Dispel Magic: Probability for the English Grad Student

[HeapThaumaturgist]

Okay ... here's my question: What's the math look like for the probabilities on a spell like Dispel Magic.

Let's take a character with 8 spells, cast by a 13th level caster, active on his person.

He gets hit with a 10th-level Targeted Dispel Magic.

The Dispeller needs a 14+ to hit a spell, so that's a 35% chance of any one spell getting hit. But what are the chances of 1-in-8 getting hit? 2? 3? 4?

We see why I never became an engineer.

--fje

 

[Hypersmurf]

The chance of spell 1 (and only spell 1) getting hit is the chance of spell 1 getting hit, times the chance of spell 2 not getting hit, times the chance of spell 3 not getting hit, etc.

35% x 65% x 65% x 65% x 65% x 65% x 65% x 65%. (or about 1.7%.)

The chance of spell 2 (and only spell 2) getting hit is the chance of spell 1 not getting hit, times the chance of spell 2 getting hit, times the chance of spell 3 not getting hit, etc.

65% x 35% x 65% x 65% x 65% x 65% x 65% x 65%. (Also about 1.7%.)

So the chance of any one (but only one) spell getting hit is 8 x 35% x 65%^7, or about 13.7%.

Similarly, the chance of exactly two spells getting hit is 28 x 35%^2 x 65%^6; three spells is 56 x 35%^3 x 65%^5, and four spells is 70 x 35%^4 x 65%^4.

Exact number of spells dispelled | Chance
0: 3.19%
1: 13.73%
2: 25.87%
3: 27.86%
4: 18.75%
5: 8.08%
6: 2.17%
7: .33%
8: .02%

-Hyp.

 

[kerbarian]

Let's take a character with 8 spells, cast by a 13th level caster, active on his person.

He gets hit with a 10th-level Targeted Dispel Magic.

The Dispeller needs a 14+ to hit a spell, so that's a 35% chance of any one spell getting hit. But what are the chances of 1-in-8 getting hit? 2? 3? 4?

As you say, the chance for each spell to be dispelled is 0.35. For exactly one spell to be dispelled, that's one success and seven failures. The chance of that is 0.35 * 0.65 ^ 7 = 0.0172. However, there are C(8, 1)* ways to arrange those successes and failures, and each one has the given probability. So the total chance for exactly one spell to be dispelled is

0.35 * 0.65 ^ 7 * C(8, 1) = 0.1373 = 14%

The probabilities for other specific numbers of spells being dispelled is:

0 -> 0.35 ^ 0 * 0.65 ^ 8 * C(8, 0) = 0.0319 = 3%
1 -> 0.35 ^ 1 * 0.65 ^ 7 * C(8, 1) = 0.1373 = 14%
2 -> 0.35 ^ 2 * 0.65 ^ 6 * C(8, 2) = 0.2587 = 26%
3 -> 0.35 ^ 3 * 0.65 ^ 5 * C(8, 3) = 0.2786 = 28%
4 -> 0.35 ^ 4 * 0.65 ^ 4 * C(8, 4) = 0.1875 = 19%
5 -> 0.35 ^ 5 * 0.65 ^ 3 * C(8, 5) = 0.0808 = 8%
6 -> 0.35 ^ 6 * 0.65 ^ 2 * C(8, 6) = 0.0217 = 2%
7 -> 0.35 ^ 7 * 0.65 ^ 1 * C(8, 7) = 0.0033 = 0%
8 -> 0.35 ^ 8 * 0.65 ^ 0 * C(8, 8) = 0.0002 = 0%

* The C(n, k) notation represents the combination n choose k.

 

[Quartz]

Ah, probability theory. The probability of any one spell being dispelled is 35% or 0.35. Therefore the probability of any one spell not being dispelled is 65% or 0.65.

So P(All dispelled) = (0.35^8) or 0.02% and P(None dispelled)=(0.65^8) or 3.19%. If your calculator won't do 8th powers, simply square three times.

So the probability of at least one spell being dispelled is 1-P(None dispelled) or 96.81% and the probability of at least one surviving is 1-P(All dispelled) or 99.98%.

Beyond this it gets interesting and I urge you to get a book on probability.

 

[Quartz]

Ah, I see others have leapt in ahead of me.

 

[HeapThaumaturgist]

Sweet. Thanks everybody.

--fje

 

[Nail]

It was the thread title that did it....ahhh, grad school days......

 

[Delta]

This calculation is called "binomial probability": http://en.wikipedia.org/wiki/Binomial_probability