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Exercise 2.66

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Let $S_{i, k}$ be the event that sum after $i$ rolls of the die is $k$ . Let $l$ denote the roll after which $k \geq 100$ . Let $X_{i}$ be the event that the die lands on $i$ .

P (S_{l, 100}) = \sum_{i = 94}^{99} P (S_{l - 1, i}) P (X_{100 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 94}^{99} P (S_{l - 1, i})

P (S_{l, 101}) = \sum_{i = 95}^{99} P (S_{l - 1, i}) P (X_{101 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 95}^{99} P (S_{l - 1, i})

P (S_{l, 102}) = \sum_{i = 96}^{99} P (S_{l - 1, i}) P (X_{102 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 96}^{99} P (S_{l - 1, i})

P (S_{l, 103}) = \sum_{i = 97}^{99} P (S_{l - 1, i}) P (X_{103 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 97}^{99} P (S_{l - 1, i})

P (S_{l, 104}) = \sum_{i = 98}^{99} P (S_{l - 1, i}) P (X_{104 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 98}^{99} P (S_{l - 1, i})

P (S_{l, 105}) = \sum_{i = 99}^{99} P (S_{l - 1, i}) P (X_{105 - i} | S_{l - 1, i}) = \frac{1}{6} \sum_{i = 99}^{99} P (S_{l - 1, i})

Thus, $S_{l, 100}$ is the most likely.

fifthist

2021-12-05 00:00

Exercise 2.66

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