Hearthpwn

Sorry folks, I'm going to disagree! The probability of Avenging Wrath killing a lone 4x minion is 46%. 

Bear with me.. Here's the grid of possibilities. 0 is a hit to the hero face. X is a hit to the 4x minion.

00000000 There is 1 way to have all hits to the hero face

X0000000 0X000000 00X00000 000X0000 0000X000 00000X00 000000X0 0000000X
8 ways to make 1 hit to the 4x minion

XX000000 0XX00000 00XX0000 000XX000 0000XX00 00000XX0 000000XX
X0X00000 0X0X0000 00X0X000 000X0X00 0000X0X0 00000X0X
X00X0000 0X00X000 00X00X00 000X00X0 0000X00X
X000X000 0X000X00 00X000X0 000X000X
X0000X00 0X0000X0 00X0000X
X00000X0 0X00000X
X000000X

28 ways to make 2 hits to the 4x minion

XXX00000 00000XXX
XX0X0000 0000X0XX
X0XX0000 0000XX0X
0XXX0000 0000XXX0

XX00X000 XX000X00 XX0000X0 XX00000X
X0X0X000 X0X00X00 X0X000X0 X0X0000X
X00XX000 X00X0X00 X00X00X0 X00X000X
0XX0X000 0XX00X00 0XX000X0 0XX0000X
0X0XX000 0X0X0X00 0X0X00X0 0X0X000X
00XXX000 00XX0X00 00XX00X0 00XX000X

X000XX00 X000X0X0 X000X00X X0000XX0 X0000X0X X00000XX
0X00XX00 0X00X0X0 0X00X00X 0X000XX0 0X000X0X 0X0000XX
00X0XX00 00X0X0X0 00X0X00X 00X00XX0 00X00X0X 00X000XX
000XXX00 000XX0X0 000XX00X 000X0XX0 000X0X0X 000X00XX

56 ways to make 3 hits on a 4x minion,, not quite dead!

XXXX0000 

XXX0X000 XXX00X00 XXX000X0 XXX0000X
XX0XX000 XX0X0X00 XX0X00X0 XX0X000X
X0XXX000 X0XX0X00 X0XX00X0 X0XX000X
0XXXX000 0XXX0X00 0XXX00X0 0XXX000X

XX00XX00 XX00X0X0 XX00X00X XX000XX0 XX000X0X XX0000XX
X0X0XX00 X0X0X0X0 X0X0X00X X0X00XX0 X0X00X0X X0X000XX
X00XXX00 X00XX0X0 X00XX00X X00X0XX0 X00X0X0X X00X00XX
0XX0XX00 0XX0X0X0 0XX0X00X 0XX00XX0 0XX00X0X 0XX000XX
0X0XXX00 0X0XX0X0 0X0XX00X 0X0X0XX0 0X0X0X0X 0X0X00XX
00XXXX00 00XXX0X0 00XXX00X 00XX0XX0 00XX0X0X 00XX00XX

X000XXX0 X000XX0X X000X0XX X0000XXX
0X00XXX0 0X00XX0X 0X00X0XX 0X000XXX
00X0XXX0 00X0XX0X 00X0X0XX 00X00XXX
000XXXX0 000XXX0X 000XX0XX 000X0XXX

0000XXXX

80 ways to make 4 hits on a 4x minion.. he's dead.

Now here's where your math is wrong. After the 4x minion is hit for the fourth time.. there is no further possibility of a fifth hit. If it were possible to hit him 5,6,7 or 8 times then you would include these possibilities in creating your simple fraction of (Outcomes where 4x minion dies/All possible outcomes = probability). This is where you are getting the 163/256 = 63.67% from and its just wrong. because 163 possibilities don't exist where he gets hit 4 times. Only 80 possibilities exist where he gets hit four times, because he cannot get hit 5, 6, 7 or 8 times.

There are only in fact 173 (1+8+28+56+80) possible outcomes. 80 of them he gets hit four times. 
80/173 = .462  A 46% chance a 4x minion will die from a 50/50 chance sequence (and SEQUENCE is the key word) of 8 hits.

Quote from equinox999 »

There are only in fact 173 (1+8+28+56+80) possible outcomes. 80 of them he gets hit four times. 

80/173 = .462  A 46% chance a 4x minion will die from a 50/50 chance sequence (and SEQUENCE is the key word) of 8 hits.

