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So I have seen comments in threads and heard comments in casts that Avenging Wrath has greater than 50% chance to kill a 4 health minion if it is alone on the board. Yesterday playing against the popular new Paladin deck I had an opponent play a wrath when I had a 3 health and a 4 health minion on board and it killed both of them.
Can someone explain to me how this math works?
Edit: For those new to the thread, turns out there is about a 63% chance of killing the 4 health minion.
It has to do with the fact that the chance of 5 or more missiles hitting the hero is more unlikely than you would think. The 4-4 scenario is much more likely than a more extreme scenario like 2-6.
Killing both my minions in the example above remains a real fluke.
Kidding. This is RNG, there's no way to predict that.
I understand that, but why is the probability greater than 50%? (or is it?
For every hit of the avenging wrath there is a chance it will hit you or one of your minions. It would take a lot of time to calculate as you'd have to do a calculation for every possible outcome and then add together the ones that kill both minions. It would just simply take too long. If someone was really passionate about this and they had some time on their hands they could do it.
I don't think it works that way.
The likelihood never changes when there are two targets.
1st hit - 50/50
2nd hit - 50/50
3rd hit - 50/50
.
.
.
and so on.
7 hits could happen to the first target, but on the 8th hit the chance is still 50/50.
Think of it this way. If you have only one creature on the bored avenging wrath is like flipping a coin on each hit until the creature dies. So if you want to calculate the probability of killing one creature with avenging wrath it would be the same as the probability of getting at least that many heads in 8 flips of a coin.
I watched Hafu's stream and Avenging Wrath killed both minions on board when the combined total HP was 8.
The math.
There's no greater or lesser chance of it hitting a given target.
Out of the 8 hits 7 of them could hit the first target, but on the 8th hit there's no greater chance of it hitting the 2nd. It's still 50/50.
Sorry if I wasn't clear, I was actually responding to the OP and the idea that there was a greater than 50% chance to kill a 4 health minion when it's the only one on the board.
Right, but for a specific outcome to happen you'd have to multiply the chance of it happening and then add all of the ones with the specific outcome together. So if you had 3 1/2 chances for something to happen it would be (1/2) ^ 3 = 0.125 rather than just 50%.
If the results of the prior hits affected the rest of them, that would be true.
In this case it isn't.
Each hit, the chance is 50/50.
The example I gave is proof of that. 7 of 8 hits hit the 1st target. The chance for it to hit the 1st target again is not miniscule. It's 50/50.
I think we are misunderstanding each other. I was going on the assumption that Semioteric was wanting to know what the probability of killing both of his minions with avenging wrath would be. Which is simply just not as easy as 1/2 or 1/3, etc. It seems like you are just saying the individual probability for one missile hitting 1/2 targets is 50%. My apologies there.
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.
I'm using this game to teach my 5 year old arithmetic.
If you're ever playing me and I'm taking a long time to make my move, that's why. :P
@betterlivinthruchemistry
8 hits against two targets, randomly determined between the two, and one of which only can only take 4 hits. After that target is 'dead' the other hits will all go to the other target if there are any left.
There are actually 5 conditions (0 hits to 4 on the Hero) which DOES make the math work out at greater than a 50% chance for the minion to die, you are correct. For those still confused, let me show you (and I'm at work, so forgive me if I'm posting this after someone else already has):
M | H (Minion vs Hero)
0 - 8
1 - 7
2 - 6
3 - 5
4 - 4 Hits
x - 5
x - 6
x - 7
x - 8
As you can see, there are 5 conditions of the 9 possible combinations where the damage to the minion is 4 or greater, resulting in minion death.
5 out of 9 is 55.5 (repeating, of course) percent.
The way it works with more minions, such as the 4 and 3 health scenario described in the OP, is similar to a multinomial distribution, but again, slightly different because of the cap on damage to minions. Because Avenging Wrath cannot "overkill" a minion, one a minion has been hit = to its health the distribution reverts to a binomial since there are only two targets.
TL; DR: Avenging Wrath with 3+ targets is tedious to calculate exactly and I don't want to program it into Excel right now.
Tempest8008, the reason it isn't 55.5% is that each of the 9 possibilities are not equally likely.
The damage can only be 4 or less than 4 because at 4 the minion is dead and can no longer take any damage.
Yes. But that also takes into account the chance the minion can take 0 hits.
0 = lives
1 = lives
2 = lives
3 = lives
4 = dies
5,6,7,8 = dies
4 lives
5 dies
So a 55.5% chance of dying. 5 out of the 9 possible combinations equals the minion dead.
I apologize here, but that is not how probability works. You are listing the possible outcomes, and out of those possible outcomes you have come to the conclusion that in 55.5% of outcomes he dies. That doesn't necessarily mean that those outcomes are equally likely.
To answer your earlier question betterlivin, there is an exactly 50% chance to kill with 7 shots.
It feels like avenging wrath puts minions on a higher priority than heroes, but then I remember that time where I cast avenging wrath when my opponent had a 1hp minion and every shot went to face.
Each time the RNG rolls in this case there is a 50% chance of hitting either target.
There is no probability calculation to make here.
If I toss a coin 1000 times and it comes up heads 999 of them, if you were to ask me the odds on my next toss of it comes up heads again I would HAVE to say 50/50. The former results do not impact the next toss. You're multiplying the chances of things happening together here, but the chances do NOT multiply together. They are distinct events.
Of the 9 possible conditions (0 hits to 8 hits) 5 of them (4,5,6,7,8) mean the minion dies. It does not matter if the minion is missed by the first 4 hits if the next 4 hit him, the condition is met. He can't be hit 5,6,7, or 8 times, so those would automatically go to the Hero. Better to look at it from the other direction. The minion LIVES if he is hit 0, 1, 2, or 3 times. There are only 9 possible combinations between the 2 targets and 4 of those combinations mean the minion lives.
There's no true probability calculation to be made here. With distinct 50/50 events occurring on each hit, you have to look the the number of combinations, not the likelihood of a given combination occurring, because it is irrelevant. Of the 9 possible things that can happen, 5 of them mean the minion dies. Each of those possibilities is equally likely.