# Need some help on my calculation

#### orichalcum

##### New Member
Basically I used a program to roll 2 dice and see the observed outcomes compared to the expected outcome.
Doing 100 rolls I should expect to see around 17 doubles.

I then added up the values for the critical value and put them in a calculator and got

Does this mean that it's unlikely to get better results than the ones I achieved? or did my test fail and was weighed heavily not by random occurrence?
I'm wondering if I made the calculation wrong in some way.

#### Dason

You would expect 16.666666666 doubles in total (don't round). You wouldn't expect that for each of the 6 possible ways to get a double though.

#### orichalcum

##### New Member
Ah. so for each expected outcome it would be 2.777 instead

#### Graric

##### New Member
n-1 degrees of freedom..so am thinking since we have to dice 12-2=10df

#### katxt

##### Member
Looking at the working above, the (E-O)^2/E has been done on 6 groups, Commonly the E values are calculated from the observed data and this calculation removes one df. If this was the case, the total of the E should equal the total of the O. However, in this case the E values are calculated from theory and the total E doesn't equal the total O so the O are independent and each of the 6 groups has its own df. 6 in total.

#### Graric

##### New Member
Looking at the working above, the (E-O)^2/E has been done on 6 groups, Commonly the E values are calculated from the observed data and this calculation removes one df. If this was the case, the total of the E should equal the total of the O. However, in this case the E values are calculated from theory and the total E doesn't equal the total O so the O are independent and each of the 6 groups has its own df. 6 in total.
Oh!okay