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# Please explain thoroughly so I could understand the procedures.

+1
1993
1
+1796

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.

Apr 8, 2015

#1
+27529
+22

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)

$${\mathtt{n}} = {\frac{{log}_{10}\left({\mathtt{0.1}}\right)}{{log}_{10}\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{12.629\: \!253\: \!136\: \!513\: \!339\: \!9}}$$

So you must have at least 13 dice.

.

Apr 8, 2015

#1
+27529
+22

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)

$${\mathtt{n}} = {\frac{{log}_{10}\left({\mathtt{0.1}}\right)}{{log}_{10}\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{12.629\: \!253\: \!136\: \!513\: \!339\: \!9}}$$

So you must have at least 13 dice.

.

Alan Apr 8, 2015