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.

I had seen this question but all the answers needed to use logarithm on a calculator. Please show me if there is another way of doing this.

Guest Feb 1, 2018

#1**+1 **

So far I have gotten (5/6)^x = 0.1, using complementary counting. I used a Calculator to solve for x and I got 13 as the final answers since you can't roll a die a decimal times.

Guest Feb 1, 2018

edited by
Guest
Feb 1, 2018

#2**+1 **

I think this is correct...but I'd like to have someone else check my answer [ and make corrections, if necessary ]

Looking at it from the other poiint of view, we are asking...."What is the fewest number of dice I need to roll so that the chances of * not* rolling a "6" are ≤ 10% ??"

The chances of not rolling a "6" with one die = 5/6

So.....the chances of not rolling a "6" with two die = (5/6)^n

So.....we need to solve this :

(5/6)^n ≤ .10

We can make this an equality

(5/6)^n = .10

Unfortunately.....I don't see a way to avoid using logs !!! [though there may be one ]

Take the log of both sides

log (/5/6)^n = log .10 and we can write

n log (5/6) = log .10 divide both sides by log (5/6)

n = log (.10) / log (5/6) = 12. 62 ≈ 13

So.... we need 13 dice to have a 90% probability of rolling at least one "6"

CPhill Feb 1, 2018