This is looking better and better to me. 

Okay will someone give me the math on Arcane Missiles ?

Would like to know the chances of killing a 3/2 when it's the only thing on board. I assume it's not just 50% considering the rest of this.

000 - Arcane missiles hits only hero
X00 0X0 00X - hits one time on 3/2 minion
XX0 X0X 0XX - hits twice on 3/2 minion
XXX is not possible so it's excluded.
3/7 possibilities he's dead. That's 43% chance.. not great.

jklsa.. in only 1/7 possibilities after killing the minion in the first two shots XX0 does the probability of hitting the hero change for the third shot. This only affects the probability of how many hits go to the hero's face, not the minion don't you think? It would stay the same as whether it is already dead or not has no effect on its own "death probability" relative to the third shot. It's a cool question.

Quote from VVatsonator »

Okay will someone give me the math on Arcane Missiles ?

Would like to know the chances of killing a 3/2 when it's the only thing on board. I assume it's not just 33.3333% considering the rest of this.

It's exactly 50%. If the first two shots kill the 3/2, the last will hit face with certainty; this scenario happens with probability 0.5^2 = 0.25. The other two possibilities are MHM and HMM (M - minion is hit; H - hero is hit) each occurring with probability 0.5^3 = 0.125. Adding this up gives 50%.

Edit: equinox999, the probability of all 7 possible outcomes are not weighted equally. The last one (XX0) has a greater chance to occur because once XX occurs the last shot is guaranteed to hit the hero.

Quote from Elitesparkle »

Quote from VVatsonator »

Okay will someone give me the math on Arcane Missiles ?

Would like to know the chances of killing a 3/2 when it's the only thing on board. I assume it's not just 33.3333% considering the rest of this.

P(X=2)=(C(3,2)+C(3,3))/D'(2,3)=50%

I know this it totally correct, just wondering what does C(x,y) mean?? Can you explain how you calculate for example C(3,2), and what is D´(x,y)??
I read your post where you kind of explained it, but it is still not clear what each term means.

Quote from Elitesparkle »

Quote from VVatsonator »

Okay will someone give me the math on Arcane Missiles ?

Would like to know the chances of killing a 3/2 when it's the only thing on board. I assume it's not just 33.3333% considering the rest of this.

P(X=2)=(C(3,2)+C(3,3))/D'(2,3)=50%

I meant to say 50%, lol. But I think its not just 50% considering everything we've seen here.

It could be FFF, FFM, FMF, MFF, MMF, MFM, FMM, but not MMM

Quote from equinox999 »

000 - Arcane missiles hits only hero
X00 0X0 00X - hits one time on 3/2 minion
XX0 X0X 0XX - hits twice on 3/2 minion
XXX is not possible so it's excluded.
3/7 possibilities he's dead. That's 43% chance.. not great.

jklsa.. in only 1/7 possibilities after killing the minion in the first two shots XX0 does the probability of hitting the hero change for the third shot. This only affects the probability of how many hits go to the hero's face, not the minion don't you think? It would stay the same as whether it is already dead or not has no effect on its own "death probability" relative to the third shot. It's a cool question.

That doesn't sound right. I think you are making the same mistake that someone in the thread earlier was making. There are 3 possible ways that it could die. Which is either (H = hit and M = Miss) 

• H H M - which has a 1/4 chance of happening (1/2 then 1/2 again)
• H M H - which has a 1/8 chance of happening (1/2 then 1/2 then 1/2)
• M H H - which has a 1/8 chance of happening (1/2 then 1/2 then 1/2)

Which means there is a 4/8 chance of a minion with 2 health dying from arcane missiles or a 50% chance.

Now as you can see not all of them are equally likely to occur so finding the possible outcomes and then dividing it by the total amount of outcomes won't get you the correct answer.

Quote from equinox999 »

000 - Arcane missiles hits only hero
X00 0X0 00X - hits one time on 3/2 minion
XX0 X0X 0XX - hits twice on 3/2 minion
XXX is not possible so it's excluded.
3/7 possibilities he's dead. That's 43% chance.. not great.

jklsa.. in only 1/7 possibilities after killing the minion in the first two shots XX0 does the probability of hitting the hero change for the third shot. This only affects the probability of how many hits go to the hero's face, not the minion don't you think? It would stay the same as whether it is already dead or not has no effect on its own "death probability" relative to the third shot. It's a cool question.

