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Mitthrawdo
20-10-2009, 00:52
How do you factor in rerolls of the dice into figuring out the chance to kill something?

For example, take a Dreadnought with two twin-linked autocannons. How do I calculate the odds of destroying an enemy Rhino (AV11), and it is not obscured.

CrownAxe
20-10-2009, 01:08
you just apply the normal roll to the chance to fail and add it to the normal results

Exp. 1 TL single shot at BS4

to hit is normally 2/3, so take the chance to fail (1/3) and apply the roll (1/3*2/3=2/9) and add it to the normal 2/3 which equals 8/9 chance to hit

Bunnahabhain
20-10-2009, 01:16
Easy. This is just basic probability, as taught to 12 year olds.

Probabality of a kill = Chance to hit x change to penetrate/glancex chance of high enough damage roll

Chance to hit = 8/9
4/6 shots hit normally, and the re-roll causes 4/6 of the misses to hit.
therefore chance to hit = 4/6 + 2/6 x 4/6 = 8/9

Change to penetrate.
S7 vs AV11. 1-3 bounce off, 4s glance, 5-6 penetrate.

Pen = 4 x 8/9 x 1/3 = 32/ 27
Glance = 4 x 8/9 x 1/6 = 16/27

We're not going to get enough hits for the weapon destroyed and immobilsed to stack, so we can ignore this

So we're now down to just the penetrating hits.

chance of destruction = number of penetraiting hitsx change of desroyed result
= 32/27 x 1/3
=32/81 = 39.4%

This is the quick and basic version. More deatiled statistical techniques are available, but they are over the top here.

solkan
20-10-2009, 01:18
That depends. Are you trying to do branching analysis, are are average or expected values good enough?

If all you want is an expected number of hits for a twin linked weapon, I think that T, the chance for the twin linked shot to hit, is just
T=P+(1-P)*P=2P-P*P where P is the probability of the regular shot hitting.

For example, if a shot hits on 3+, P=2/3. So, the probability of the twin linked shot is 2P-P*P=4/3-4/9=12/9-4/9=8/9.

If you have more shots, just multiply the number of shots times T and that'll provide the expected number of hits.

Mitthrawdo
20-10-2009, 01:37
Telion Mathhammer:
His chance of hitting would be 86.1%
(5/6)+(1/6)*(1/6)
Would that be correct?

Bunnahabhain
20-10-2009, 01:43
nope.
chance of hitting = 5/6 + 1/6 x 5/6 = 35/36

The other way of doing it is:
1 - ( change of missing x change of missing) = chance of hitting.

EDIT. Doh. My maths is right, memeroy of telions stats not so...

Mitthrawdo
20-10-2009, 02:04
Telion is BS6 and his weapon isn't twin-linked.

PhalanxLord
20-10-2009, 02:07
The simplest way?

Here's a simple formula: 1-(1-P)^N= the chance of something happening 1 or more times.

P is the probability of the event and N is the number of times it has a chance of occuring.

So for your dual twin-linked autocannons: They have a 8/9 chance of hitting, a 1/3 chance of penning, and a 1/3 chance of destroying for a total of 8/81 chance to destroy the rhino per shot. You have 4 shots so the total probability to destroy the rhino is 34%.

Edit: Mitthrawdo is correct with Tellion. Its 5/6+1/36= 31/36.

darker4308
20-10-2009, 02:09
http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html

Alessander
20-10-2009, 03:43
we're such nerds. i ♥ these forums :D

ntin
20-10-2009, 05:17
Some reason I cannot link an image?

http://img17.imageshack.us/img17/3214/dieodds.png

I cooked up a quick chart to help the less math inclined people.

It is a given that there is a 1/6 chance of rolling of rolling any face on a single die or 6 unique combinations.

When there are two die in use means 36 unique combinations when rolling two die together (1/6 * 1/6 = 1/36).

For a model with BS:6 to miss it would have to roll a 1 on the first die and then a 6 on the second die to hit. The highlighted combinations are results that will still miss with the reroll. We can see there are 5 ways that a BS:6 model can miss so 36-5 = 31 and we can divide that by the total number of combinations 31/36 to get the probability of a BS:6 model to hit.

Tsear
20-10-2009, 06:11
As everyone has pointed out, it's easy to figure out the odds of you hitting. If you want an expected value of how much you will destroy, just take the odds you'll destroy it and multiply it by your shots. If you want the actual probability, find the probability P that you'll destroy it, subtract it from one, take it to the power N (number of shots), then subtract one from that again. I.E.

Probability of destruction = 1-(1-P)^N

As Phalanx Lord pointed out.

we're such nerds. i ♥ these forums

This is extremely easy math. Bunna is right when he says 12 year olds can do this. Take some responsibility for your education, guys. Learn some basic math, some basic physics, basic philosophy...

kardar233
20-10-2009, 06:33
I'm in grade 10, and regularly calculate combats in both 40K and Fantasy; I often run the numbers on character matchups just for fun.

Don't let anyone tell you you can't do it; it's easy once you get the hang of it. I've done a full Orks versus Zerkers points match, each round all the way to annihilation in less than 5 minutes calculator-less.

Lord Solar Plexus
20-10-2009, 13:26
This is extremely easy math. Bunna is right when he says 12 year olds can do this. Take some responsibility for your education, guys. Learn some basic math, some basic physics, basic philosophy...

Yes, really, how does anyone dare ask such extremely easily answered questions here instead of learning the answers! Some posters will be extremely easily annoyed by having to explain the obvious over and over!

Kulgur
20-10-2009, 15:01
Don't let anyone tell you you can't do it; it's easy once you get the hang of it. I've done a full Orks versus Zerkers points match, each round all the way to annihilation in less than 5 minutes calculator-less.

You can do it, but it's folly to rely on it 100% as what you're getting is what will happen on average, just because that's the most likely outcome, doesn't mean that's the outcome you will get

CrownAxe
20-10-2009, 17:06
You can do it, but it's folly to rely on it 100% as what you're getting is what will happen on average, just because that's the most likely outcome, doesn't mean that's the outcome you will get

The point of mathhammer isn't to make predictions, it's to calculate typical efficiency of a unit and compare it to the efficiency of other units

It's stupid people who think otherwise

Ianos
20-10-2009, 17:36
The point of mathhammer isn't to make predictions, it's to calculate typical efficiency of a unit and compare it to the efficiency of other units

This only happens though under very specific circumstances. I think of math-hammer as more of a risk/expectancy calculator. It provides me with the knowledge, of which unit should engage what and when.