I am a self-confessed math idiot. Can someone help me out with some dice pool probabilities?
I am looking for the chance of rolling either a 5 or 6 in a d6 dice pool. The number of dice in the pool will range from 3-8.
So, for example, if someone rolls 5d6 what is the chance of at least one of those dice being a 5 or 6?
Not nearly as bad as the card question !
The odds of rolling a five or six is 1/3 - so the odds of NOT rolling a five or six is 2/3.
Independent probabilities multiply: the odds of not rolling any fives or sixes in 5d6 is (2/3)(2/3)(2/3)(2/3)(2/3)=32/243 = ~1/8
So the odds of rolling at least one five or six in 5d6 is 7/8.
For 3d6, it would be 1-(2/3)(2/3)(2/3) =19/27
- James
Hey James,
Can you explain "independent probabilities"? I see that if you multiply 1/3 by itself 5 times, you don't get the total 7/8, so presumably that's dependent, but why is that dependent and the not-rolling-the-number thing is independent?'
Chris Lehrich
Hi Chris,
QuoteCan you explain "independent probabilities"? I see that if you multiply 1/3 by itself 5 times, you don't get the total 7/8, so presumably that's dependent, but why is that dependent and the not-rolling-the-number thing is independent?'
In a word, no.
1/3 to the power 5 is not 7/8, but 1/3 to the power 5 is not the probability of getting one or more 5 or 6's either. It is the probability of getting exactly five 5 or 6's, so you wouldn't expect it to be the same as 1 - p(getting no 5's or 6's).
I'm not at all sure where you get your ideas about dependent/independent from here, so I'll give the full explanation:
A variable is
independent if it
is not affected by the other results. The standard example here is tossing a coin. On any given coin toss the chance of getting a head is
always 0.5 - regardless of whether the last result was a head or a tail, or if the last hundred results were heads.
A variable is
dependent if it
is affected by the other results. The standard example here is a pack of cards. Suppose you're after the ace of spades - on the first draw you have a 1/52 chance of getting it. If you don't get it on the first draw you're chance of getting it on the second is 1/51, if you did get it on the first your chance of getting it on the second is 0. This is
only true if you're not replacing the cards after each draw.
In roleplaying games you only need to remember this:
All systems using dice are independent.
Most systems using cards are dependent,
unless you replace and shuffle after every single draw.
Clear?
To go a little farther with this, if you ask, what's the probability of rolling exactly one 5 or 6 in five dice? You might think the answer is:
(1/3)(2/3)(2/3)(2/3)(2/3) = 16/243
but that's just the probability of the first die rolling a 5 or 6, and none of the others doing so; just like in the card example, you have to consider the number of permutations of rolls that result in the outcome you're interested in. Since there are, by inspection, five different ways of rolling exactly one 5 or 6 on five dice, you multiply the probability above by 5, for a net probability of 80/243.
- James
Also, my reply in Brain hurts, help with math (http://www.indie-rpgs.com/viewtopic.php?t=9975) might be useful to you.
Quote from: Jack Aidley... A variable is independent if it is not affected by the other results. The standard example here is tossing a coin. On any given coin toss the chance of getting a head is always 0.5 - regardless of whether the last result was a head or a tail, or if the last hundred results were heads ...
While I agree with most of what Jack said, including the fact that coin tossing is the classic example of independence, it might not be the case that the chance of getting a head is always one half.
see The Not So Random Coin Toss (http://www.npr.org/features/feature.php?wfId=1697475)