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Very bad luck or...?: 7/17/2012 00:05:43

RvW 
Level 54
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Not to mention that nobody is going to post a thread "I went first in turn 2, 3, 5 and 7, while my opponent went first in the other 6 turns", even though that's exactly as unlikely.

Human intuition positively sucks at estimating probabilities and statistics. Just ask any RPG-gamer: "a fifty-fifty chance never happens, a one-in-a-million is a sure thing!"... That's not just a joke, it really feels that way. Partly because of confirmation bias and partly because human brains are no good at all with big numbers:
  • Confirmation bias -- you don't remember all those one-in-a-millions which failed (because you already knew they'd fail, but the one which you actually made... you're still talking about it years later.
  • Big numbers -- without a calculator and in at most ten seconds, answer the following question. How long is a million seconds: two days, two weeks, two months or two years?

There's a half documentary, half infotainment (but they did get the documentary part completely right!) by Derren Brown, called The System. It nicely demonstrates "no matter how small the chance, if you try often enough, sooner or later it will happen". If you'd like, you can watch it online at http://www.youtube.com/watch?v=9R5OWh7luL4 (for the people who are strictly anti-gambling: (1) it's just an example, (2) believe me, after having watched it entirely, you'll agree that this in no way promotes gambling... if anything, it is discouraged).
Very bad luck or...?: 7/17/2012 00:25:30


Richard Sharpe 
Level 59
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Artham, the luck counter ignores first-turn or similar aspects of luck as we see it. It is solely a counter of the armies (or percentage thereof) gained or lost by the 'diceroll'

Having bad dice luck but getting lucky in first-turns can be more advantageous than what Red saw
Very bad luck or...?: 7/17/2012 02:04:58


{RSP} Nike
Level 4
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sorry but why did he pick europe to start with?
Very bad luck or...?: 7/17/2012 16:22:07

Paul
Level 16
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Because nobody would expect him to pick it. Take your enemy by surprise.:)
Very bad luck or...?: 7/17/2012 16:32:56


szeweningen 
Level 60
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RvW, what you say at the beginning of your post is a horribly inaccurate use of probability... You don't seem to have the slightest grasp of what probability is, please check your facts before you post...
Very bad luck or...?: 7/17/2012 16:50:09

Dr. TypeSomething 
Level 3
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I don't see how this is inaccurate (but please correct me if I am wrong):

"Not to mention that nobody is going to post a thread "I went first in turn 2, 3, 5 and 7, while my opponent went first in the other 6 turns", even though that's exactly as unlikely. "

If you were to calculate the chance of going first in exactly turns 2, 3, 5, and 7 and the opponent goes first in turns 1, 2, 4, 6, and 8 it is exactly the same probability of going first every turn. He's not saying the same probability of going first 50% of the time, he is saying going first in exactly that pattern. It should be P(A AND B AND C AND D AND ... 12th number) for both scenarios.

It's like people scoff at the notion that is possible to win the lottery by picking numbers 1,2,3,4,5,6 even though it is just as likely to win as any other combination of numbers.
Very bad luck or...?: 7/17/2012 16:52:57


Richard Sharpe 
Level 59
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Dr Typesomehitng is correct. The odds of getting any one particular string are identical. It is when order is immaterial that the odds start changing. The difference between combination and permutation.
Very bad luck or...?: 7/17/2012 17:09:32

Dr. TypeSomething 
Level 3
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Ah yes, those were the words I was looking for. Thanks!

Any permutation is equally as likely (because going first each round is a binary event each time) but combinations vary quite a bit, with going first 50% the most likely and going first or last every time the least likely. There are many permutations that make up going first half the time while only one that leads to going last every time. So while Red clearly had an unlikely event, it is equally as unlikely as any other permutation, such as going first only in the exact rounds stated by RvW.
Very bad luck or...?: 7/17/2012 17:17:09

RvW 
Level 54
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@szeweningen:
I meant (like Dr. TypoSomething says), but rewritten in standard notation, that P(H,T,T,H,T,H,T,H,H,H) = 0.5^10 = P(H,H,H,H,H,H,H,H,H,H). Is it possible you simply misunderstood my post? Or would you like me to make a stricter distinction between the terms "probability" and "statistics"; I admit I'm sometimes a little lax in that respect?
Very bad luck or...?: 7/17/2012 17:32:57


BearsBeetsBattlestarGalactica
Level 20
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I completely agree with RvW. The only reason you perceive it as unlucky is because it is notable.
Very bad luck or...?: 7/17/2012 17:59:20


J Russell Mikkelsen 
Level 4
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Oh dear god. Math nerds have taken over this thread. We are doomed.
Very bad luck or...?: 7/17/2012 21:10:21


szeweningen 
Level 60
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RvW, that is exactly why I said it was wrong...

The point was that we don't compare "luck" on specific permutations, as it was mentioned every one has the same probability. The question at hand is "how probable it is to get the right coinflip 5 of 12 times compared to getting the right coinflip 0 of 12 times".

