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Author | Frontier 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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