- 06 Apr '16 16:49

Note: There are links to discussion of this poser at*Originally posted by JS357***If you choose one of the following answers to this question at random, what are the chances you will be correct?**

A) 25%

B) 50%

C) 33.3333...%

D) 25%

Please explain why your answer is correct.

http://blog.tanyakhovanova.com/2011/11/a-probabilistic-paradox/ - 07 Apr '16 17:11 / 3 editsSince the percent values listed aren't relevant, I get rid of the confusion they offer by substitution like this:

a) x

b) y

c) z

d) x

..actual answer

....x..y..z

.a 1 0 0 = 1/3

.b 0 1 0 = 1/3

.c 0 0 1 = 1/3

.d 1 0 0 = 1/3

Okay, that's a matrix, with the columns labeled (x,y,z) and the rows (a,b,c,d).

The first column assumes x is the correct answer, and each row under that column shows what happens when x is correct. Do the same then for each column.

Total each row. They all add up to 1 chance out of 3 tries to guess the correct answer.

I guess the answer is 1/3.

I keep thinking that because one answer appears twice, that should affect the odds. But it seems that no matter which of the three possible answers is correct, and no matter which of the four possible choices I select, my chances of choosing the correct answer are 1 out of 3. Probability is not very intuitive.

Now I wanna go see that discussion link. - 09 Apr '16 23:43 / 2 edits

It depends how we "choose at random".*Originally posted by JS357***If you choose one of the following answers to this question at random, what are the chances you will be correct?**

A) 25%

B) 50%

C) 33.3333...%

D) 25%

Please explain why your answer is correct.

If we randomly choose from (A, B, C, D) then there is no solution.

However if we randomly choose from the available answers

(25%, 33.3%, 50% ) then the answer is C. 33.3%

So in a written test I'd go for 'C'

On a chess forum I'd plump for