# multinomial distribution , interpretation

• February 21st 2011, 11:31 AM
jickjoker
multinomial distribution , interpretation
1 . If a fair die is to be tossed 12 times , the probability of getting 1,2,3,4,5
and 6 points exactly twice each is

(apologies cant find decent latex software)
0.166 = fraction 1 in 6

(12! / 2!2!2!2!2!2! ) (0.166*2 ) (0.166*2) (0.166*2) (0.166*2) (0.166*2) (0.166*2) = 0.00344

question 1

does the formula used account for any order of the 2 of each ,meaning
there is no order
3 3 , 4 4 , 6 6 , 5 5 ,1 1, 2 2 is the same as
1 1, 2 2, 3 3, 4 4, 5 5 , 6 6 and so on , it accounts for all the possible combinations of the 2
of each ?.........

if so ...

question 2
what would be the formula for the exact order
the first is 1 , second is 1 , the third is 2 , the fourth is 2 etc in perfect running order?

• February 21st 2011, 11:49 AM
matheagle
take away the factorials....

$\left({1\over 6}\right)^{12}$ by independence.
• February 21st 2011, 12:13 PM
jickjoker
struggling to understand what you mean.
• February 21st 2011, 03:47 PM
matheagle
P(first is 1)=1/6 times P(second is 1 )=1/6 times P(third is 2)=1/6 times...
I'm using independence and looking at the 12 tosses without using any formulas.
the 12 choose 2,2,2,2,2,2 gives you all the rearrangements, here you aren't aking for that.
• February 21st 2011, 08:15 PM
jickjoker
1 * 12 = 2176782336
---
6

so a 1 in 2176782336 chance