# What are the odds for the student to be right? [duplicate]

If a student answers this questions randomly, without thinking, what are the odds he will get the question right ?

• 25% chances
• 1 in 2 chances
• 75% chances
• 1 in 4 chances

EDIT : This question is not a duplicate since the choices differ.

## marked as duplicate by Oray, boboquack, Mike Earnest, Rubio♦Sep 18 '17 at 1:16

• Dear Arthur Attout, perhaps this puzzle would get re-opened if you edit it so say that you have a non-paradoxical solution in mind that doesn't reach beyond what is stated. This might have been misapprehended as being a certain Zen-like puzzle that has made the rounds. – humn Sep 18 '17 at 16:53

The odds for a random choice to be correct are...

...0 to 1.   In other words it is impossible because none of the choices can be correct as each results in a paradox.   After all, the puzzle does not state that any offered choice is actually correct.

About the choice “25%”:

Were 25% correct then the choice “1 in 4” would also be correct. That would make 2 correct choices out of 4, so the probability would actually be 50%, not 25%.

About the choice “1 in 2”:

Were this choice correct, a random guess would choose this 1⁄4 of the time, giving a probability of 1 in 4, not 1 in 2.

About the choice “75%”:

As with the choice “1 in 2,” were this choice correct, the probability would actually be 25%, not 75%.

About the choice “1 in 4”:

This is the same as the choice “25%,” so the probability cannot be 1 in 4.

Strictly speaking:

It depends on the students probability distribution. Furthermore the question asker needs to know this distribution in order to correctly mark the submission! As it stands the question asker seems to think that they know the students probability distribution is a uniform one yet they have not stated this in the question.

A uniform distribution will (as pointed out by humn) result in them never getting the question correct.

With a non-uniform distribution, as known by the question asker, the results may differ. For example if their distribution were to pick the options with chances $\frac{1}{12}, \frac{1}{12}, \frac{3}{4}, \frac{1}{12}$ respectively (or any other distribution with $\frac{3}{4}$ weight on the "75% chance" option) then they would indeed be correct 75% of the time. Note that even though skewed, this is still, technically, random.

The chance is zero, under the assumption that there is a probability. If the probability were 25% / 50% / 75%, then one / two / three answers would be correct, but looking at the answers, we would find that two / one / one answers are correct, so 25%, 50%, 75% are all wrong answers.

This means none of the answers is correct, which means the probability of getting the right one is zero, which is the only solution consistent with the answers given.