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| Author | Merging hunts |
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numbers numbers numbers.....
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Guys, you have it all wrong. The chance is 50% - you either have a merged hunt, or you don't have it XDDD
Oh I just had to write this ^^
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Svens conclusion sounds logical, 5% is about the frequency I get co-hunts. As for 2Pv2AI, 1/400 sounds logical too me | When 10 people start hunt at the same time ... first question is, how the mechanism exactly works: When 4 people get "take to heels" message, then if 2 couples are created or just one. Let's assume that 2.
Solution must go this way:
We must start bye 0.05(probability that YOU get a merg hunt(=MH))
now we have 9 players left.
1, Probability of sample that odd number out of 9 players gets "take to heels" message(TTHM)
- if odd number out of 9 players get TTHM, then you get MH 100%
exactly 1 player TTHM 9*0.05*0.95^8=0.297
exactly 3 players TTHM 84*0.05^3*0.95^6=0.00077
exactly 5 players TTHM 126*0.05^5*0.95^4=3.2*10^-5
exactly 7 players TTHM 36*0.05^7*0.95^2=2.5*10^-8
exactly 9 players TTHM 0.05^9=1.95*10^-12
altogether 0.298
notice that cases 3,5,7,9 are insignificant
1, Probability of sample that zerro or even number out of 9 players gets TTHM
- if 0,2,4,6 or 8 out of 9 players get TTHM, then you get MH with probabilities 0, 2/3, ...
exactly 2 players get TTHM 36*0.05^2*0.95^7=0.063
we can ignore cases 4,6,8, because their probabilities are insignificant like in cases 3,5,7,9.
Now let's get it all together.
0.05*(0.298+2/3*0.063)=0.05*0.03426=0.017, which is 1.7%
That means that you get MH once in 59 hunts, and if you get TTHM, then once in 3 hunts.
This result is relevant for 10 players starting a hunt at the same time. But, there are let's say 1000 players online, everyone does let's say one hunt per hour, and let's assume that server waits 10 seconds during TTHM for response of another TTHM player.
Then it's 3(*) players starting hunt in every 10 second interval (1000/360), not 10, so our result(now without explanation) is 0.00475, so 0.475%, so once in 211 hunts.
(*)But I must revise it, because it's incorrect to work with average here, so the final estimate is about on merging hunt out of 100 hunts.
But who knows, how long the waiting interval for another player exactly is :-) | | Interesting fact - if server waits not 10, but 15 seconds for another player response, you get a merging hunt once in 70 hunts. It depends a lot on the waiting time. | Guys, you have it all wrong. The chance is 50% - you either have a merged hunt, or you don't have it XDDD
Oh I just had to write this ^^
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yep. what the chance to meet a hydra at the street?
50/50
nice logic | 19+
That is the formula I came up with. Although I had to figure out the factorial coefficient, but I had the same answer.
22
The coefficient you are multiplying are wrong. For 1 it is 10!/(9!*1!)=10, 3 it is 10!/(7!*3!)= 120, and 5 it is 10!/(5!*5!)=252. | Guys, you have it all wrong. The chance is 50% - you either have a merged hunt, or you don't have it XDDD
Oh I just had to write this ^^
1|2
yep. what the chance to meet a hydra at the street?
50/50
nice logic
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Or better yet, if I believed there was a 50% chance that I will get bit by a shark when I go to the beach(either it does or doesn't), I definitely would not be going to the beach at all. | #25
Dont forget that I have just 9 players in that part ... coeffs are correct. | Gee, I just knew I should've put even more smiles in my post *rolls eyes*
Or signed it with "jk". Seems like it's the only way for some people to tell if a post is a [quoted!] joke. Oh well, I'll keep it mind for future reference. | 27
Sorry, we were all talking about 10 people and you switched to 9, I missed that part.
28
perhaps I should have put smiley faces or signed it with "jk". | #29
I am talking about 10 people, too, but "you" is one of them.
The question doesn't stand "What is the probability that 2 out of 10 people get merging hunt.", but "What is the probability that YOU get a MH, while 10 people are starting hunt at the time(I assume, you are one of them)"
So first condition is that YOU get TTHM(0.05) and then it's about the other 9 ... | 29 Perhaps you should've.
[Player banned by moderator Shebali until 2009-02-25 17:31:54 // Initiating off-topic.] | to #22
You want to say that if that if i just start my own hunt in random place and random time i have grater chance to get merged with another hunt than if we all start hunt in one place and one time??
How it could be? I will go to place Mistrey cost and start hunt alone when none will be there then merging hunt is more possible than in crowded place?? | 30
I understand what you're saying now. I read the question as what is the probability that there is a merge hunt out of 10 friends. While yours is the probability that you and one of your 9 friends get a co hunt.
