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Twin-Linked Mathhammer...
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Old 11 Mar 2009, 02:42   #1 (permalink)
Kroot Shaper
 
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Default Twin-Linked Mathhammer...

So my opponent and I were wondering the other day: If you're firing a twin-linked weapon that gets two shots (Rapid-fire pulse rifle at under 12") or say you have two guys both firing twin-linked weapons at once with the same BS, can you roll two dice at once, either for one weapon's hit then the others, or for both weapon's "first shot" then roll the misses with second dice rolls? Does it matter? I'm pretty sure you can't just roll 4 dice at once and take two hits... but I think the other two ways are equivalent.

Put more succinctly:

Is it statically equivalent, when firing two shots of a twin-linked weapon at the same target, to roll

A) Two dice representing the first attempt and potential re-roll of the first "shot," counting as a "hit" if at least one of them "hits," then rolling two more dice in the same fashion to represent the second "shot."

and

B) Two dice representing the first attempt for each of the two "shots," and then re-rolling any "missed" dice (potentially none, one, or both).


And as long as they are equivalent for the two shots, does the BS matter? a 3 vs a 4? or?

I feel like "A" is equivalent to "B" in all respects statistically (likelihood of hitting twice, once, or none). On the other hand, I feel it is NOT equivalent to just roll 4 dice and count any two "hits" as your two "hits."

Also, if you add a third "shot" (Say a squad of 3 crisis suits firing twin-linked pulse rifles each) is the equivalence expandable by rolling 3 dice at once to represent first shots then re-rolling any misses, or just rolling 3 sets of 2 dice at a time? I think this is also true but don't know enough math to prove it.

Thoughts? Proof?

Thanks!
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Old 11 Mar 2009, 03:04   #2 (permalink)
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Default Re: Twin-Linked Mathhammer...

If you want to roll both the first roll and the reroll together, in order to save time, they MUST be separate sets of dice from other twin-linked weapons.

Why?

Well, let's take 2 Broadsides, firing their Railguns at a target. First, we roll them separately.

Let's say you roll 3 and 1 for the first broadside, with a 4 and 6 for the second broadside. The first broadside's railgun misses(both the first attack and reroll would miss), and the second one hits (but only once, since the other roll was the reroll IF the first one failed).

But what if those exact same results had been rolled as 4 dice, without separating the shots? You'd have 2 Railgun hits.

Suffice to say that you can do both rolls and rerolls together, but each shot MUST be separated from all others, because otherwise you might make successes out of failures.
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Old 11 Mar 2009, 03:11   #3 (permalink)
Shas'Ui
 
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Default Re: Twin-Linked Mathhammer...

For the first question, it is illegal to do anything but B. A would be if the weapon was, instead of twin linked 2, assault 4. Also, it makes a large difference percentage wise. Scenario A at BS3 should get 2 hits, while scenario B at the same BS should get 1.5 hits. The math behind this is fairly simple. 4 shots * 50% = 2 hits, While 2*(.5 + (.5(1-.5))) = 2*( .5 + (.5 * .5)) = 2*(.5 + .25) = 2 * .75 = 1.5.
For the second question, the answer is a bit more complex. Let's take BS2, it has a 33% chance of hitting, ergo a 66% chance of missing. The chance of the twin linked hit hitting is 33% (the chance of hitting) of 66% (the chance of missing). This can be done by multiplying .33 by .66 and adding the result to .33. The result is .55 or 55%, meaning gun drones with twin linked BS2 have a better chance of hitting than fire warriors at BS3 (50% chance). For twin linked BS3, it would be 50% of 50% (25%) plus 50% which equals 75%. To put it simply (kinda), the equation for twin linked accuracy is as follows.
N * (D + (D * (1-D))) = C
where D is the decimal form chance of hitting, N is the number of shots, and C is the number of hits.

Note: I think I misunderstood your second question. I thought you wanted to know how to calculate the accuracy of a twin linked BS. However, the answer I gave should allow you to figure out the difference between "C" in the twin linked formula (above) and the normal "to hit" formula (below).
N * D = C
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Old 11 Mar 2009, 03:27   #4 (permalink)
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Default Re: Twin-Linked Mathhammer...

Actually, Leatherback, his (A) situation works fine.

I misinterpreted his post too.

Yes, Grorx, they are equivalent.

No, Leatherback, his (A) situation would not statistically result in 2 hits. Look at it more closely, and you'll see there's a 1/4 chance of missing for each shot at BS 3, with a 3/4 chance of hitting (since he's specifically called the shot a hit if at least 1 of the rolls "hits", which means 2 "hits" is still 1 hit).





Edit: Let's work on that proof.

Normally, with TL weapon shots, you make the first roll. If it hits, the shot hits. If it misses, you roll to hit again. If that hits, the shot hits, if it doesn't, you miss.

The only time you miss with a TL weapon is when you miss with both die. Simple enough.

Let's say you roll using (A) scenario, with 2 die rolls at the same time.

For each die, you have the chance to hit and the chance to miss. Because we've already established that the only time a TL shot misses is if you miss both shots, and so long as we abide by the rule that those 2 die rolls count as a hit only if at LEAST 1 dice hits, those two dice accurately represent whether or not a TL shot hits.
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Old 11 Mar 2009, 03:39   #5 (permalink)
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Default Re: Twin-Linked Mathhammer...

Right. What you said, Unusualsuspect. At most with (A) or (B) you can only get 2 hits. I just wanted to make sure that, regardless of (A) or (B), you stand the statistically same chance of rolling a total of 2 hits vs 1 hit vs no hits. I guess option (B) would end up being faster occasionally, in the 25% chance you hit both on the first roll for a BS3 shot.
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Old 11 Mar 2009, 03:43   #6 (permalink)
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Default Re: Twin-Linked Mathhammer...

I mix them up when I play.

If I have 2 or more TL shots to fire, I use option (B), but if I just have a single TL shot to fire (Like a monat or Target Lock Broadside), I use option (A).

While equivalent, however, know that the game tells you to always use option (B), so if you aren't playing with your friend, its best to stick with (B) so as to avoid confusion and assertions of cheating. It really doesn't save THAT much time either way.
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Old 11 Mar 2009, 22:00   #7 (permalink)
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Default Re: Twin-Linked Mathhammer...

CBA to read all of the mathhammer posts, but...

Single shot weapon with BS3 and twin linked has 75% chance of hitting.

Do a tree diagram. It's true.

First roll: 50% hit. 50% miss

Second roll: 50% of 50% = 25%

50 + 25 = 75
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Old 11 Mar 2009, 22:19   #8 (permalink)
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Default Re: Twin-Linked Mathhammer...

best way is to say right red dice are 1 twin liked shot and black are another. But you couldn't roll 4 red and pull two hits from it as pointed out.
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