This sounds right, thanks

Quote from VVatsonator »

Quote from equinox999 »

000 - Arcane missiles hits only hero
X00 0X0 00X - hits one time on 3/2 minion
XX0 X0X 0XX - hits twice on 3/2 minion
XXX is not possible so it's excluded.
3/7 possibilities he's dead. That's 43% chance.. not great.

jklsa.. in only 1/7 possibilities after killing the minion in the first two shots XX0 does the probability of hitting the hero change for the third shot. This only affects the probability of how many hits go to the hero's face, not the minion don't you think? It would stay the same as whether it is already dead or not has no effect on its own "death probability" relative to the third shot. It's a cool question.

This sounds right, thanks

It might sound right on the surface, but it's not how probability works. If you scroll up you will see that it is a 50% chance.

Quote from Tehstool »

Quote from VVatsonator »

Quote from equinox999 »

000 - Arcane missiles hits only hero
X00 0X0 00X - hits one time on 3/2 minion
XX0 X0X 0XX - hits twice on 3/2 minion
XXX is not possible so it's excluded.
3/7 possibilities he's dead. That's 43% chance.. not great.

jklsa.. in only 1/7 possibilities after killing the minion in the first two shots XX0 does the probability of hitting the hero change for the third shot. This only affects the probability of how many hits go to the hero's face, not the minion don't you think? It would stay the same as whether it is already dead or not has no effect on its own "death probability" relative to the third shot. It's a cool question.

This sounds right, thanks

It might sound right on the surface, but it's not how probability works. If you scroll up you will see that it is a 50% chance.

So why is it not a 50% chance to kill a 4hp minion on board with Avenging Wrath?

Quote from Elitesparkle »

Quote from Moonkonnig »

Quote from Elitesparkle »

Quote from VVatsonator »

Okay will someone give me the math on Arcane Missiles ?

Would like to know the chances of killing a 3/2 when it's the only thing on board. I assume it's not just 33.3333% considering the rest of this.

P(X=2)=(C(3,2)+C(3,3))/D'(2,3)=50%

I know this it totally correct, just wondering what does C(x,y) mean?? Can you explain how you calculate for example C(3,2), and what is D´(x,y)??
I read your post where you kind of explained it, but it is still not clear what each term means.

Sure.

C(n,k)=n!/(n-k)!/k!
C(3,2)=3!/(3-2)!/2!=3*2/2=3

D'(n,k)=n^k
D'(3,2)=3^2=9

Do you need more?

Ok, so those are the formulas.

It´s clear that D´(2,3) is all the possible outcomes. It´s still unclear what exactly is C(3,2)=3 and C(3,3)=1=C(3,0),, so for example why did you chose C(3,3) and not C(3,0) which give the same answer, what does the 3 represent??

EDIT: corrected C(3,3)=1=C(3,1) to C(3,3)=1=C(3,0)

Agreed.

00 0X X0 XX

After 2 shots
1/4 he's not hit .25
2/4 he's hit once .50
1/4 he's dead already .25

After the third shot
of the 1/4 not hit - 50% will get hit (.5X.25) = .125 hit once .125 not hit
of the 2/4 hit once 50% will get hit (.5x.5) = .25 hit twice .25 hit once
of the 1/4 hit twice already - no change 100% (1x.25) .25 hit twice
total
.125 not hit
.375 hit once
.5 hit twice

Quote from VVatsonator »

So why is it not a 50% chance to kill a 4hp minion on board with Avenging Wrath?

Because with 8 missiles and a 4 health minion there are a lot more ways that the missiles could go compared to 3 missiles with a 2 health minion. You aren't just doubling it because the missile count is different.

Quote from Tehstool »

Quote from VVatsonator »

So why is it not a 50% chance to kill a 4hp minion on board with Avenging Wrath?

Because with 8 missiles and a 4 health minion there are a lot more ways that the missiles could go compared to 3 missiles with a 2 health minion. You aren't just doubling it because the missile count is different.

Right, right, thanks for the clarification. Believe it or not I have taken statistics at university, oh my how I have forgotten everything XD

Quote from VVatsonator »

Quote from Tehstool »

Quote from VVatsonator »

So why is it not a 50% chance to kill a 4hp minion on board with Avenging Wrath?

