brainteaser!

[jimmiebarton]

"Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale. What's the fewest number of times you have to use the scale to find the heavier ball?"

don't google the answer and be a spoiler

disclaimer: i got answer in just under 20 min

 

[Baron]

1?

 

[jimmiebarton]

1?

no. try again!

explain your reasoning ?

 

[heypa]

If you are lucky in two. If not then 3

 

[jimmiebarton]

If you are lucky in two. If not then 3

yep here...
http://www.businessinsider.com/how-to-solve-the-classic-heavier-ball-interview-brainteaser-2015-7

 

[heypa]

I made a mistake. As soon as i posted the answer I knew luck or not only two were required. Too lazy to correct it . Total time about a minute.

 

[Max E.]

1?

I agree, theoretically you could fluke out and grab the heavier ball first, so the fewest times it could take is 1. Based on the wording of the question the fewest number of times you could find the heavier ball in is 1.

 

[jimmiebarton]

I agree, theoretically you could fluke out and grab the heavier ball first, so the fewest times it could take is 1. Based on the wording of the question the fewest number of times you could find the heavier ball in is 1.

No, that's silly Luck is not a reliable method of discovery.

 

[heypa]

Baron is correct. The answer to the question is really 1. My answer is an engineers certainty answer.

 

[MoreLeverage]

For the minimizing the worst case to find it,

Log3(#balls), rounded up

Add 1 if you don't know the odd one is heavier or lighter.

I still like Baron's answer best, but you won't get lucky like that with the above algorithm.