### Brain Teasers

# Terminate Average

Math
Math brain teasers require computations to solve.

There is a scoring system in which the highest is 4, and the lowest is 1, and you may only vote whole numbers.

Imagine that six people voted before you, and they all voted the same thing, however you do not know what they rated. You vote seventh, and your two friends behind you would vote whatever you did.

If you are the one in the end tallying the scores, and you want the average to be a simple, terminating decimal number, what would your best bet to vote for be?

Imagine that six people voted before you, and they all voted the same thing, however you do not know what they rated. You vote seventh, and your two friends behind you would vote whatever you did.

If you are the one in the end tallying the scores, and you want the average to be a simple, terminating decimal number, what would your best bet to vote for be?

### Answer

The format of the answers will be...[Previous 6], [Your Choice], [Average of all 9]

1, 1, 1

1, 2, 1.3(cont)

1, 3, 1.6(cont)

1, 4, 2

2, 1, 1.6(cont)

2, 2, 2

2, 3, 2.3(cont)

2, 4, 2.6(cont)

3, 1, 2.3(cont)

3, 2, 2.6(cont)

3, 3, 3

3, 4, 3.3(cont)

4, 1, 3

4, 2, 3.3(cont)

4, 3, 3.6(cont)

4, 4, 4

If you choose 1, then you have a 2/4 chance of having a terminating end average.

If you choose 2, then you will have a 1/4 chance of having a terminating average.

If you choose 3, then you will have a 1/4 chance of having a terminating average.

If you choose 4, then you will have a 2/4 chance of having a terminating average.

Your best bets are choosing 1 or 4. Now, you can still vote based on whether you liked it or not. If you did, then you choose 4. If you didn't, choose 1. If you choose 1 or 4, it won't have a difference on whether the average is terminating or not.

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## Comments

A simple one, though it involved a few calculations. I got the same answer, except that I wrote down the total of votes of each of the Group of 6 and Group of 3 for the 4 different vote selections, adding these Group totals (resulting in 16 combinations) and then dividing each of the 16 combinations by 9.

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