Copyright ©2008 by Paul Niquette All rights reserved. 

ere is a simple version of the bar challenge with N = 3. Just for fun, consider...
He takes the first slip out of the bowl, looks at it, and is immediately confronted with the question, "Is that first number the largest of the three?" A yes answer would be a pure guess, with a oneinthree chance of being correct, and a no answer would have two out of three chances of being correct. Thus, YDB will surely answer no, rejecting the first number. If you then disclose that the first number happens to be the largest of the three, YDB loses the bet. Otherwise, if the first number happens not to be the largest, YDB does not win the bet but instead gets to see the number on a second slip chosen at random. When that happens, YDB must answer the question, "Is that second number the largest of the three?" Now, if he sees that second number happens to be smaller than the first number, he will obvously answer no, which is not a pure guess. He then knows for sure that he will get to see the third number, which, being the last, must be the largest, giving YDB the win. If the second number happens to be larger than the first number, YDB continues to confront the same question, "Is that second number the largest of the three?" We will show that a yes answer has three out of four chances of being correct, while no, rejecting the second number, will give YDB a one out of four chance for winning. ophisticated
solvers will observe that for the bar challenge,
there are
N!
(Nfactorial)
arrangements, or permutations  sequences of disclosures. For N
= 3, there are a total of
N! = 6 permutations as follows:
321, 312, 231, 213, 123, 132. Let us now use all six sequences
for disclosure tabulated below to develop a Strategic
Policy on behalf of YDB...
Accordingly, in the bar challenge with N = 3, on average YDB will WIN three out of six wagers, assuring "even money"  a onetoone wager in the Land of Odds. Without invoking ESP at all, YDB's probability for winning p(WIN) = 3/6 = 1/2, which is significantly higher the 1/3 for a pure guess. The outcomes suggest a simple policy for YDB: Always answer No for the first slip, then, if the second number is larger than the first always answer Yes, accepting the second number. Note: YDB has no need to see the third number. By using, say, a coinflip to make the select/reject decision for Slip #1, YDB would expect to have a 5050 chance to win the 321 and 312 permutations but also a 5050 chance to lose 231, 213, 123, 132 permutations, never having an opportunity to see the number on Slip #2.It is easy to carry out the analysis of YDB's prospects for N = 4, wherein N! = 24 permutations, as follows: 4321 ~ LOSE,
4312 ~ LOSE,
4231 ~ LOSE,
4213 ~ LOSE,
4132 ~ LOSE,
4123 ~ LOSE
Using an extended
Strategic
Policy, the number of winning wagers for YDB
turns out to be 11 out of 24, which may not be an even bet, but still better
than 1/4 for the Pure Guess Policy.
You are invited to try your hand at extending the Strategic
Policy and confirming that for N
= 5, YDB wins
50 out of 120 permutations (the table for N
= 5 on this page is available for your convenience). As expected,
the larger the value of N the smaller will be YDB's
probability of winning p(WIN).
Those factorials do get huge fast, making the analysis of each permutation quite tedious  eventually impossible for the Challengee vs Challenger puzzle, inasmuch as... For half the standard deck of cards, N! = 25,852,016,738,884,976,640,000, which is greater than the number of subatomic particles in the known universe.By the way, the limit is 51 playing cards for selection by Gee from a full deck. If Gee lets N = 52, Ger merely waits until the King of Spades turns up and takes Gee's money. Optimal Stopping Problem More than 10,000 websites explore this fascinating class of problems under a variety of names, Secretary Problem being most common. Others deal with finding optimum gasoline prices, communications switching algorithms, space missions  hey, even optimizing web searches. The bar challenge imposed a kind of "sudden death" upon YDB, which would be administered by you, having godlike knowedge available for instant disclosure. An equivalent outcome arises from the more common formulation called the "Cutoff Rule" (CoR), in which YDB says no to the first K slips with no immediate consequences if one of the passed up numbers happens to be the largest. Then starting with the Kth slip, YDB ascertains whether that slip happens to be the largest number seen so far. If not, YDB continues to the next and the next, stopping only when a slip has the largest number of those written on the unfolded slips and takes the consequences if that number is not the largest of all the numbers in the bowl. he Challengee vs Challenger puzzle is designed as a twohanded card game, thereby allowing a sequence of wagers without waiting in between to think up and write down numbers on slips of paper. Like the bar challenge, the game narrative might feature "Carnac the Magnificent" and something else. The number of cards selected by Gee is not specified by Ger. That might seem to necessitate Ger to devine the exact size of the stack, adding to the mystery. However, we will show that the solution is the same for all values of K! Exclamation point or a factorial symbol, your choice. The solution to the puzzle may be derived directly from the "Cutoff Rule" (CoR) appearing in many web references... Set K = N / e (integer value), where e = 2.718281828, the base of the natural logarithms.For increasing values of N, the probability for Ger's success converges to an asymptote of... p(WIN) > 1 / e = 0.367879441, which is better than the 2to1 odds proposed by Ger.


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