My son is studying for his Ohio 3rd Grade Achievement Assessment test. He brought home this practice test and I'm confused. Take a look:

Basically, the test is asking which number will most likely come up next and gives three choices. My kid chose the sucker's bet, the one Vegas hopes you choose. The number with the least amount of roll HAS to come up next... right?

The correct answer is that the die is weighted and that the number six is more likely to come up.
Either way, this is wrong.

I think perhaps this is actually testing the parents to see who brings it up to the teacher's attention.

## 6 comments:

Really hoping the test author was beaten with a stats textbook and then fired.

Independent events people - Google it!

Oh my, two graduate degrees and I would still fail a 3rd grade standardized test!

What? This may be the worst question I've ever seen on one of these tests, and I've seen some awful ones.

Below I've worked out the problem for those of you who spent your high school years at home alone with your math books instead of out on dates.

The results, though, are pretty straightforward. In a sequence of rolling a fair die 25 times, the odds of getting exactly 8 sixes is about 3%. Unless you come from a land with lots of crooked dice, I think this isn't much evidence of foul play.

It's worse than that, though. How many third graders will actually think about the probability of getting 8 sixes in 25 rolls? No, most who get it "right" will do so for the wrong reason. They'll go the opposite of Holy Juan, thinking that because six has come up so much it must be lucky. That's as bad as picking three because it's due.

Suppose the question were "Jordan wished real hard for sixes and this was her result. If she keeps wishing real hard for sixes, what's the most likely number to come up next?"

I think this is a horrible problem, not because crooked dice are impossible, but because this problem actually fosters those easily-built misconceptions about "lucky" and "unlucky" numbers.

Math Alert! I'm going to use this opportunity to get all geeky about math. Forgive me.

The odds of rolling a certain number N of sixes in 25 rolls of a fair die is a bit tricky for third graders to calculate. Here's what the formula looks like:

[(1/6)^N x (5/6)^(25-N)]x 25!/N!(25-N)!

The first term is just the "chance of rolling a six" term. The next is just the "chance of rolling something else" term. The stuff that comes after that just says that we don't care about the order - whether you rolled N straight sixes or the sixes were all spread out doesn't matter, so there are various ways for the sixes to be found in the list.

OK, when you insert 8 for N, you come up with 0.029, or around a 3% chance. So there's a 3% chance of getting 8 sixes in a list of 25 rolls of a fair die. That's not much to conclude that this die is weighted.

By the way, it's actually worse than this, since there's no reason ahead of time to suspect the six of being the weighted face. The odds of any number coming up 8 times is actually 6 times this result, or 18%. So if I come across a random list of 25 die rolls, around 18% of the time I'd expect to find one of the numbers represented 8 times, if the die was fair. Since I think the null assumption has to be that the die is fair, a result that we expect 1 time in 5 isn't much evidence to the contrary.

Holy Shit! When I was in the third grade (1961) we were just beginning to memorize the multiplication tables. Which I failed at. What the hell is this?

I miss you Steve Whitt.

Aren't all of these choices incorrect? Each number still has a 1/6 chance of being rolled. No matter what happened before, the odds are still 1/6 for all numbers.

I'm confused.

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