How many of you can get this right? (Part 2)

[Steroids]

no xD someone answered 7, 6.8, 10  so i assumed they didnt read well...

I will try to solve it on logical way (since im so lazy to look trough books lol)
Might not be correct but
ok since crit hit is 2 dmg and has 50% chance to crit max number of crit hits required to kill mob is 5. Ofc minimum is 0 if lets say u dont crit at all.
So average will be 2.5
k bai

That's pretty logical.

[엔젤ღ]

no xD someone answered 7, 6.8, 10  so i assumed they didnt read well...

I will try to solve it on logical way (since im so lazy to look trough books lol)
Might not be correct but
ok since crit hit is 2 dmg and has 50% chance to crit max number of crit hits required to kill mob is 5. Ofc minimum is 0 if lets say u dont crit at all.
So average will be 2.5
k bai

ok ok I'm very sorry :c

[アイドンノウくん]

ok this is gonna be random, 3.888

Aww, close but not right...But it would be great if you could provide your argument/explanation, as this thread is manly for discussion purposes.

[アイドンノウくん]

As far as my luck goes, 0.

Well, in this question, you only have 0.0976563% chance to get 0 critical hits until you kill the monster.

[アイドンノウくん]

no xD someone answered 7, 6.8, 10  so i assumed they didnt read well...

I will try to solve it on logical way (since im so lazy to look trough books lol)
Might not be correct but
ok since crit hit is 2 dmg and has 50% chance to crit max number of crit hits required to kill mob is 5. Ofc minimum is 0 if lets say u dont crit at all.
So average will be 2.5
k bai

People answered 7,6.8,10,... because they answered the question in Part 1 (see the first thread). The correct answer for Part 1 is 6.8887.

For part 2, you are only partially right. The possible number of critical hits are 0,1,2,3,4,5. However, the probability for each of these number is not all equal i.e some of the numbers has greater chances than others. You are assuming all 0,1,2,3,4,5 have equal chances. That's why your finally answer is not correct. But you listed out all the possible number of critical hits correctly. 

[アイドンノウくん]

That's pretty logical.

Logical doesn't always mean correct. Lol 

[Feone]

This is the question for Part 2:

Assuming you deal 1 damage to a monster every time when u attack, your accuracy is always 100%, your critical rate is 50%, and you deal 2 damage when you deal a critical hit. If the monster has 10 hp, how many critical hits will occur on average until you kill the monster?



--------------------------------------------------------------------------------------------------------------------------------
This is the question for Part 1:

Assuming you deal 1 damage to a monster every time when u attack, your accuracy is always 100%, your critical rate is 50%, and you deal 2 damage when you deal a critical hit. If the monster has 10 hp, how many hits do you need on average to kill the monster?

Answer: 6.8887 times

Also, someone has got it right (see reply #14).

If the expected average amount of hits is 6.888, with a 50% critrate, then the expected amount of crits would be 0.5*6.888 = 3.444

A slight alteration of the formula will find this one too. Rather than counting hits it'll count crits only.

Let E(m) be the expected amount of crits remaining to reach n starting at a sum of m

E(n) = 0
E(n-1) = 0.5
E(n-x) = (1/2)*(E(x-1)) + (1/2) * (1+E(x-2))

E(n-10) = 3.444

[Bre.]

Holy fck this is why I skip school classes..

[Feone]

Holy fck this is why I skip school classes..

Recursive functions fall under calculus if I'm not mistaken, you're safe from these in school!

[Lena]

lol. So much smart ppl.

[ᴰᴱᴬᴰ†]

Too many replies for such boring thread sir

sorry.

[Shirai]

omg wtf is this math about PW how dare you bring math here

this cant be allowed

This. ^

[アイドンノウくん]

If the expected average amount of hits is 6.888, with a 50% critrate, then the expected amount of crits would be 0.5*6.888 = 3.444

A slight alteration of the formula will find this one too. Rather than counting hits it'll count crits only.

Let E(m) be the expected amount of crits remaining to reach n starting at a sum of m

E(n) = 0
E(n-1) = 0.5
E(n-x) = (1/2)*(E(x-1)) + (1/2) * (1+E(x-2))

E(n-10) = 3.444

Yes, you are right. A short cut way to do this is using 0.5*6.8887 which gives 3.4443.
There are actually a number of ways to solve this problem, either by hand or by computer. 

[アイドンノウくん]

Too many replies for such boring thread sir

sorry.

Lol. I actually find it interesting, but I am the one who started the thread, so it makes sense. 

[Resolve]

lol. So much smart ppl.

yes much smart very intelligent