Melody

Asker guest:

Please reinstate the original address.

If questions are repeats they should always be linked with the original.

Thanks Alan :)

This site might help...

Happy New Year to you too CU

Happy New Year to all forum users.    

Thanks Guest,

I am glad that we are now in agreement. That is always comforting.     

Thanks for all that information, I do appreciate your efforts.

But I do not think that is correct.

The way i have done it, 

Is to think of every dice as being distinguishable.  Like they are all different colours so

if x and y are different (3,4) is different from (4,3)

So what I think is,

You have found how many combinations there are that satisfy the conditions given that the dice are indistinguishable.

then

You have divided by by the total number of ways WHEN the dice are different from each other, (distinguishable)

So I think you have mixed identical dice in the numberator with different looking dice in the denominator.

So I still think my answer is more likely to be correct than your is.

in how many ways can we distribute 6 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

1:

The twins get 3 pecies each and the others get none.        1 Way

 2:

The twins get 2 pieces each and one of the others gets 2 pieces                3 ways

The twins get 2 pieces each and 2 of the others get 1 piece                        3 ways

3:

The twins get 1 piece each and one of the others gets 4 pieces               3 ways

The twins get 1 piece each and the others get 3,1, and 0 pieces              3!=6 ways

The twins get 1 piece each and the others get 2,2, and 0 pieces              3 ways

The twins get 1 piece each and the others get 2,1, and 1 pieces              3 ways

4.

The twins get 0 and one of the others gets 6 pieces               3 ways

The twins get 0 and the others get 5,1, 0                                6 ways

The twins get 0 and the others get 4,2, 0                                6 ways

The twins get 0 and the others get 4,1, 1                               3 ways

The twins get 0 and the others get 3,3, 0                                3 ways

The twins get 0 and the others get 3,2, 1                                6 ways

The twins get 0 and the others get 2,2, 2                                1 ways

So that is  1+3+3+3+6+3+3+3+6+6+3+3+6+1 = 50

I could easily have made a careless mistake but my logic is sound.

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

My answer does not agree with yours guest.  Generally I have a lot of faith in the output of people's computer programs.  

I certainly do not have full confidence in my answer but the logic you have used to justify your programmng output does not make sense to me.

I think we need another adjudicator.  

Can I see you code please. It probably will not make sense to me, my programming skills are way our of date, but you never know maybe I wills see somthing there.  It is a good idea to include your code for these answers anyway.

Here is my logic:

1,1,2,2,x,y,

possibilities

x and y are both the same.     4*6!/2!2!2! = 4*90 = 360  ways

x and y are different.             4*3*6!/2!2! = 12*180 =   2160 ways

Total of 360+2160 = 2520  ways

Sample space = 6^6=46656

Prob = 2520/46656 = 35/648 = approx    0.054

Atroshus,

If you think an answer is wrong then politely say so.  

It is not something you should report formally.