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| Author | Merging hunts |
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#60
"then there are 5% of 5% chances, that means 1% chance for them"
No no, carefully ;-) 5% of 5% is 0.05*0.05=0.0025, so 0.25%, as Shebali wrote before. But I believe we are solving problem "how often do you get merged hunt?". But we didn't get to calculation, because we don't really know how the mechanism works and of course we can do some estimation, but as for me, I am too lazy to do that :-P | | Sorry, Sven wrote it, Shebali had another theory :-P | #59:
"Do you mean X by [the number of starting hunts with "ready to merge" flag in the time window of 15 seconds from 1500 players online]?" yes
"Do you mean Y by [the number of flagged friends (from 9)]?" yes
You got them correctly with one correction: I made Y+X and not X+Y. :) But, in this case, the convolution is commutative, so, it doesn't matter.
Well, when 9 friends are pressing "attack" in the same time window, that's something independent on the casual hunts started in the same time window. Reciprocal is valid as well. In some other words, the 9 friends are "cheating" the casual hunts and that makes them independent on the rest of the statistical system. But, you are right, they still can be connected by the rule of merging. For that, I introduced a little trick. I left X and Y to be independent from the point of view of the two binomial distributions, but I connected them after that by the hypergeometric distribution (by distribution I mean probability function) which allows me to keep their independence as characteristics, but to connect and sort by successful/insuccessful event (another reason was for ensuring that the game is choosing two flagged friends in a row and 0 casual hunts).
Now, generally, if given two function f and g, by f convoluted with g is defined (in non-strictly mathematical terms) the "shaping" of f over g pattern to create a new pattern. So, my definition of "a convolution between the binomial and hypergeometric distributions" was in that understanding.
Actaually, my whole logic started in the oposite way. I constructed the hypergeometric pmf (probability mass function) and after that I asked each X and Y to be computed by the binomial pmf's. And, because we have two independent conditions (2<=Y<=9 and 0<=X<=N), I made the double sum.
I hope I was clear because I am enough tired to write you in my native language instead of English. Let's pm each other if you want to ask me more and let's post only the outcome of our discussion (for the sake of the length of this thread). | #60:
Sorry, I am not mathematician, I am physicist working in Computational Physics. :P | hmm ok, you are right 5% of 5% is 0.25% my mistake.
And about frequency, Yesterday I had a hunt of 81 bowmen (my alternate character, imsunny3), which was joined with another guy. The next hunt again was 81 bowmen, that too was joined. I think thats just too much co-incident, could be ever better if that got joined with the same guy having same monsters.. lol | #61:
That's why I said the both need to be flagged (5%x5%). Otherwise you cannot apply that idea of 0.25%. |
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