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
re(2): regarding posted solution by Cory Taylor)
As you have mentioned Cory in your comment that as events pass the probabilities change and that is what I have shown in my proof giving the formula for the probability of the occurrence of the highest number in a slip whose position is greater than 'k' depends on 'k' thus showing that the probabilities change, but as mentioned in the solution, that each of the first (i – 1) slips is equally likely to be the one with the highest number on them.
Then we obtain that: P{W | B(i)} = {k/(i – 1)}, if i is greater than k.
Also, we have the following results:
P(W) = (k/N)[{Summation from i = (k+1),(k+2),….,N}1/(i – 1)]
or, P(W) = (k/N)[{Summation from j = k,(k+1),(k+2),….,(N – 1)} (1/j)]
re(3): regarding posted solution