Also, we could even complicate things by locations. As far as I know, I believe you have to be in the same location to get a shared hunt. So if you hunt in a relatively obscure location the probability of you getting a shared hunt will be lower. Just as a more crowded area would give you a higher chance to hunt. As for the interval, I'm pretty sure I've seen it waiting for at least 30 seconds. I need to time it the next time it happens though. | OK:
Now i will explanation why I wanted to know about this probability.
As from previous posts cleared that if I win hunt in single than next time they will be 30% stronger but if you get merged hunt you get +2 GH exp and the predators will stay at the same lvl.
So according to this if I will be able to get all my hunts merged they will never increase in strength even when I lvl up. So and all the hard hunts will be gone.
I know there are no way to get all them merged but if I and another 100 players will start to hunt in same place in same time there will be hunts (+/- 1 lvl) which could be merged, so I wanted to know what is probability that someone (Not me) will get his hunt merged. | #32
No, I said that in case of 10 players it's 1,7%, in case of 3 players 0.475%.
But you reminded me of the fact, that I didn't take location into consideration. 2 hunters must be in the same location to get grouped, and in addition it seems they can't be more than one lvl different. That significantly decreases the probability of MH (I will not count it exactly again) and brings me to idea, that it all works differently. Maybe if one player get TTHM, then the server looks for another player just starting hunt and join him authomatically, without 5% ... | #33
Good, we understand each other :-)
I had TTHM two times today and it was 15 and 14 seconds ...
#34
it differs depending on how many pairs out of 100 can be merged, I definitely won't count it today :-D | I’ve been asked to calculate the exact probability of getting a merged hunt.
One thing is clear: I can’t do it, because I don’t know the exact mechanism, i.e. how players are picked (only admins know that).
So, I have to make some assumptions. And my answer will be correct only if those assumptions are correct.
Assumptions (they just are my guesses):
1) when a player starts a hunt a random experiment is performed: with probability p (ex, p = 0.05) a player is marked as “ready to merge”.
2) This experiment (i.e. marking) is done independently of other players (i.e. it doesn’t matter how much players is trying to hunt and how much of them are already marked)
3) status “ready to merge” lasts some maximum time interval (ex, 3 sec), or shorter if player merges earlier.
4) if there is already a player with status “ready to merge” then the both players starts a merged hunt.
5) if there is no player with status “ready to merge” then he waits some time (ex, 3 sec) and if no player becomes “ready to merge” during that time then no merging if performed and he starts a hunt alone.
6) so, there is always no more than 1 player with status “ready to merge”, because as soon as there are 2 players with “ready to merge” status they are instantly merged and begin a hunt (and they become unmarked).
Question: what is the probability to merge if you go to hunt?
Answer: there is no exact answer, because we don’t know the distribution of the number of players in those 3 sec (we also don’t know how that distribution changes in time, i.e. we have a random process here with unknown distribution – unless someone will post the exact algorithm of how this merging is performed then we should be able to find that distribution and then we will calculate exact probabilities).
But if
7) there always are so many players that a player with status “ready to merge“ doesn’t wait too long, i.e. he always merges with someone if he is marked.
Then the probability to merge is exactly p ( = 0.05).
Question: Let’s say 1)-7) are correct. What is the probability to merge with my friend if we both go to hunt at the same time?
Answer: I can’t give you the exact answer, because I don’t know how the algorithm works (it can be done in many ways).
If there is already waiting one player then one of you will merge with him (even when you both join exactly at the same time only one of you will be first).
Probability of event “merge with friend when you both hunt at the same time” = P(you are market) * P(your friend is marked)*P(there is no player waiting) = p*p*q
8) if assumption 7) is correct, i.e. there are really many players trying to hunt then approximately q = 0.5 (this is also an assumption)
So, if 1)-8) are correct (and p = 0.05) then p*p*q = 1/800.
Actually I doubt that those assumptions are correct, but I can’t say anymore (until I get more information). | My opinion would be to use a convolution between the binomial and hypergeometric distributions. I did that and I found that the probability is in a range of about 5% to about 7%. The first number is coming considering the computing the maximum number of starting hunts with "ready to merge" flag in the time window of 15 seconds from 1500 players online and the second is coming from the probability for at least 2 of you (9 friends) to be flagged as "ready to merge" in the same time (minimum number of other starting hunts with the "ready to merge" being considered 0).
If you want the proof, I can give it, but it implies you have some math knowledge. But I can give you the concept to see if I am wrong.
