My DiariesA Lurid Tale of Obsession, Depravity, Wits and Attempted WitWed Mar 3, 1999Days 10 to 19 used car inventory, and rambling about winning a 3 race series. |
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<<-- Used Car Information -->>Copyright © 1999,2000, the author/owner of the following ==> page <==.
Last night I discovered that between day 10 and day 19 inclusive you can buy a "lime green 2tone" S13 '88 Silvia Q's 1800cc for an ultra-cheap Cr3450. Spend all the extra money on weight reduction, and it can win the Sunday Cup, although it has a hard time straight out of the box. After that the only turbo option (sort of annoying that it does not have the usual Cr4000 turbo option, but only a "spec 1" for Cr12500) turns it into quite a respectable little money-maker able to sweep the poles as well as the races in the Sunday Cup. |
Strangely, the first I won the Sunday Cup with this car, the competitors connived to do the most amazing job of distributing the other spoils. I went into the last race leading 18 to 9, i.e. unable to lose, and winning 27 to 15. |
I realized that 14 is the minimum number of points needed to win the Sunday Cup, but that is extremely unlikely to happen. |
e.g. |
6 4 4 = 14 9 3 1 = 13 9 3 1 = 13 9 3 1 = 13 6 4 2 = 12 6 2 2 = 10 |
Or |
6 6 2 = 14 9 3 1 = 13 9 3 1 = 13 9 3 1 = 13 6 4 2 = 12 4 4 2 = 10 |
Various other combinations are theoretically possible, including ones where the winner with 14 wins one race. |
The thing is, Gran Turismo usually seems to have one "slowpoke" in any series who never gets more than 2 points in a race. So you have to do better than above to have a chance of winning. |
It's relatively easy to prove that you cannot win such a 3-race series with only 13 points. In fact, the above charts demonstrate it; you could take one point from two of the 13's and give them to the last-placed "10", but there's one more 13, plus the point you didn't win, so the best you could do is a three-place tie for 1st at 13, which, I have come to suspect, the game always awards to the computer competitor. |
E.g. |
6 4 3 = 13 9 3 1 = 13 9 3 1 = 13 9 2 1 = 12 6 4 2 = 12 6 4 2 = 12 |
More simply, the total number of points available is 3*(9+6+4+3+2+1) = 75. 75/6 is 12.5; competitors can only get integer numbers of points, so, for competitors to get less than 13, they must get 12. 12*5 is 60, which leaves 15 points required for you to win to stop the other competitors exceeding 12. So if you get only 13, there are two other points the other competitors must get so at least one of them must get more than 12. Any mathemeticians out there, feel free to submit a rigorous proof. |
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Copyright © 1999,2000 the author/owner of the following ==> page <==.