problem 66

2.6.7 problem 66

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}) = 99 \sum i = 94 P (S_{l - 1, i}) P (X_{100 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 94 P (S_{l - 1, i})

P (S_{l, 101}) = 99 \sum i = 95 P (S_{l - 1, i}) P (X_{101 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 95 P (S_{l - 1, i})

P (S_{l, 102}) = 99 \sum i = 96 P (S_{l - 1, i}) P (X_{102 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 96 P (S_{l - 1, i})

P (S_{l, 103}) = 99 \sum i = 97 P (S_{l - 1, i}) P (X_{103 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 97 P (S_{l - 1, i})

P (S_{l, 104}) = 99 \sum i = 98 P (S_{l - 1, i}) P (X_{104 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 98 P (S_{l - 1, i})

P (S_{l, 105}) = 99 \sum i = 99 P (S_{l - 1, i}) P (X_{105 - i} ∣ ∣ ∣ ∣ S_{l - 1, i}) = \frac{1}{6} 99 \sum i = 99 P (S_{l - 1, i})

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

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