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Expected Value


New member
Jan 19, 2011
In a multiple-choice test, each question has five options. Students will get 5 points for each correct answer; lose 2 point for each incorrect answer; and receive no points for unanswered questions. A student does not know the correct answer for one question. Is it to her advantage or disadvantage to guess an answer? Show your calculations for expected value.

My work so far:
Failure to pick the correct answer is 4 to 5.
Success in picking the correct answer is 4 to 5.

Odds against:
P(not picking the correct answer)/P(picking the correct answer)
4/5 divided by 4/5=
4/5 * 5/4=

So my guess would be that it wouldn't matter either way. Is this correct?


Staff member
Apr 12, 2005
Generally, the point of such schemes is to make the examied indifferent toward guessing. This neither encourages nor discourages guessing.

I might go about it this way, since you entitled the Post "Expected Value"

E[Not Guessing] = 0*1.00 = 0.00

E[Guessing] = 5(0.20) + (-2)(0.80) = -0.60

It appears to be advantageous to avoid guessing.

An interesting question always comes up. What if the examined can rule out one response?

E[Not Guessing] = 0*1.00 = 0.00

E[Guessing] = 5(0.25) + (-2)(0.75) = -0.25 -- Nope. Still bad.

Maybe we can rule out 2?

E[Not Guessing] = 0*1.00 = 0.00

E[Guessing] = 5(0.33) + (-2)(0.67) = +0.31 -- Now, we're talking!