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The probability of getting a 4 on one roll of the die is 1/6.

The probability of having to roll the die twice to get a 4 is 5/6×1/6=5/36, because there is a 5/6 probability of failing to get a 4 on the first roll and a 1/6 probability of getting 4 on the second roll.

Failing to get 4 on 10 rolls is (5/6)¹⁰ but getting 4 on the next roll is (5/6)¹⁰(1/6)=5¹⁰/6¹¹=0.02692 approx. Let P=5¹⁰/6¹¹.

Not getting 4 until the 12th roll is (5/6)¹¹(1/6)=0.02243=P×5/6, and so on. We need to sum these probabilities from 11 to infinity:

P(1+5/6+(5/6)²+(5/6)³+...)=P/(1-(5/6))=6P=(5/6)¹⁰=0.1615 approx, 16.15%.

by Top Rated User (680k points)

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