Concept:
Let n be the number of friends who start in the same time a hunt. Let N be the number of random hunters (coming from 1500 online players, starting their hunts in the time window of 15 seconds). The final probability for two friends to join in a combined hunt is given by the formula:
P(n,N) = double_sum_over_i_and_k (B(i,n) x B(k,N) x H(i,k))
where:
- P is the final probability;
- i is from 2 to n;
- k is from 0 to N;
- B(a,b) is the probability of having exact "a" successful events (hunts with the "ready to merge" flag on) from total "b" number of tries (binomial distribution);
- H(i,k) is the probability of two friends to join the same hunt out of i friends and i+k total hunts started in that time window (all the hunts here are considered with the "ready to merge" flag on; hypergeometric distribution).
Concept explanation:
The number of friends is independent on the casual players who start their hunts in that time window, so, the probability for at least 2 players to start a combined hunt is given by the convolution (in this case, direct product) of the probability of at least 2 friends to be flagged "ready to merge" and none or any number up to the maximum allowed casual hunts in that time window to be flagged as "ready to merge". If no casual hunt is flagged as "ready to merge", then the probability to hunt with one of your friends is given by the probability for at least you and one of your friends to be flagged with "ready to merge" state (if nine friends, the probability is about 7%). If at least one casual hunt is flagged as "ready to merge", then the probability to have 2 players to start a combined hunt must be convoluted (in this case, multiplied) with the hypergeometric distribution in which the constrains are given by 2 friends out of how many are flagged as "ready to merge" in that combined hunt and 0 casual hunts from how many are flagged as "ready to merge".
The sum over i is because we need _at_least_ 2 friends flagged as "ready to merge", but they can be 3, 4... 9. The sum over k is because we can have any of the number inside the range of [0; N] (N is the maximum allowed hunts in that time window, dictated by the number of players online).
Let me know if I made any conceptual mistake in thinking the problem. :) | Hm... Let me just point out one thing... I've been looking over the math here just briefly and thought a little about some issues.
Let's start with the "Take to their heals"-deal. I am not sure about this, but I have never seen a red exclamation-mart TttH. And since people have complained about having HG-records "stolen" from them in this way, I have come to the conclusion that only one person needs the Ttth.
Another thing, and this may be covered in any of the above posts. I could have missed it. If person A has a 5% chance to recieve a joint hunt, then person B has 100% chance to recieve it. Person A will need a partner after all. | #39:
"I have come to the conclusion that only one person needs the Ttth."
"If person A has a 5% chance to recieve a joint hunt, then person B has 100% chance to recieve it. Person A will need a partner after all."
Actually, every player has 5% (0.05) chances to have a combined hunt. Once two players have the flag "ready to merge" on (e.g., they reached their 20th uncombined hunt in the same time) and they start their hunts in a time window of 15 seconds, then the hunts are merged.
The frequency of hunting for one player is in the interval [1/3600;1/2400] hunts/second (one hunt every 40 or 60 minutes). So, the maximum casual hunts per 15 seconds time window for 1300 online players is about 8 (33 per minut if you want to check if this result is correct in the "current battles"; in my computations I took the rounding toward infinity, meaning 33 hunts divided by 4 intervals of 15 seconds is 8.125 and I took it 9 to find the lowest limit for my computed probability for the given problem from the beginning of the thread). Also, here, I took the highest hunting frequency, 1 for every 40 minutes (for the same reason).
In this moment, disregarding the region distribution of the players, the 3rd and lower levels players and the combat level distribution of the players, the probability to get at least 2 casual hunts merged in one is given by the binomial distribution where you cut away the first 2 terms (the first term is giving the probability of not having any hunter with the flag "ready to merge" on, while the second places the same flag on only one hunter, so, no combined hunt):
B(x>=2,N=9,p=0.05) = (N!/((N-2)!2!))p^2(1-p)^(N-2) + (N!/((N-3)!3!))p^3(1-p)^(N-3) + ... + (N!/((N-N)!N!))p^N(1-p)^(N-N)
where x is the number of hunters to be flagged "ready to merge" and N! = 1*2*3*...*N (N factorial; 0! = 1! = 1). So, in this case the probability to have at least one merged hunt is about 7% (7.12114...%).
But this is the maximum limit for having a casual combined hunt, and this probability doesn't care which hunters has joined the merged hunt. E.g., if you have one friend with you pressing attack in the same time, this distribution doesn't tell you what is the probability to be with your friend in that combined hunt. Actually, it doesn't tell you if you will be in a merged hunt. It just tells you what are the chances for two hunters to get the flag "ready to hunt" on in the same time (actually, the same time window). |
This topic is long since last update and considered obsolete for further discussions.
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