Expert Answer to "If you roll a single die, what is..."

Seelakant6vr

Answered question

2022-02-11

If you roll a single die, what is the expected number of rolls necessary to roll every number once?

stefjumnmt

Beginner2022-02-12Added 14 answers

P [all numbers thrown] = 1 - P [1, 2, 3, 4, 5, or 6 not thrown]

P[A or B or C or D or E or F]

= P [A] + P [B] + \dots + P [F] - P [A and B] - P [A and C] \dots . + P [A and B and C] + \dots

Here this is

P_{1} = 6 \cdot {(\frac{5}{6})}^{n} - 15 \cdot {(\frac{4}{6})}^{n} + 20 \cdot {(\frac{3}{6})}^{n} - 15 \cdot {(\frac{2}{6})}^{n} + 6 \cdot {(\frac{1}{6})}^{n}

P = P_{1} (n) - P_{1} (n - 1)

= 6 \cdot {(\frac{5}{6})}^{n - 1} (\frac{5}{6} - 1) - 15 \cdot {(\frac{4}{6})}^{n - 1} (\frac{4}{6} - 1) + \dots

= - {(\frac{5}{6})}^{n - 1} + 5 \cdot {(\frac{4}{6})}^{n - 1} - 10 \cdot {(\frac{3}{6})}^{n - 1} + 10 \cdot {(\frac{2}{6})}^{n - 1} - 5 \cdot {(\frac{1}{6})}^{n - 1}

The negative of this is our probability.

\sum n \cdot a^{n - 1} = \sum (\frac{d}{d a}) (a^{n})

= (\frac{d}{d a}) \sum a^{n} = (\frac{d}{d a}) (\frac{1}{1 - a}) = \frac{1}{{(1 - a)}^{2}}

\Rightarrow E [n] = \sum n \cdot P

[all numbers thrown after n throws]

= \sum n \cdot ({(\frac{5}{6})}^{n - 1} - 5 \cdot {(\frac{4}{6})}^{n - 1} + \dots

This is helpful 0 Flag

rossegues03b

Beginner2022-02-13Added 11 answers

Think of it like six mini-games. For each game, we roll the die until we roll a number that hasnt

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