+1836 I have an unlimited supply of standard 6-sided dice. What's the fewest number of dice that I have to simultaneously roll to be at least 90% likely to roll at least one 6? You may use a calculator to help you with the computations if you like -- in fact you'll almost certainly want to -- but your final answer should be a positive integer, and you should explain how you got it.
+33654 Probability of no sixes in rolling n dice = (5/6)n
Hence the probability of at least one six is 1 - (5/6)n
For this to be 90% we must have 0.9 = 1 - (5/6)n
Subtract 0.9 from both sides and add (5/6)n to both sides: (5/6)n = 0.1
Take logs of both sides and use the fact that logab = b*loga
n*log(5/6) = log(0.1)
n =log(0.1)/log(5/6)
n=log10(0.1)log10(56)⇒n=12.6292531365133399
So you must have at least 13 dice.
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+33654 Probability of no sixes in rolling n dice = (5/6)n
Hence the probability of at least one six is 1 - (5/6)n
For this to be 90% we must have 0.9 = 1 - (5/6)n
Subtract 0.9 from both sides and add (5/6)n to both sides: (5/6)n = 0.1
Take logs of both sides and use the fact that logab = b*loga
n*log(5/6) = log(0.1)
n =log(0.1)/log(5/6)
n=log10(0.1)log10(56)⇒n=12.6292531365133399
So you must have at least 13 dice.
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