Because with 8 missiles and a 4 health minion there are a lot more ways that the missiles could go compared to 3 missiles with a 2 health minion. You aren't just doubling it because the missile count is different.

Right, right, thanks for the clarification. Believe it or not I have taken statistics at university, oh my how I have forgotten everything XD

You're not completely off base though. If Avenging Wrath had 7 missiles instead of 8, the chance to kill a lone 4 hp minion is also 50%.

Quote from VVatsonator »

Right, right, thanks for the clarification. Believe it or not I have taken statistics at university, oh my how I have forgotten everything XD

Yeah sorry. I don't think I explained it well.

I have calculated the exact probability of killing a lone 4 health minion with avenging wrath, taking into account that the hit chance to the hero becomes 100% when the minion has taken four damage:

Result=0.6171875 = 61.71875%

Quick "resume" of the calculation:

P=( 0.5^8 )*32 + (0.5^7 )*19 + (0.5^6)*10 + (0.5^5)*4 + (0.5^4)*1 = 0.6171875

Here are results of my code applied to simple examples. Maybe this can help understand them for some of you.

Arcane Missiles, one 2hp minion dies - 50%

[0,1,1] 12.5%
[1,0,1] 12.5%
[1,1,0] 25%

Avenging Wrath, one 4 hp minion - 63.671875%

[0,0,0,0,1,1,1,1] 0.390625%
[0,0,0,1,0,1,1,1] 0.390625%
[0,0,0,1,1,0,1,1] 0.390625%
[0,0,0,1,1,1,0,1] 0.390625%
[0,0,0,1,1,1,1,0] 0.78125%
[0,0,1,0,0,1,1,1] 0.390625%
[0,0,1,0,1,0,1,1] 0.390625%
[0,0,1,0,1,1,0,1] 0.390625%
[0,0,1,0,1,1,1,0] 0.78125%
[0,0,1,1,0,0,1,1] 0.390625%
[0,0,1,1,0,1,0,1] 0.390625%
[0,0,1,1,0,1,1,0] 0.78125%
[0,0,1,1,1,0,0,1] 0.390625%
[0,0,1,1,1,0,1,0] 0.78125%
[0,0,1,1,1,1,0,0] 1.5625%
[0,1,0,0,0,1,1,1] 0.390625%
[0,1,0,0,1,0,1,1] 0.390625%
[0,1,0,0,1,1,0,1] 0.390625%
[0,1,0,0,1,1,1,0] 0.78125%
[0,1,0,1,0,0,1,1] 0.390625%
[0,1,0,1,0,1,0,1] 0.390625%
[0,1,0,1,0,1,1,0] 0.78125%
[0,1,0,1,1,0,0,1] 0.390625%
[0,1,0,1,1,0,1,0] 0.78125%
[0,1,0,1,1,1,0,0] 1.5625%
[0,1,1,0,0,0,1,1] 0.390625%
[0,1,1,0,0,1,0,1] 0.390625%
[0,1,1,0,0,1,1,0] 0.78125%
[0,1,1,0,1,0,0,1] 0.390625%
[0,1,1,0,1,0,1,0] 0.78125%
[0,1,1,0,1,1,0,0] 1.5625%
[0,1,1,1,0,0,0,1] 0.390625%
[0,1,1,1,0,0,1,0] 0.78125%
[0,1,1,1,0,1,0,0] 1.5625%
[0,1,1,1,1,0,0,0] 3.125%
[1,0,0,0,0,1,1,1] 0.390625%
[1,0,0,0,1,0,1,1] 0.390625%
[1,0,0,0,1,1,0,1] 0.390625%
[1,0,0,0,1,1,1,0] 0.78125%
[1,0,0,1,0,0,1,1] 0.390625%
[1,0,0,1,0,1,0,1] 0.390625%
[1,0,0,1,0,1,1,0] 0.78125%
[1,0,0,1,1,0,0,1] 0.390625%
[1,0,0,1,1,0,1,0] 0.78125%
[1,0,0,1,1,1,0,0] 1.5625%
[1,0,1,0,0,0,1,1] 0.390625%
[1,0,1,0,0,1,0,1] 0.390625%
[1,0,1,0,0,1,1,0] 0.78125%
[1,0,1,0,1,0,0,1] 0.390625%
[1,0,1,0,1,0,1,0] 0.78125%
[1,0,1,0,1,1,0,0] 1.5625%
[1,0,1,1,0,0,0,1] 0.390625%
[1,0,1,1,0,0,1,0] 0.78125%
[1,0,1,1,0,1,0,0] 1.5625%
[1,0,1,1,1,0,0,0] 3.125%
[1,1,0,0,0,0,1,1] 0.390625%
[1,1,0,0,0,1,0,1] 0.390625%
[1,1,0,0,0,1,1,0] 0.78125%
[1,1,0,0,1,0,0,1] 0.390625%
[1,1,0,0,1,0,1,0] 0.78125%
[1,1,0,0,1,1,0,0] 1.5625%
[1,1,0,1,0,0,0,1] 0.390625%
[1,1,0,1,0,0,1,0] 0.78125%
[1,1,0,1,0,1,0,0] 1.5625%
[1,1,0,1,1,0,0,0] 3.125%
[1,1,1,0,0,0,0,1] 0.390625%
[1,1,1,0,0,0,1,0] 0.78125%
[1,1,1,0,0,1,0,0] 1.5625%
[1,1,1,0,1,0,0,0] 3.125%
[1,1,1,1,0,0,0,0] 6.25%

