How Do You Calculate the Probability of a Sum in n Dice Rolls?

Click For Summary
To calculate the probability of achieving a sum of r from n dice rolls, one must determine the number of successful outcomes divided by the total possible outcomes. A probability distribution can indeed be created to assess probabilities for sums greater than, equal to, or less than r. The discussion emphasizes the importance of combinatorial methods and suggests revisiting basic concepts in combinatorics to understand the calculations better. Generating functions can be particularly useful, as they allow for the determination of coefficients that represent the number of ways to achieve specific sums. This approach provides a more generalized framework for solving such probability problems.
Astudious
Messages
61
Reaction score
0
How does one calculate the probability of a sum of r in the dice rolls of n dice? Can a probability distribution be written for something like this, to calculate the probability of a sum greater than r, greater than or equal to r, equal to r, less than or equal to r, less than r, etc.?

I do not want to simply brainstorm solutions leading to a sum of r, unless it is easy to generalize from there to n dice (which, from my consideration of n=2 and n=3, doesn't seem like the case - I need a more general way of thinking about it).
 
Physics news on Phys.org
How does one calculate the probability of a sum of r in the dice rolls of n dice?
Same way you calculate any probability - the number of ways it can happen divided by the total number of things that can happen.

Can a probability distribution be written for something like this, to calculate the probability of a sum greater than r, greater than or equal to r, equal to r, less than or equal to r, less than r, etc.?
Yes.

I do not want to simply brainstorm solutions leading to a sum of r, unless it is easy to generalize from there to n dice (which, from my consideration of n=2 and n=3, doesn't seem like the case - I need a more general way of thinking about it).
The more general way is above, and going through specifics does generalize eventually - it's just not so obvious. Perhaps revisit combinatorics?
 
Simon Bridge said:
Same way you calculate any probability - the number of ways it can happen divided by the total number of things that can happen.

Yes.

The more general way is above, and going through specifics does generalize eventually - it's just not so obvious. Perhaps revisit combinatorics?

What should I use to revisit combinatorics properly? I've read the Schaum's Outline but it doesn't seem to cover problems of this standard.
 
Schaum probably has enough information to get you going - you only need the basic concepts: the bit before "combination" and "permutation" notation is defined.
OR you can just http://www.mathpages.com/home/kmath093.htm and see how others have done it...
 
Last edited by a moderator:
There is a general method related to the combinatorics of the problem. Consider the product:

(x + x^2 + x^3 + x^4 + x^5 + x^6)^3

If you multiply it out and combine like terms then, for example, the coefficient of x^4 gives the number of sequences of 3 die rolls that produces a sum of 4.

This method is an example of using "generating functions" in combinatorics The term "generating function" has differnt meanings in different branches of math, so to search for the topic on the web, you should use a more specific search that the keywords "generating function". There are many examples on the web of using this type of generating function as a step in solving problems like "How many ways can you pay a dollar debt if you have 3 quarters, 5 dimes, and 6 nickels".
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

Similar threads

  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
1K
  • · Replies 41 ·
2
Replies
41
Views
5K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 20 ·
Replies
20
Views
9K
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K