Dice

Mar 2010
60
1
What is the probability of throwing exactly two 3x a two and 2x a three with five normal dice? Order is not to be respected (eg 22233=32322)
The possible combinations for n=6 and k=5, if order is to be neglected and repetition (as the task is about dice) is possible, is the formula (n+k-1 // k) which equals 720 possibilities. Thus the probability for axactly two threes and three twos is 1/720.
The solution is 10 //7776 though. Where is my mistake?

Thanks
 

Plato

MHF Helper
Aug 2006
22,455
8,631
I disagree with your suggested answer. You must consider order.
If we throw a die five times there are \(\displaystyle 6^5\) different possible outcomes(5-tuples).
Of those there are only 10 that contain exact two 3's and three 2's.
So what is the probability?
 
Mar 2010
60
1
But why do you have do consider order when 33322 is the same as 32223 - I still dont get that. 6^5 possibilities would be right if there was a difference between e.g 21 an 12...
 

Plato

MHF Helper
Aug 2006
22,455
8,631
I still dont get that. 6^5 possibilities would be right if there was a difference between e.g 21 an 12...
Once again: order must be considered.
The pair \(\displaystyle (1,2)\) is different from the pair \(\displaystyle (2,1)\).
If the question were, "what is the probability of tossing a one and a two?"
The answer is \(\displaystyle \frac{2}{36}\) because there are two ways out of thirty-six to do it.

From a content point of view there are only 21 different outcomes.
But the difference is that all contents do not have the same probability.