crippled Avenging Wrath that shots only 7 times (for when they print "spell damage -1" ;-)), one 4hp minion - 50%

[0,0,0,1,1,1,1] 0.78125%
[0,0,1,0,1,1,1] 0.78125%
[0,0,1,1,0,1,1] 0.78125%
[0,0,1,1,1,0,1] 0.78125%
[0,0,1,1,1,1,0] 1.5625%
[0,1,0,0,1,1,1] 0.78125%
[0,1,0,1,0,1,1] 0.78125%
[0,1,0,1,1,0,1] 0.78125%
[0,1,0,1,1,1,0] 1.5625%
[0,1,1,0,0,1,1] 0.78125%
[0,1,1,0,1,0,1] 0.78125%
[0,1,1,0,1,1,0] 1.5625%
[0,1,1,1,0,0,1] 0.78125%
[0,1,1,1,0,1,0] 1.5625%
[0,1,1,1,1,0,0] 3.125%
[1,0,0,0,1,1,1] 0.78125%
[1,0,0,1,0,1,1] 0.78125%
[1,0,0,1,1,0,1] 0.78125%
[1,0,0,1,1,1,0] 1.5625%
[1,0,1,0,0,1,1] 0.78125%
[1,0,1,0,1,0,1] 0.78125%
[1,0,1,0,1,1,0] 1.5625%
[1,0,1,1,0,0,1] 0.78125%
[1,0,1,1,0,1,0] 1.5625%
[1,0,1,1,1,0,0] 3.125%
[1,1,0,0,0,1,1] 0.78125%
[1,1,0,0,1,0,1] 0.78125%
[1,1,0,0,1,1,0] 1.5625%
[1,1,0,1,0,0,1] 0.78125%
[1,1,0,1,0,1,0] 1.5625%
[1,1,0,1,1,0,0] 3.125%
[1,1,1,0,0,0,1] 0.78125%
[1,1,1,0,0,1,0] 1.5625%
[1,1,1,0,1,0,0] 3.125%
[1,1,1,1,0,0,0] 6.25%

This one is interesting. Killing one 2hp minion with Arcane Missiles is 50% but killing two 1 hp minions is 61.(1)%.

[0,1,2] 5.555555555555556%
[0,2,1] 5.555555555555556%
[1,0,2] 8.333333333333334%
[1,2,0] 16.666666666666668%
[2,0,1] 8.333333333333334%
[2,1,0] 16.666666666666668%
61.111111111111114%

Quote from LightsOutAce »

The math doesn't take long at all; with one minion on board it is binomial distribution. To simplify the formula since I don't know how to type "m chose n,": Chance of killing lone 4 health minion = 1 - P(5 or more hits to the face) = 1 - .5^8 -  8 * .5^8 - 28 * .5^8 - 56 * .5^8 = 63.7%

*It is slightly different in that the minion can only be hit x times, where x is its health, but for purposes of determining probability of killing a lone minion you can use binomial.

This. I don't even know why the discussion continued after this post.