number of ways of ending at position n. For example, where n = 3, there is one way to end at positions -3 and 3, and three ways to end at positions −1 and 1 (ways to reach position – 1: 0, −1, 0, −1, or 0, 1, 0, −1, or 0, −1, −2, −1). [As an aside, the number of N-step walks ending at position a - b is given by the binomial coefficients of (a + b). For N = 4, we have (a + b)² = aª + 4a³b + 6a²b² + 4ab³ + b4. The coefficient () gives us the number of ways of ending at a - b where a counts as an increment (to the right) in the walk and b as a decrement (to the left).] For this problem, write count_random_walks(N, nbr_tries) that returns a dictionary that maps n to the number of N-step walks that ended at n, out of a total of nbr_tries many walks. For example (note your result will probably vary slightly from this): counts = count_random_walks (3, 10000) for c, v in counts.items (): print (f'{c}: {v}') -3: 1227 3: 1251 -1: 3766 1: 3756 Let pN(n) be the probability that an N-step walk ends at position n. Some examples for small n: • Po(0) = 1 • p₁(−1) = p₁(1) = // • P2(−2) = P2₂(2) = 1; P2(0) = // • P3(−3) = µ3(3) = {}; P3(−1) = P3(1) = ²³ Generally, PN(n) depends on the number of ways of ending at position n. For example, where n = 3, there is one way to end at positions -3 and 3, and three ways to end at positions −1 and 1 (ways to reach position – 1: 0, −1, 0, −1, or 0, 1, 0, −1, or 0, −1, −2, −1). [As an aside, the number of N-step walks ending at position a - b is given by the binomial coefficients of (a + b). For N = 4, we have (a + b)² = aª + 4a³b + 6a²b² + 4ab³ + b4. The coefficient () gives us the number of ways of ending at a - b where a counts as an increment (to the right) in the walk and b as a decrement (to the left).] For this problem, write count_random_walks(N, nbr_tries) that returns a dictionary that maps n to the number of N-step walks that ended at n, out of a total of nbr_tries many walks. For example (note your result will probably vary slightly from this): counts = count_random_walks (3, 10000) for c, v in counts.items (): print (f'{c}: {v}') -3: 1227 3: 1251 -1: 3766 1: 3756

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