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AuthorFrontier Ursary Knockback
What's the ratio of doing a knockback with Frontier ursary if it covers all the 7 steps?

Regards
70%
1 step = 10% of being knocked
thanks and 1 more thing:
If you use the double speed rune and then hit a creature while covering at least 10 steps does that mean it is necessary to knockback the creature?
Yes
thanks
NP
Isn't it 1 bonus iteration for each step n?

and the percentage would be just like any other † Triggering special abilities?

so 10% could be correct, but only in a very specific setting given the hp of the attacker and the defender, and so on..


it's a bit like asking how often do guardians stun (n=1), which you have the formula but no fixed percentage. But in this case, it might be possible to find the answer for n=7 expressed in terms of the answer for n=1?
#4

no, it wouldn't add up like that,
suppose the formula spits out the percentage p (the formula would be for step n=1)

for demonstrative purpose, i'm going to use n=3 steps
then the final chance would be
p + (1-p)p + (1-p)(1-p)p

the first term being the chance of success in the first iteration
the second term is when the first iteration fails (1-p), and you get another p chance of success there
and so on...

it's a geometric progression with ratio 1-p < 1, so it converges
check this with the formula for the finite sum of a geometric series and you should see that this number never reaches one unless p is 1


and i don't think this p can be 1, but do take a look at the formula in the about the game page, i forgot exactly what it is.

An answer to this question can only be given in terms of p (n=1)


If p = 0.1 = 10% and the panda took 10 steps
then you'd be summing up the first 10 terms in the above series
this should give your panda about 65.1% in the end

If p = 20% with same 10 steps
still summing up first 10 terms but with different p (depending on the hp of the attacking/ defending stack, blah blah blah..)
then this should give your panda about 89.2%


If you are curious, the formula would be
p * (1- (1-p)^n ) / (1 - (1-p))
Which conveniently simplifies to 1 - (1-p)^n
remember where p is the number you get after using the formula for † Triggering special abilities found on the about the game page

please do correct me if I am wrong, but this is how I always interpret what knockback means
ok got it. thanks a lot:)
closed by Lord hpsim (2013-03-07 15:41:51)
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