User talk:Yvain

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Puddings

  • you've made a at Awwww, Yeah that 72 is the average number of puddings you need to start with to complete the trophy. this is fairly counter-intuitive. as a student of mathematics you will appreciate that the requirement to show your workings is not superfluous. any chance you could include them here? --Evilkolbot 12:56, 28 October 2008 (UTC)
    • I suspect he's factoring in the fact that, when you fight a black pudding, it is apparently guaranteed to drop two more black puddings. However, even given that, you would have to have a combat rate greater than 50% to break even - if you eat 100 puddings, only 35 of them will be fights, so you'll end up with only 70 afterwards. Going by my own math, you would need at least 213-226 black puddings (with average RNG) in order for the replenishing effect to be enough to get you to 240 fights. --Quietust (t|c) 15:22, 28 October 2008 (UTC)
      • This isn't too hard to calculate, at least if you'll idealize a bit. When attempting to consume N puddings, you get 2*p*N puddings left over on average, with p the probability of fighting a pudding (so .7*N if the .35 figure is correct), and with half of that amount corresponding to fights (p*N). You then attempt to eat those 2*p*N puddings, resulting in (2p)^2*N puddings left, half of which correspond (new) fights (2*p^2*N; total of p*n+2*p^2*N). If we ignore potential issues with fractional (between 0 and 1, in particular) puddings, summing over all possibilities yields a geometric series, the sum of which is (N/2)(1/(1-2p)) fights. Plugging in p=.35 and setting the equation equal to 240, we get N=144. So I'm guessing Yvain used this method, but forgot to divide by 2 to get the correct number of fights. --Flargen 16:42, 28 October 2008 (UTC)
        • Are you sure about that math? I worked out the geometric series to Np(2p)x with x going from zero to infinity (expanding as Np + 2Np2 + 4Np3 + 8Np4 + ...) with p = 0.35, and setting 240 equal to the sum of the first 50 elements (beyond which they are only a tiny fraction) gave me N = 205+5/7 (and summing N(2p)x for the total number of puddings consumed then gave me 685+5/7, 35% of which happens to be equal to 240).--Quietust (t|c) 17:38, 28 October 2008 (UTC)
        • Some quick searching (in particular, here) gave the formula for the sum of Np(2p)x to be Np/(1-2p) (which also gives the same result of N = 205+5/7 as I got above), not (N/2)(1/(1-2p)) as you came up with. --Quietust (t|c) 17:57, 28 October 2008 (UTC)
          • Oh, whoops. I did make a mistake. I forgot I was summing from 1 to infinity, not 0 to infinity, which means I should multiply my result by 2p. So my formula should be (N/2)*2p/(1-2p), which is the same as Np/(1-2p), as you state. The sum of p^k from k=0 to infinity is just 1/(1-p) for |p|<1, which is what we were using. So I guess Yvain made this mistake and the not dividing by 2 mistake. --Flargen 18:50, 28 October 2008 (UTC)

My Maths

Given x puddings to start with you will get .35x fights so .7x items/new fights. These will then in turn give .7*.7x items/new fights. These will then in turn give .7*.7*.7x items/new fights. Etc. So we need to sum .7^n from n = 0 to infinity. Sum of GP's formula gives us x/(1-0.7) = x/0.3. Setting this equal to 240 gives x/0.3 = 240 --> x = 72. Ah I see my problem. I counted total number of puddings not total number of fights. The number of fights will be x/0.35 So 205+5/7. My apologies. I'll change the page to 206. Yvain 10:22, 29 October 2008 (UTC)