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I have tried to structure that 2A, 2B ticket structure manually and it takes a lot of doing (and end up with lots of tix). But it can shave some bucks off the ticket cost. Especially if you use a lot of B's. Take a mythical series. We can forget about AA and AC's. Leg #1 A - 1,2 B - 3,4 Leg #2 A - 1 B - 2,3 Leg #3 A - 1,2 B - 3 Leg #4 A - 1,2 B - 3 Total AB cost on a buck is 4 X 3 X 3 X 3 = $108 To me, preferred play is as follows. AAAA 2 X 1 X 2 X 2 = $8 BAAA 2 X 1 x 2 X 2 = $8 BBAA. 2 X 2 X 2 X 2 = $16 BABA 2 X 1 X 1 X 2 = $4 BAAB 2 X 1 X 2 X 2 = $8 ABAA 2 X 2 X 2 X 2 = $16 ABBA 2 X 2 X 1 X 2 = $8 ABAB 2 X 2 X 2 X 1 = $8 AABA 2 X 2 X 1 X 2 = $8 AABB 2 X 2 X 1 X 1 = $4 AAAB 2 X 2 X 2 X 1 = $8 Total AB Cost = $108 Total Multi Ticket Cost = $96 % Savings = 11% The dollars don't seem like much, but reducing ticket cost by 11% is nothing to sneeze at. Figure if total tickets were $175 with additional AA and AC covers, it's still almost 7%. Also, when there are more B's than example the % savings can get quite a bit higher. Tough to zig zag. But personally, I have never hit BBBB or ABBB. Anyone take it to this extreme? Last edited by pmacdaddy : 01-17-2008 at 12:05 PM. |