How to answer 9-choice questions with Hexagon no.7

Ricchan

I’ve come across a very interesting scheme how to use Ricchan’s Hexagon no.7, a six sided pencil, for answering 9-answer-choice questions. Not surprisingly, it has been created by a Japanese so I’ll translate it for you, though you might understand the pattern from the answer table. It’s quite simple but very interesting. And actually, I think it might even be useful but I can’t think of any purpose now.

Ricchan

Ricchaaaaaan! “Hexagon no.7 can’t be used for a 9-choice question test orz” That’s what you think, isn’t it?

But rest assured!

There’s a good way how to answer 9-choice question with a six sided pencil.

Listen…

Ricchan

You have to roll the pencil twice!

If you roll the pencil once, you get 6 possibilities. That way, you cannot answer 9-choice questions. But! If you roll the pencil twice, you get 36 possibilities. 6 x 6 = 36 possibilities. And 36 = 9 x 4; therefore, if you roll the pencil, it will be possible to derive answers for 9-choice questions.

Ricchan

For example…

If the first throw is (1 or 2) [3 or 4] {5 or 6}, the answers may be (1, 2 or 3) [4, 5 or 6] {7, 8 or 9}.

If the second throw is (1 or 2) [3 or 4] {5 or 6}, the answers may be (1, 4 or 7) [2, 5 or 8] {3, 6 or 9}.

The combination of both throws will lead you to one of the nine answer choices.

If you put it into a table, it’ll look this way (see below).

The table in the picture above

If you do it thus, each answer choice, 1 through 9, will have the same probability: 4/36 = 1/9.

Ricchan

In case you roll the Hexagon no.7 once, the accuracy is 60%. Therefore, the result of a simple thought process is that the accuracy in case of 2 throws is 60% x 60% = 36%. If you chose one of the nine answers by intuition, the probability of the answer being correct would be 1/9 ≃ 11%. Therefore, the former way you get 3 times the feeling of security.

If you sit a 100 point exam and answer half of the answers with your own capabilities and the other half using the Hexagon no.7, the statistically expected result is 50 + 50 x 36% = 68 pts. If you answered 60 percent on your own, the expected result is 60 + 40 x 36% = 74.4 pts.

With this, you’ll pass and get accepted, won’t you?

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Simple use of combinatorics and probability theory. Quite interesting, isn’t it?

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