Wonderlic for All

[B1GTide]

BIG wins!

I never thought about splitting into three sets. Well done!

I will say that my first thought was 4 ball sets, then I realized that 3 was faster unless you got lucky with the 4.

 

[TexasBama]

That’s not the way the question read.

The goal is to identify the lightest ball, using a scale that compares the weights on two platforms...The scale doesn’t to determine how much ball weighs....only which side of the scale has the heavier load.

I stand by the premise: You could get lucky and ID the lightest ball with only one weighing. You could also gamble on the second round and simply pick one of the remaining four balls.

But if you weren’t dead lucky on the first round, and didn’t gamble on the second, you can do it with 100% certainty with no more than three weighings.

Rgr that. I didn’t see the question on my mobile. If you did groups of 3 like above them one weighing to eliminate 2 sets of 3s. Then second weighing you’d know. So min one max two

 

[4Q Basket Case]

Rgr that. I didn’t see the question on my mobile. If you did groups of 3 like above them one weighing to eliminate 2 sets of 3s. Then second weighing you’d know. So min one max two

Actually, as I think about it more, BIG’s solution gives up the possibility of a lucky outcome on the first round in order to get the certainty of two rounds.

Even so, from a risk management perspective, his solution is unquestionably the best.

My solution had (.1111 x 1) + (.8889 x 3) expected reps, or 2.7778.

His solution has a 100% chance of 2.0 reps. The only way you’d choose my solution is if you absolutely positively had to have a one-hit wonder.

 

[The Ols]

I would do it with sets of 3. It is possible to do it in two weighings with sets of 3.

Weigh two sets of 3. You know then which set of 3 has the lighter ball. It is either the lighter of the two sets on the scale, or the remaining set if those 2 sets are equal.

Now you have the set of 3 with the lighter ball - weigh two of the balls. The lighter ball is either the ball which weighs less, or is the remaining ball if the other two balls are of equal weight.

Very well done B1G!!!
*Edit...I would have said that too, but I ran out of time!!!

 

[NoNC4Tubs]

I 100% agree with that approach. They also asked me a "puzzle question" and I had to talk out my thoughts. Weird! It went something like this:

If you had 9 billiards exactly the same size and weight except one which was just slightly lighter...and an old scale (think scales of justice), what is the least amount of times you could use the scales to find the lighter ball?

Talk about being put on the spot!!!

Now everybody wants the answer, right???

I'm guessing two times. The first time narrows it down to five balls and then the second time leaves a ball out and if the scales balance, then the odd ball out is the Cue Ball.

 

[Obieone42]

I would do it with sets of 3. It is possible to do it in two weighings with sets of 3.

Weigh two sets of 3. You know then which set of 3 has the lighter ball. It is either the lighter of the two sets on the scale, or the remaining set if those 2 sets are equal.

Now you have the set of 3 with the lighter ball - weigh two of the balls. The lighter ball is either the ball which weighs less, or is the remaining ball if the other two balls are of equal weight.

Bingo!!! We have a winner! I didn't get it right in the interview, though.

 

[NoNC4Tubs]

Bingo!!! We have a winner! I didn't get it right in the interview, though.

Hmmmm.

How was my answer wrong?

 

[The Ols]

Hmmmm.

How was my answer wrong?

Yours is hopefully in two...B1G's IS in two...
Edit: Your answer is correct, your theory is not consistent in it's return...

 

[Obieone42]

Hmmmm.

How was my answer wrong?

You are correct, if you said two. His was just the first one I saw.