Suppose three different numbers are chosen independently at random. What is the probability that the third lies between the first two?
When the third number is chosen it
appears to me to have an equal possibility of being the largest number, the
lowest number or the number which is of a value between the first two. It
therefore seems to me that it has a 2 to 1 against chance of being the number
which lies between the first two chosen. That, it seems to me, is the
Not written in mathematical notation, I
know, and you write of probability rather than possibility, but makes sense to
Very stimulating and I look forward to
seeing the solution.
Yes that’s right. If the numbers in the order they are chosen are a, b, c then taking them in order of size there are six equally likely possibilities
abc, acb, bac, bca, cab, cba
Of these six the last choice, c, is in the middle twice. So the probability we are after is 2/6 = 1/3.