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).
There is a best solution to the problem, but as in the previous post, I can't remember exactly what it is (although I have a better memory than my predecessor). There is a mathematical formula for picking the largest member is a set of known size, when following the bounds of this problem, and it gives a suprislingly high probability. The method is something like this (but not exactly);
let x=roundup(n/e), where e=2.7181281828...
view and disard the first x slips, making a note of the highest total seen so far.
View the remaining slips and accept the first one that beats the highest in the first set.
If my memory serves correctly, this will give a probability of around 30% to get the largest number.
This is, of course, not quite correct, as I can't verify that it is e you should be using to derive x, and that other details of the method could be in error, but the process does exist.
And further, because this strategy is for numbers in general (not distinct positive integers), this strategy could be bettered by the user, based on keeping track of the numbers as they come (for example, if the largest number encountered in the first x slips is still less than n (by more than 1), then just look for the first slip of n or greater. Also, coming to the second last slip, it would be a judgment decision on weather to accept this slip or the last one, which would be based on the users gleaned knowledge of the numbers used and possible numbers remaining.
partial solution
