We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
2484
1
avatar+1802 

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

Best Answer 

 #1
avatar+28134 
+24

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
avatar+28134 
+24
Best Answer

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

17 Online Users

avatar
avatar