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Abomb23
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Sum of dices problem

November 28th, 2006, 3:54 pm

I ran into this next problem and I am having hard time getting a final answer for it for every m :A dice is rolled and summed over and over , What is the probability That the sum will be "m" , "m" is a positive integerMy problem starts after the numer 6 , as I start to loose options , as I can't use 1 dice for higher numbers then 7 , same goes for 2 dices after 13 and so on.Any1 got a soultion ?
 
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MCarreira
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Sum of dices problem

November 28th, 2006, 6:41 pm

Please look at the thread "fair dice" it may help.Regards.
 
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Abomb23
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Sum of dices problem

November 28th, 2006, 8:02 pm

hi , tnx for the reply , I went trough that topic but I'm Think it's a bit over my head in the other topic m was a known number (m=15) and also u used 3 dices.In this problem I need to find a probabilty of any "m" (postive integer) for any number of dices that will form that sum (m)So for insatnce to create the number 5 then :p(5) = 1*(1/6) + 4*(1/6^2) + 6*(1/6^3) + 4*(1/6^4) + 1(1/6^5) = so meaning I have 1 option to get to 5 in 1 roll > (5)I got 4 options to get to 5 in 2 rolls > (14)(41)(23)(32)I got 6 options to get to 5 in 3 rolls > (311)(131)(113)(221)(212)(122)I got 4 options to get to 5 in 4 rolls > (2111)(1211)(1121)(1112)and i got 1 more options to get to 5 with 5 rolls > (11111)so meaning in "worse" case I will get to sum "m" with m rolls (getting 1 in each roll)Problem i have is that every time i pass 6n+1 numbers (n=1,2....) like 7 and 13 I'm lossing options as I can't create 7 with 1 die , same with 123 with 2 dice , So i'm looking for either :a soltion with a sum of sevrel Mulplexers that create the final probabilty for is p(m)ora soltion with some sort a Recorse that depand on the result of the previous probabilty (with knowing ofc that p(1)=1/6)Tnx in advance for any helpp.s - sorry for poor english
Last edited by Abomb23 on November 27th, 2006, 11:00 pm, edited 1 time in total.
 
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mhughes
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Sum of dices problem

November 28th, 2006, 8:24 pm

I think the answer you're looking for is that the number of ways to get a sum of n with m rolls of a dice is . You can see this by looking at the generating polynomial. The number of ways is the coefficient of x^n in the expansion of (x+x^2+x^3+x^4+x^5+x^6)^m. But this is (x * (1-x^6) / (1 - x))^m, which you can expand as a product and get the resulting sum. There may be a nice way of simplifying this sum, but if there is I'm not sure what it is. Of course, with n > 6m, it simplifies to be 0.
 
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MCarreira
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Sum of dices problem

November 28th, 2006, 10:21 pm

Not sure if this exactly the answer you need, but if you consider n throws of a dice, and then build the cumulative sum, the frequency distribution in that cumulative sum array can be approximated as an uniform distribution between 0 and 3.5n ( frequency 1 / (3.5n) ); this approximation gets better the larger the n is ; and you are correct in that the probability increases from 1 to 6, then it decreases from 6 to 7, and decreases again from 12 to 13.
 
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Abomb23
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Sum of dices problem

November 30th, 2006, 2:19 pm

QuoteOriginally posted by: mhughesI think the answer you're looking for is that the number of ways to get a sum of n with m rolls of a dice is . You can see this by looking at the generating polynomial. The number of ways is the coefficient of x^n in the expansion of (x+x^2+x^3+x^4+x^5+x^6)^m. But this is (x * (1-x^6) / (1 - x))^m, which you can expand as a product and get the resulting sum. There may be a nice way of simplifying this sum, but if there is I'm not sure what it is. Of course, with n > 6m, it simplifies to be 0.I don't think I quite understand how you got to this formula , I was able to solve to problem with a recorse function like this :http://img296.imageshack.us/img296/610/ ... of6.jpgBut my professor wants me to find the genreal soltion without a recorse ,Which might be what You replyd to me . meaning some sort of C(n over m) and/or P(n over m) Which I have no idea where to start with. any1 can see how I can get to that formula and can showme the way for it ???
Last edited by Abomb23 on November 29th, 2006, 11:00 pm, edited 1 time in total.
 
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mhughes
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Sum of dices problem

November 30th, 2006, 3:50 pm

Google search for "generating function". Your generating function is (x+x^2+x^3+x^4+x^5+x^6)^m. You want to find the coefficients of this generating coefficient, which can be done by decomposing it in the way I suggested (or some other way), and writing each as a power series (which is only really difficult for the 1/(1-x)^m part). Then you need to calculate the product of these power series, and find an expression for the coefficient of x^n.Alternatively, you can avoid power series and use only polynomials by decomposing it as (x*(1+x+x^2)*(1+x^3))^m, though I think this would yield double sums instead of single sums (unless it can be simplified of course).
Last edited by mhughes on November 29th, 2006, 11:00 pm, edited 1 time in total.
 
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fn075425
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Sum of dices problem

December 4th, 2006, 10:44 am

QuoteOriginally posted by: mhughesGoogle search for "generating function". Your generating function is (x+x^2+x^3+x^4+x^5+x^6)^m. You want to find the coefficients of this generating coefficient, which can be done by decomposing it in the way I suggested (or some other way), and writing each as a power series (which is only really difficult for the 1/(1-x)^m part). Then you need to calculate the product of these power series, and find an expression for the coefficient of x^n.Alternatively, you can avoid power series and use only polynomials by decomposing it as (x*(1+x+x^2)*(1+x^3))^m, though I think this would yield double sums instead of single sums (unless it can be simplified of course).I think this function (x+x^2+x^3+x^4+x^5+x^6)^m assume you roll the dice for m times, if you want to use this function, you need add the x^0;The solution for this problem I believe is simple:the expection value of one dice is 3.5 (1/6*1+1/6*2+1/6*3.....)so if we want to get m, the total number we roll should be m/3.5
 
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karakfa
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Sum of dices problem

December 5th, 2006, 5:35 pm

well, that doesn't give you the probability.check this old post for a related Q.