riddler-castles/castle-solutions: 957
Data source: https://github.com/fivethirtyeight/data/blob/master/riddler-castles/castle-solutions.csv
rowid | Castle 1 | Castle 2 | Castle 3 | Castle 4 | Castle 5 | Castle 6 | Castle 7 | Castle 8 | Castle 9 | Castle 10 | Why did you choose your troop deployment? |
---|---|---|---|---|---|---|---|---|---|---|---|
957 | 0 | 9 | 0 | 0 | 0 | 26 | 30 | 35 | 0 | 0 | We have to keep in mind, our goal is to beat other people, not randomness. My feeling is that most of the analytical riddler minds will modify proportional distribution, giving slight edges to certain castles to try to win them by slight margins, as this seems like the optimal plan. So let's turn that on it's head, and beat a lot of people who smoothly allocate their points. There are 55 total points, so 23 total points win. There are many ways to get this with only three castles, but let's keep in mind people will tend to try to do sneaky things to steal high number castles (particularly #9, as that seems "sneaky" to ignore 10 and steal 9). My first reaction was: just win 7,8,9. Put all your points in and win those. This gives 24 and a sure win. But again, 9 seems like a very highly contested castle. So I decided instead, 6,7,8,2. Surely 2 and 6 should be more guaranteed than 9! Now just how to distribute. Well, I should mirror how others will be distributing their points here. (obviously 25 troops to each could lose me 7,8 somewhat frequently). While it seems like I MUST win all four to win, many people will likely assign 0 to some castles, so tie points may come into effect. So even losing 6 can be repaired by a tie in 10 and 3. So I aim to get 23 total points, so let's assign proportionally: {0,2/23*100, 0,0,0,6/23*100, 7/23*100, 8/23*100, 0, 0} = {0,9,0,0,0,26,30,35,0,0}. I need to win every one of these four I've chosen (unless other people elect 0 on some castles... very possible?), but I think in the long run, I've overvalued weird castles that aren't likely to be beaten in general. |