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Let us fix a value ‘k’, where k lies between 0 (inclusive) and N and consider the strategy that rejects the first k slips and then accepts the slip thereafter that has a number greater than the maximum among the ones observed on each of the first k slips that have been rejected. Let W be the event that the slip with the highest number is chosen and the person wins when this strategy is employed. Also, for i = 1,2,….,N, let B(i) be the event that the slip with the highest number is in position ‘i’. Exactly one of the events B(i), i = 1,2,….,N, must occur and so, upon conditioning on which event does occur, we obtain:
P(W) = [Summation from i = 1,2,….,N] P{W | B(i)}P{B(i)}
or, P(W) = (1/N)[Summation from i = 1,2,….,N] P{W | B(i)}
The final equality follows because the slip with the highest number is equally likely to be in any of the N positions and so, P{B(i)} = 1/N, for all i. Now since we are employing the strategy of rejecting the first k slips, it follows that there is no chance of obtaining the slip with the highest number if it is among the first k.
Consequently, P{W | B(i)} = 0, if i is less than or equal to k.
On the other hand, if the slip with the highest number is in position i, where i is greater than k, then this slip will be selected if the highest number on the first (i – 1) slips is among the first k.
However, conditional on the slip with the highest number being in position i, it is easy to see that all possible orderings of the other slips remain equally likely, which implies that each of the first (i – 1) slips is equally likely to be the one with the highest number on them.
Hence we obtain that: P{W | B(i)} = {k/(i – 1)}, if i is greater than k.
From the previous results we have:
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)]
Now, if we select (k/N) to be equal to (1/e), where e = 2.718281828459045
(approximately), then we have P(W) = 1/e
That is, the strategy that lets approximately the fraction (1/e) of all the slips go by and then accept the first one thereafter that is the greater than the maximum number yet observed (among the first k slips that have been rejected) has probability equal to (1/e) which is approximately equal to 0.37 of obtaining the slip with the highest number.
So, the best strategy is to reject the first k (= N/e) slips and then accept the slip thereafter which will bear the number greater than the maximum number observed in the first k slips rejected. |
