A box is filled with 'N' slips of paper. On each slip of paper is written some positive integer (note that any positive integer may appear on the slips - not just the integers from 1 to 'N'). The integers do not necessarily appear in any sequence or pattern. Each of the slips has a different integer on it, so there is just one slip with the greatest integer.
A person who has no prior knowledge of which numbers appear on the slips - but who does know that there are 'N' slips - is to blindly pull slips from the box one by one. The person looks at each slip, then either agrees to accept that number (of Rupees) and quit or decides to go on and choose another slip.
Note that the person looks at each slip as he/she proceeds, and then decides whether to quit or to go on. That person can go forward, but cannot go back. If no choice is made by the time the 'N'th slip is reached, then the person must accept the number (of Rupees) on the 'N'th slip.
Does there " EXIST " a 'Best Strategy' for the person ? If " YES ", then what is that strategy ? (Here the term " Best Strategy" means that the person will get the greatest amount of Rupees).
(In reply to solution
The probability of getting the highest slip for a given r is then:
p(r) = ((n-r)/n)(1/(n-r))(r/r + r/(r+1) + ... + r/(n-1)) = (r/n)(1/r + 1/(r+1) + ... + 1/(n-1))
Unfortunately, the number of terms differs from one value of r to the next, so you can't differentiate with respect to r.
However, a program with evaluates p for various n and r, shows that the maximum r for a given n is always either n/e truncated or rounded to the next integer. It is not consistent in this regard. For example, for n=40, n/e = 14.72, and the maximum probability of success is for r=15 at which p = 0.375743. (at r=14, p=0.375693). However for n=135 n/e=49.66, but the maximum prob of success occurs at the truncated r=49, with p=0.370217, rather than at the rounded 50, where p=0.370214.
Posted by Charlie
on 2003-03-19 16:22:39