hewhorocks wrote:
Ok let us see the actual total chance to get your one spell.
With 3 dice there are 216 distinct results. 116 of these either have a 6 or have a double (or both.) With 3 dice the chance of getting the single spell you want is a little better than 53.7%
I'm sorry, you're just wrong. The total probability that you get the spell you wanted is the probability of you rolling it added to the probability of you rolling none of the number you want but at least one double from the rest of the dice now the actual maths involved in this second bit are a bit more complicated but I can demonstrate to you how to work them out
1st part p(that you roll at least one six) = 1 - (5/6)^3
2nd part p(roll a double not including a six) = (5/6)^3 * (1-(5/5)*(4/5)*(3/5))
Now I need to explain how I got that second bit... First it's the probability of you rolling no sixes multiplied by "1 minus the probability of you rolling no doubles" here I'm doing the same trick as earlier as it makes things much simpler
So the probability that I roll no doubles without rolling a six is 5/5 (because the first dice has nothing else it could be the same as
multiplied by 4/5 because the second dice cannot be equal to the first dice
multiplied by 3/5 because the third dice cannot be equal to either of the first two
this simplifies for rolling n dice to give (5/6)^3 * (1-FACT(5)/(FACT(5-n)*5^n)
Factorials are a mathematical way of writing 5*4*3*2*1 for example fact(100) would be 100*99*98.... all the way down to 1 so they can be used (especially it probabilities) to simplify a lot of really horrible looking maths
John Rainbow wrote:
+1 for the maths lesson.
Were you referring to me or hewhorocks? I actually trained as a maths teacher though giving a lesson over the internet is never easy...
I'm not sure if Jwg20 worked it out in the same way there are other ways of working it all out... like for instance you could do 1-(5/6)*(4/6) * (3/6)
This is basically worked out by saying the probability of the first dice NOT giving me either a six or a double is 5/6 the probability of the second dice NOT giving me a 6 or a double is 4/6 and the probability of me not getting a double or a six from the third dice is 3/6 both these methods of working it out result in the same answer at the end, which more than anything confirmed for me that I haven't made a mistake... they should both simplify to give exactly the same thing so general result for n dice here would be written as
1-FACT(5)/(fact(5-n)*6^n)
Hope this helps
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