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mortiferum
06-08-2010, 19:55
A couple of questions for you maths genius's out there. I'm having difficulty in deciding how many dice I should roll during the magic phase for powering / dispelling and my chances of generating a IF, Miscast.

Can anyone help me with the following -

1. Has anyone generated / calculated and tabulated the probabilities of rolling a total number on 1, 2, 3, 4, 5 and 6 dice? For example the probability of rolling a 12 (double six) on two dice is 1 in 36 or 2.8%.

2. Whats the probabilityof rolling atleast 2 6's on 2, 3, 4, 5 and 6 dice?

Please provide the formulas used to calculate.

Hope you guys can help

Morty

Justicar Valius
06-08-2010, 20:01
Somebody made a chart showing the chances on eher but I can't remember which thread :cries:

give me some time and I'll update this post with 3 dice, 4 dice, 5 ice then 6 dice.

For 3 dice I think this is it (note I am open to admit I am wrong and am often)

216 combos, 11 double 6s, 5.0925% for 3 dice

mortiferum
06-08-2010, 20:46
I think I found the post you are talking about...

Justicar Valius
06-08-2010, 20:49
Nope thats not it, had a table showing chances of rolling a double 6, maybe i was imagining it.

oh and for 4 dice:

1296 combos, 157 double 6s, 12.114% of a double 6

again, cave eats from last post apply to this as well.

Gloryseeker
06-08-2010, 21:14
The easiest is the probability of a number of dice being a particular number. For n dice, this is 1/(6n). For example, the probability for throwing 5 dice and getting them all sixes is 1/(65) = 1/(6x6x6x6x6) = 1/7776 = 0.000128 or 7776 to 1.

if you throw three dice, what is the chance of one of the dice getting less than five? This means that one of the dice (at least) will be a one, two, three or four. Now this is quite hard to work out directly. What we can do is work out the opposite. The chance of one dice not getting less than five means the chance of it getting a five or six, and this is 2/6, or 1/3. But we need all dice to get this. This is 1/3*1/3*1/3 or 1/27. This was the opposite of what we want, so we must take it away from one. 1 - 1/27 or 26/27 which is about 96%.

Therefore 2 6's on 4 dice is 1/6*1/6*5/6*5/6 = 69%

yay for A level statistics

Jetty Smurf
06-08-2010, 21:20
The easiest is the probability of a number of dice being a particular number. For n dice, this is 1/(6n). For example, the probability for throwing 5 dice and getting them all sixes is 1/(65) = 1/(6x6x6x6x6) = 1/7776 = 0.000128 or 7776 to 1.

yay for A level statistics

That's all well and good, but he asked for the chance of rolling double 6's on N dice (where N is between 2 and 6 (inclusive)).

I believe it was Avian who had the chart. It was on his website (or so I seem to recall).

http://www.avianon.net/calculations/casting.php

*NOTE* - This is currently out of date at the time of this post (as it refers to 7th edition, which had a slight variance in regards to miscast/IF and did not include +x for wizard of level x casting the spell(s))

Lord Inquisitor
06-08-2010, 21:32
Sitting comfortably class? Mathhammer is in session.

What you're looking for is the probability of rolling two or more 6s when you roll X dice. Enter binomial theorem. Bunch of formulas that involve angry letters with exclamation points.

You don't want formulas! Okay then, there are a bunch of binomial calculators out there on the interwebz. Here's (http://faculty.vassar.edu/lowry/binomialX.html) my favourite.

So in the formula "n" is the number of trials (translation = the number of dice you're rolling), "k" is the number of successes required (translation = 2, the number of 6's required for an irrisistible), and "p" is the probability of success (translation = 1/6 chance of rolling a "6").

So if you want to know the probability of irrisistible on 4 dice, put in n=4, k=2, p=1/6 (nice thing about this calculator is that it accepts fractions). Then what we're interested is "p:2 or more out of 4" and that's 0.13 or 13%.

Now you can make your own table. :) The calculator is great for all sorts of mathhammer calculations.

Gork or Possibly Mork
06-08-2010, 21:47
Don't know if this helps or not but here's the thread I believe jusicar was talking about.