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The Café Lover's Problem, PART 2 Poor Marcello, can't sleep over the issue. He wants to increase dramatically the probability of the man and the woman meeting before 98 weeks have passed. He knows he cannot attract more than randomly 0-6 patrons at any one time. He enlists the help of a Mathematician who proves to him that variants of both his and David's solutions actually can't decrease the probability of them meeting. The Mathematician also shows him that if n=# of tables, then p = 1/(28n - 14) where n>3 and 1/98 if n<=3. The Mathematician also shows that subtracting a chair makes no difference to p if n>4, and p= (n-1)/49(2n-1) if n<=4. Therefore he shows them that 98 weeks is the optimum, given that their clientele won't vary - and that they needn't bother re-arranging the shop over the issue. Marcello wakes in the middle of the night and thinks he has a solution: offer half price coffee's to couples that come into the shop. Let's say for the sake of the problem that this will increase the number of couples by 25%, and that couples will occupy both chairs of a table. Let's also say that if the man and the woman are sitting not opposite from each other, this deal will increase their chances of talking and pretending to be a couple by 1% in order to take advantage of the deal, but not if they're both on outer tables - and only if either the man or the woman can sit opposite each other, either on their own tables or on a vacant table. Has Marcello hit on a way to reduce the 98 weeks? By how much? Hint: the distributions 0-6 patrons rather than being random are now calculated to be 0.1215, 0.1215, 0.1366, 0.1366, 0.1544, 0.1544, 0.1751. You may not need this to solve the problem. If you do, double check. For background to this problem, please read the original PART 1 thread. This has also been posted there on p2.