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Probability of getting a sum of 13by karthik666
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#1
Oct2613, 06:29 AM

P: 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. 


#2
Oct2613, 06:35 AM

P: 427

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



#3
Oct2613, 06:40 AM

P: 11

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



#4
Oct2613, 06:42 AM

P: 427

Probability of getting a sum of 13
Just extend the ideas then. How about a 9 with 3 dice?



#5
Oct2613, 06:46 AM

P: 11

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



#6
Oct2613, 08:32 AM

P: 427

Why not?



#7
Oct2613, 10:03 PM

P: 11

it will take more than halfan hour,but in Exam it won't entertain such silly tasks!!



#8
Oct2613, 10:40 PM

Mentor
P: 15,147

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
Oct2613, 11:52 PM

P: 11

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


#10
Oct2713, 12:13 AM

Mentor
P: 15,147

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".
Regarding your "Lie", no, that is not the answer I obtained. 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
Oct2713, 10:31 AM

P: 11

Your Approach is wrong,and This eventually leads to Obvious Wrong Answer!!
The Answer is:25/162. 


#12
Oct2713, 10:40 AM

P: 427

OK, and how did you get that number?
Can you explain why DH's approach fails? 


#13
Oct2713, 10:46 AM

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P: 15,147




#14
Oct2713, 02:19 PM

P: 11

http://math.stackexchange.com/questi...ngasumof13
This Link Clearly Shows you all work, I'm very happy that i Proved That Your Physics Forum Mentor is wrong!! @D H Mentor:First You work out the question then post appropriate Comment,PLEASE DON'T MISGUIDE a STUDENT. Thanks For your Concern!!! 


#15
Oct2713, 02:51 PM

Mentor
P: 15,147

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.
The answer is not 25/162. The correct answer is 35/324. Here's a link that gets that correct answer: http://www.wolframalpha.com/input/?i...faces+total+13. 


#16
Oct2713, 03:30 PM

P: 11

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


#17
Oct2713, 03:44 PM

Mentor
P: 15,147

It's obviously wrong. Just calculate (x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4}. (Better: Ask Mathematica or Wolfram Alpha to do that for you.) The x^{13} 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 x^{13} 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. 


#18
Oct2713, 05:37 PM

Admin
P: 23,530

Speaking of brute force:



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