There several ways to get a ten.
Now you have to count all of those.
See if I missed any.
POST SCRIPT.
Here is a neat trick.
Look at this webpage.
Look for the coefficient of .
That tells us how many ways the the four dice add to 10.
Hi everyone,
I have the solution to a problem, but I am nore sure how the answer was arrived at, or whether it is 100% correct. The problem involves tossing 4 dice, and finding the probability of the sum being 10.
The solution to this problem is ((9 choose 3)-4) / 6^4.
Any ideas as to how this solution was derived? Any insight is appreciated.
There several ways to get a ten.
Now you have to count all of those.
See if I missed any.
POST SCRIPT.
Here is a neat trick.
Look at this webpage.
Look for the coefficient of .
That tells us how many ways the the four dice add to 10.
Hello, KelvinScale!
I have the solution to a problem, but I am not sure how the answer was arrived at,
or whether it is 100% correct.
The problem involves tossing 4 dice, and finding the probability of the sum being 10.
The solution to this problem is: .
Any ideas as to how this solution was derived?
Any insight is appreciated.
I have a theory . . . see if it makes sense.
Place ten objects in a row, leaving a space between each pair.
. . -
Select 3 of the 9 spaces and insert "dividers".
So that: . . represents the roll [3,2,1,4]
There are: . ways to place the dividers.
But these include the division: . -
. . which represents [1,1,1,7] which is not permitted.
There are four such divisions: [1,1,1,7], [1,1,7,1], [1,7,1,1], and [7,1,1,1].
Hence, there are: . .ways to get a sum of ten on four dice.
And there are: . possible outcomes.
Therefore: .