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4/6 = 2/3

8/12 =2/3

This proves that 2 meltaguns have the same chance of hitting once at a target, than one meltagun.

Is this wrong?

your meant to add the fractions

2/3 for one melta

2/3 + 2/3 for 2 melta

4 possible sides can hit, 2 possible sides will deny the hit. The total is 6.

4/6

With two dice, there are 8 possibilities of a successful hit, 4 of a miss.

8/12


If you add 2/3 and 2/3 you get 4/3. That's a 1.3 (period 3) chance of hitting... meaning about 133.333333 %... =.=

Your math is off.
Firing two meltaguns does not grant you a 133% chance of hitting, since it is possible to roll a 1 or a 2 on both dice.
2/3 (or 8/12) is the chance of both meltaguns hitting (I think).
The chance of at least one of them hitting is 2/3 + 2/3*1/3, or 8/9.

... No, you are not correct

first common sense, having 2 chances to hit is better than having 1 chance to hit. Now for some math. This will take a while so I'm totally going to get ninja'd like five times (12:48 typing)

ballistic skill 4 gives us a 2/3 chance of hitting.
So the odds of 1 melta gun hitting is 2/3.

Also the addtion is correct, think about it like this, if you fire 2 shots you'll expect to land 1.33333333 hits.
your options are
0 hit (1/3*1/3) = 0.111111111
1 hits (2/3*1/3)*2 = .444444444
2 hit (2/3*2/3) = .44444444444

statistically this will work out so that you can expect
2*0.444 + 1*0.444+ 0*0.111 or 1.33333 hits.
Obviously you can't get that many hits in a turn, so what this means is that over the course of many turns, if you totaled up the number of hits from 2 melta guns and divided by the number of turns, thats what it would average into.

More specifally to what you want. In order to find out the odds of getting at least one hit from 2 melta guns we must figure out the odds of it not happening 2 times in a row, and subtract from one.

So we figure out the odds of it missing,

(1-2/3)

We are firing 2 shots!

(1-2/3)^2

So thats the odds of it missing twice in a row. so the odds of getting at least 1 hit is

1-(1-2/3)^2 = .888888

This is much better than 2/3...

akaean is correct.

http://en.wikipedia.org/wiki/Combination
http://en.wikipedia.org/wiki/Permutation
http://en.wikipedia.org/wiki/Probability

the three fundamental bases of math-hammer

akaean wrote:Also the addtion is correct, think about it like this, if you fire 2 shots you'll expect to land 1.33333333 hits.
your options are

Landing 1.3333 is impossible.

akaean wrote:statistically this will work out so that you can expect 1.33333 hits. Obviously you can't get that many hits, so what this means is that over the course of many turns, if you totaled up the number of hits from 2 melta guns, thats what it would average into.

If you add up the above options you will get 9/9, which is 100%, for obvious reason. If you totaled up the number of hits from two meltaguns, it would average EXACTLY with the number of hits of one meltagun, and equal the average of one million meltaguns.


If you add the number of dice, the possibility of it hitting does not change, since you are not changing anything.

You are adding 4 possible sides of it hitting, but you are also adding 2 possible sides of it not hitting, therefore the possibility stays the same. 8/12 = 4/6

wrong.

you need a table like the one below


---------1----------2------------3----------4----------5-----------6

1----1, 1--------1, 2--------1, 3------1, 4-------1, 5------1, 6

2----2, 1--------2, 2--------2, 3------2, 4-------2, 5------2, 6

3----3, 1--------3, 2--------3, 3------3, 4-------3, 5------3, 6

4----4, 1--------4, 2--------4, 3------4, 4-------4, 5------4, 6

5----5, 1--------5, 2--------5, 3------5, 4-------5, 5------5, 6

6----6, 1--------6, 2--------6, 3------6, 4-------6, 5------6, 6


this shows all the possable results when rolling 2 dice.


the possability of 2 BS4 melta gun's shots are

11% chance of both missing

44.4% chance of only 1 hitting

and a 44.4% chance of both hitting


a single melta guns chance of hitting is 33% miss, 66% chance of a hit.

2 melta guns have a 89% chance of getting at least one hit.

so doubling the number of melta guns increases your hit chance by 23%

The equations I gave you were correct.

I'm going to try explaining the expected results again

Obviously each melta gun will have a 2/3 chance of hitting, What you are stating however is that by firing both melta guns at a tank you have exactly the same chances of landing a hit as if you had only fired a single melta gun.

That is counter intuitive, and the mathhammer as I detalied above explains why. Lets shoot some melta guns for 6 turns with statistical average for accuracy

(gun one, gun 2)
Turn 1 (miss, hit) = 1 melta hit
Turn 2 (hit, miss) = 1 melta hit
Turn 3 (hit, hit) = 2 melta hits
Turn 4 (miss, miss) = 0 melta hits
Turn 5 (hit, hit) = 2 melta hits
Turn 6 (hit, hit) = 2 melta hits

gun one hit 4 times and missed 2 (2/3)
gun two hit 4 times and missed 2 (2/3)

there were 8 total melta hits, over 6 turns = 1.3333333

So over the course of (infinite) number of turns 2 melta guns will give you an average of 1.33333 hits per turn. This in no way guarentees anything (indeed turn 1+2 had only a single hit while turn 4 had 0 hits!) the 1.33333 is merely the total number of hits / the total number of turns.

As for the odds of getting at least one hit

1-(1-2/3)^2 is the correct forumla for the reasons I detailed above.

EDIT:

Nulipuli2 wrote:If you add up the above options you will get 9/9, which is 100%, for obvious reason. If you totaled up the number of hits from two meltaguns, it would average EXACTLY with the number of hits of one meltagun, and equal the average of one million meltaguns.

Of course what you are forgetting is that we need to multiply the number of hits by each part of the equation, which means it no longer adds up to 1!
.44444+ .44444+ .11111= 1 Yes this is true however when we want to number of expected hits we are multiplying each fraction by the hits they give

2 hits * .444 odds + 1 hit * .444 odds + 0 hits * .111 odds = 1.333333 hits on AVERAGE from 2 melta guns....

Makes sense now.