What is the expected rolls of a die to see all sides at least once?

Probability Level pending

Suppose you have a fair die. (Most of them are these days.)

You roll it once and observe a 3. You roll it again and observe a 1. You roll it yet again and get a 4. In three throws you’ve observed 3 of the possible sides.

What if you keep going?

What are the expected rolls that it takes to see all six sides at least once rounded off to the nearest integer?

12 13 14 15

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1 solution

Pop Wong
Jan 2, 2021

If a coin that flips heads with probability p , q = 1 p p, q = 1-p , then the expected trials to get a head will be

E(trials to get head) = p + 2 q p + 3 q 2 p + 4 q 3 p + . . . = p [ ( 1 + q + q 2 + q 3 + . . . ) + ( q + q 2 + q 3 + . . . ) + ( q 2 + q 3 + . . . ) + . . . ] = p [ 1 1 q + q 1 q + q 2 1 q + . . . ] = p 1 q [ 1 + q + q 2 + . . . . ] = 1 1 q = 1 p \begin{aligned} \text{E(trials to get head)} &= p + 2qp + 3q^2p + 4q^3p + ...\\ &= p [ ( 1+q+q^2+ q^3+...) + ( q + q^2+q^3 +...) + (q^2+q^3+...) + ... ] \\ &= p [ \cfrac{1}{1-q} + \cfrac{q}{1-q} + \cfrac{q^2}{1-q} + ...]\\ &= \cfrac{p}{1-q} [ 1 + q +q^2 + ....] \\ &= \cfrac{1}{1-q} \\ &= \cfrac{1}{p} \end{aligned}

Let X i X_i be the state that total i i faces have not been rolled out yet, the probability of getting any i i faces i 6 \cfrac{i}{6}

E ( X 6 + X 5 + X 4 + X 3 + X 2 + X 1 ) = E ( X 6 ) + E ( X 5 ) + E ( X 4 ) + E ( X 3 ) + E ( X 2 ) + E ( X 1 ) = 6 6 + 6 5 + 6 4 + 6 3 + 6 2 + 6 1 = 6 [ 10 + 12 + 15 + 20 + 30 + 60 ] 60 = 147 10 = 14.7 15 \begin{aligned} E(X_6+X_5+X_4+X_3+X_2+X_1) &= E(X_6) +E(X_5) +E(X_4) +E(X_3) +E(X_2) +E(X_1) \\ &=\cfrac{6}{6} + \cfrac{6}{5}+ \cfrac{6}{4}+\cfrac{6}{3}+\cfrac{6}{2}+\cfrac{6}{1} \\ &=6\cfrac{[10+12+15+20+30+60]}{60} \\ &=\cfrac{147}{10} \\ &= 14.7 \approx 15 \end{aligned}

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