09242012, 21:13  #91  
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I swear you must never have played any form of lottery or gambled in any sense of the word. 


09242012, 21:16  #92 
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Pretty sure multiple people multiple times have emphasized the part of independent events.
You need help with adding fractions together so I thought I'd be a nice guy and give you some help. So here. http://www.mathsisfun.com/fractions_addition.html Last edited by thehempknight; 09252012 at 01:07. 

09242012, 21:22  #93  
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09242012, 21:27  #94  
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Just toes a coin and find out if you will get all head in a row if you do that is called luck
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09242012, 22:14  #95  
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Here i go again. So the chance of failing to compose 3 times in a row is 12,5%, but succeding is after failing is also the same. Win Win Win Win Win Lossl Win Lossl Win Loss Win Win Loss Loss Win Loss Win Loss Win Loss Loss Loss Loss Loss So if you allready fail to compose 2 times all the other options are gone except the two whic starts with loss loss that allready hapened. Win Win Win Win Win Lossl Win Loss Win Loss Win Win Loss Loss Win Loss WinLoss Win Loss Loss Loss Loss Loss So when there are only 2 posible outcomes left what are the chance of one of them hapening? The only way you could have higher chance of succeding after failing to compose if it is programed this way but this have nothing to do with math and knowing TQ you would only have lower chance after succeding too much. So if system detects that you are lucky next time when it will say you have 50% succes rate it will be 40% because TQ are greedy
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09242012, 22:56  #96 
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Instead lets all go and buy 100 +1 stones and chart our success vs failure vs progress on item. Then we can all report back here and laugh at the liars who make out like their numbers consistently math the probability factor. The second thing tought in statistics is that no matter how linear a path you make for yourself, the numbers are rarely ever right. May as well buy 10 BigMacs from 10 different McDonalds and ponder as to why it is they all taste the same.


09242012, 22:59  #97  
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09252012, 00:39  #98  
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1. Plain luck (most likely in this case) 2. TQs programing which determines your luck by result of previous actions. So when system show that you have 50% succes rate it consider that its acctualy 40% or 60% (showing false info) which is posible but have nothing do do with math.
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09252012, 01:41  #99  
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Within the limits placed already previously in the thread of adding +1s to a +0 or +1 item which gives 10/20 progress of 50% chance to succeed with quick compose..... The chance of any given try is 5050. Meaning any time will give you a fail or success at even rates in the long run. What the original issue was, was what was the probability to fail three times in a row. Which you agree is 12.5%. Now you are trying to say we are implying there is only a 12.5% chance of the last attempt to fail. Which I haven't seen anyone even say. It is why we always said 50% or 5050 for each attempt. Quote:



09252012, 02:33  #100  
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If i where the programmer of TQ i would say
"Its still based on your luck" Quote:
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09252012, 02:41  #101 
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^
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09252012, 02:59  #102 
JesusFreak
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this thread is full of luls.
Okay first, I needa explain the 50/100 + 50/100 + 50/100 thing in a way that people would understand. Half + half + half right? Half a pizza + half a pizza + half a pizza = 3 halves of a pizza = 1 full pizza and one half pizza. ie. 1.5 pizzas. 1/2 + 1/2 + 1/2 = 3/2 = 1.5 pizzas! not 3/6 = 0.5 pizzas! ~~~~ Okay, next, the probabilities. We're looking at a succession of independent bernoulli trials. ie. A binomial trial, with n=n. In the case of using 3 +1 stones, we're looking at n=3. So yes, as people have said, you can get Win Win Win Win Win Loss Win Lossl Win Loss Win Win Loss Loss Win Loss Win Loss Win Loss Loss Loss Loss Loss More simply: Pr(3 wins) = 3C3 * 0.5^3 * 0.5^0 = 12.5% Pr(2 wins) = 3C2 * 0.5^2 * 0.5^1 = 37.5% Pr(1 win) = 3C1 * 0.5^1 * 0.5^2 = 37.5% Pr(0 wins) = 3C0 * 0.5^0 * 0.5^3 = 12.5% The expected number of wins this way would be: 0.125*3 + 0.375*2 + 0.375*1 + 0.125*0 = 1.5 Which makes sense cause using 3 stones with 50% chance each, we would expect 50% of 3 to work = 1.5 So if you use 3 stones, on 3 +1 items, you would expect to get 1.5 +2 items. Or, if you do that twice, 6 +1 stones on 6 +1 items, you would expect to get 3 +2 items. The other way to do this, would be to have one +1 item, and 3 +1 stones, and then try the three +1 stones until you succeed. ie. Win first time = 0.5 Lose, then Win = 0.5*0.5 = 0.25 Lose, then Lose, then Win = 0.5*0.5*0.5 = 0.125 Lose, then Lose, then Lose = 0.5*0.5*0.5 = 0.125 The expected number of stones you would use in this case is 1*0.5 + 2*0.25 * 3*0.125 + 3*0.125 = 1.75 stones. And the expected rate of success would be 1*(0.5 + 0.25 + 0.125) = 0.875 So on average, this way, you would use 1.75 stones for every 0.875 successes. ie. 2 stones for one success. Same thing. Essentially, there isn't a trick. In the long run, your wins will balance out the losses and you'll have spent the same amount as if you were to just buy +1 stones and compose them normally. The only way to win consistently, is if you're "lucky".
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09252012, 03:45  #103  
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09252012, 03:46  #104  
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Last edited by Planes v1.0b; 09252012 at 03:48. 


09252012, 03:56  #105  
I<3Food
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but fcuk y'all this guy is right.
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