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Farsight Studios
The Pinball Arcade / Farsight Studios
(An attempt at) The top 40 TPA players from leaderboard scores
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<blockquote data-quote="vikingerik" data-source="post: 150820" data-attributes="member: 3745"><p>Wow, neck and neck on me and AntonR. But the timing just missed some games I just played tonight, took #1 on Harley and #3 on Whirlwind.</p><p></p><p>As for the dropoff formula, here's some brainstorming. Make it so that every doubling in rank is equal. So the difference between places 1-2 is equal to the difference between places 2-4, 4-8, 8-16, and so on. In other words, every time you improve yourself past half the players above you, that counts equally. Going from #100 to #50 is the same as going from #80 to #40, or #30 to #15, or #4 to #2, or any other such combination. This feels nicely elegant mathematically.</p><p></p><p>This also has the advantage of scaling out to however far down the leaderboard you want to go while always working the same way. (It would work best going to a power of 2 like 128 or 512 rather than 100 or 500.)</p><p></p><p>So suppose you want place #1 to have 100 points and to rank down to place #512. There are 9 doublings in that range, since 2^9 = 512. The closed-form expression is (1-log2(RANK)/9) * 100.</p><p></p><p>Edit: actually, a power of 2 isn't necessary. You can rank down to any place N with this expression: (1-log(RANK)/log(N)) * 100.</p></blockquote><p></p>
[QUOTE="vikingerik, post: 150820, member: 3745"] Wow, neck and neck on me and AntonR. But the timing just missed some games I just played tonight, took #1 on Harley and #3 on Whirlwind. As for the dropoff formula, here's some brainstorming. Make it so that every doubling in rank is equal. So the difference between places 1-2 is equal to the difference between places 2-4, 4-8, 8-16, and so on. In other words, every time you improve yourself past half the players above you, that counts equally. Going from #100 to #50 is the same as going from #80 to #40, or #30 to #15, or #4 to #2, or any other such combination. This feels nicely elegant mathematically. This also has the advantage of scaling out to however far down the leaderboard you want to go while always working the same way. (It would work best going to a power of 2 like 128 or 512 rather than 100 or 500.) So suppose you want place #1 to have 100 points and to rank down to place #512. There are 9 doublings in that range, since 2^9 = 512. The closed-form expression is (1-log2(RANK)/9) * 100. Edit: actually, a power of 2 isn't necessary. You can rank down to any place N with this expression: (1-log(RANK)/log(N)) * 100. [/QUOTE]
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The Pinball Arcade / Farsight Studios
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Farsight Studios
The Pinball Arcade / Farsight Studios
(An attempt at) The top 40 TPA players from leaderboard scores
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