That is pretty much the only possible mathematic model we could operate on, and in that model for example (H,T,T,H,T,H,T,H,H,H) is indistinguishable from (T,T,T,T,H,H,H,H,H,H,). When we compare how "lucky" or "unlucky" we were we should not distinguish turns from each other, because on every one of them (bar OP cards) the coinflip is the same. In mathematics it is described as binomial distribution (related to Bernoulli) and is the most basic tool used when stydying cases like that:

http://en.wikipedia.org/wiki/Binomial_distribution

Please try to use the probability mass function to calculate both occurances (Red's example and yours), you should quite clearly see the difference.
Very bad luck or...?: 7/17/2012 21:35:26

RvW 
Level 54
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Ah, then I see where we misunderstood each other; I actually did mean the exact order (not merely number) of occurrences. Hence, "H,T,T,H,T,H,T,H,H,H" is distinct from "T,T,T,T,H,H,H,H,H,H" (maybe I should've written it as "P(F1=H, F2=T, F3=T, F4=H, F5=T, F6=H, F7=H, F8=H, F9=H, F10=H)" instead?).

BTW, I'm slightly disappointed nobody mentioned the prime numbers...
Very bad luck or...?: 7/17/2012 21:49:16


szeweningen 
Level 60
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Yes I understand what you meant, but it just made no sense at all to view first orders in specific permutations. If you want to analyse/measure being "lucky" or "unlucky" which are just simple words for "deviation from expected value" then we cannot distuinguish turns between themselves...
Very bad luck or...?: 7/17/2012 22:02:11


szeweningen 
Level 60
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Also since you requested prime numbers and I am very much into number theory...

Here you go, a closed formula for n-th prime number ;)

Very bad luck or...?: 7/17/2012 22:30:05

[16] Jasper 
Level 52
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I don't think you do understand what he meant. What he was saying was something along the lines of "it all depends on what you perceive as luck" (which is quite close to what you are trying to correct him to). To show this, he gave a random example that was just as likely, yet nobody would care about at all - showing that the human mind only finds things unlikely in special cases ("unlucky" cases, cases with a pattern to them, homogenous cases).

So basically, you are saying the exact same thing (except for the part you claim he's wrong and he claims he is right).
Very bad luck or...?: 7/17/2012 22:46:20


szeweningen 
Level 60
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...
...
...
I really have no strength to explain the basics of buldin probability models...

We want to observe who gets first order right? Since it's a coinflip we expect both players to come out on top half of the time (bar OP cards). So when do we say one player was unlucky? When the outcome deviates far from the expected outcome... With perspective he provided there was no luck at all, with his perspective no outcome would be unlucky or lucky. So basically what I am saying is if we are using words "lucky", "unlucky" we mean something that does not have to be described with great detail but basically means an anomally, a deviation from expected outcome. Of course here we are talking globally, about first orders through the whole game so it is quite clear how we should implement that into our model if we want to analyse it. Of course if we go into the game deeper we might call lucky many other things when we compare it to the outcome of the whole game. Anyway in case of that global approach to first orders I think the case is closed...
Very bad luck or...?: 7/18/2012 01:50:50

RvW 
Level 54
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Jasper, actually me and szeweningen do indeed agree. I was just talking more in general, while he was still looking at "luck".
From his point of view the order of events is irrelevant (we just want to know how many times a given player got first turn, "lucky"), while from my point of view (talking more in general, semi-ignoring how the discussion got started) the order of events is relevant. That's a huge difference.



szeweningen, regarding that (black on dark-grey, gotta love transparent backgrounds :p ) formula for primes, the part after the multiplication looks very weird...
The innermost part is "1 + floor(k/i) - k/i" (with k and i both strictly positive). Unless I'm mistaken "floor(k/i) - k/i" will always be somewhere between 0 (inclusive, if i divides k) and -1 (exclusive). That puts "1 + floor(k/i) - k/i" in the range (0, 1], after flooring it will be either 0 or 1.
Because the final iteration has i=k there will be at least one case where it's a 1 (good, wouldn't want to divide by zero!), so the total sum is at least 1. Then we take the reciprocal and floor again... giving either 1 (if all other terms came out 0) or 0 (if at least one other term came out 1).
Since terms are 1 when i divides k..., does that second part simply say "if prime (k) return 0 else return 1"...!? While I do understand the appeal of closed formulae, my (computer science) heart weeps at the ridiculously inefficient (not to mention horribly roundabout and utterly unreadable) way of expressing it. (For some reason, I have to think about GEB now. :) )
Still, can you link the paper (??) where you found that beast? I'm very curious about the explanation the "inventor" him/herself gives. So far, the "best" result I knew about primes is that there are approximately n/ln(n) primes smaller than n; I "remember" (probably incorrectly) having been taught that such a formula as you just gave is supposed to be impossible...
Very bad luck or...?: 7/18/2012 04:26:53


J Russell Mikkelsen 
Level 4
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Please oh please tell me you guys get paid to do this. Nobody talks about math this much, or takes it this seriously, who doesn't get paid.
Very bad luck or...?: 7/18/2012 05:58:28


szeweningen 
Level 60
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Hehe, well, that is a tricky formula. Of coourse there is no useful formula for n-th prime number that'd help in computation of them, all formulas like that are based upon some tricks that basically do what you said (if p prime return sth, if p prime return sth). There are many more examples, probably the easiest one to see comes from the Wilson's theorem:
For every n


The formula I gave has a longer and a bit more convoluted explanation, but follows similar lines (try the last one). Anyway you remember correctly, there is no useful formula, the ones existing have no real mathematical or practical value.
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