# Probability of getting a sum of 13

1. ### karthik666

11
Obtain the probability of getting a sum of 13, when four fair dice are
rolled together once.

if we do by just calculating all possible values of sum,then it will take more time.
so,we can do above problem as Multinomial Coefficents of sum
i.e.,x1+x2+x3+x4 = 13 where,1<= xi <= 6,for all 1<= i <= 4.
i got stuck that form the above equation how can we obtain no.of.possible ways to get sum = 13.

Thanks for Help.

Last edited: Oct 26, 2013
2. ### sjb-2812

431
Do you know how to calculate the probability of getting, for instance a 4 when two dice are rolled?

3. ### karthik666

11
yes,it's easy,but here we have to find when four dices rolled together..

4. ### sjb-2812

431
Just extend the ideas then. How about a 9 with 3 dice?

5. ### karthik666

11
your approach is correct upto 3 dice,but u cannot do for 4 dice

431
Why not?

7. ### karthik666

11
it will take more than half-an hour,but in Exam it won't entertain such silly tasks!!

### Staff: Mentor

I just did it three different ways and it took less than a couple of minutes, total.

What do you get using your half hour long approach?

9. ### karthik666

11
may be u r saying Lie,probaility(saying a Lie)= 0.5..
ok,what is the answer that you got?

### Staff: Mentor

Aside: Stop using text speech. It's against our rules. Sentences start with a capital letter and end with one period. "u" is not a word. It's "you". The same goes for "r". Use "are".

You mentioned you know how to solve the probabilities with three dice rather than four. Imagine one of the four dice is red, the other three are white. The red die can come up as 1, 2, 3, 4, 5, or 6, each with probability 1/6. If it comes up 1, the white dice need to total to 12. What's the probability of rolling a sum of twelve with three dice? How do you combine that with the probability that the red die came up as a one? Now do the same for the red dice being a two (with the white dice now summing to 11), the red die being a three (with the white dice now summing to 10), and so on. This should not take half an hour to solve.

11

12. ### sjb-2812

431
OK, and how did you get that number?

Can you explain why DH's approach fails?

### Staff: Mentor

That most assuredly is not the answer. Show your work and we can tell you where you went wrong.

11

### Staff: Mentor

In this case it's that other site that is doing the misguiding. He calculated incorrectly. BTW, using the probability generating function is one of the methods to which I was alluding in post #8.

16. ### karthik666

11
What is wrong in his method althrough i checked correctly
PS: we can use Generating Functions also to solve this question!!

### Staff: Mentor

It's obviously wrong. Just calculate (x+x2+x3+x4+x5+x6)4. (Better: Ask Mathematica or Wolfram Alpha to do that for you.) The x13 term is not 200. It's 140.

Calculating that polynomial is a rather hard way to solve this problem, and it's also overkill since all you want is the x13 term. If you have a tool such as Mathematica, that it is overkill is somewhat irrelevant. Sometimes brute force just works. Note: A brute force expansion of that polynomial was not one of the methods I used.

### Staff: Mentor

Speaking of brute force:

Code (Text):
borek@raspberrypi ~/python $cat dices.py total = 0 i13 = 0 for d1 in range(1,7): for d2 in range(1,7): for d3 in range(1,7): for d4 in range(1,7): total += 1 if d1+d2+d3+d4 == 13: i13 += 1 print i13, total borek@raspberrypi ~/python$ python dices.py
140 1296

11

### Staff: Mentor

There's nothing wrong with using generating functions if you do so correctly. Use them incorrectly and you have elegant nonsense. For example, your first link is nonsense, but rather elegant nonsense. Use it correctly and you'll find that the number of ways to obtain a sum of 13 with a roll of 4 fair 6-sided die is ##- \binom 4 0 \binom {-4} 9 + \binom 4 1 \binom {-4} 3 = \binom 4 0 \binom {12} 9 - \binom 4 1 \binom 6 3##. It looks like your first link omitted the first factor from each term.

Another way to use the generating function f(x)=x+x2+x3+x4+x5+x6 is to realize that f(x)4 is (f(x)2)2. It's fairly easy to compute f(x)2, and with this you can read off the pairs that will form the x13 term. Not near as elegant, but you are worried about what to do during a test. Oftentimes brute force wins over elegance in a testing situation.