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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.

Feb 1, 2018

#1
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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.

Feb 1, 2018
edited by Guest  Feb 1, 2018
#2
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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"

Feb 1, 2018
#3
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Ok, Thank You!

Guest Feb 1, 2018