17 November 2013

Summing Dice Throws

What is the probability of getting a particular sum after a dice is thrown n times??

During the 1st terminal exam(class 12), I came across a problem in probability wherein a die is thrown twice and finding the probability that the sum obtained is 6. Well the solution was arrived at by finding out the sample space(I actually hate that method, just computation no logic). So naturally a question arose, what if the die is thrown n times........

It preoccupied me the whole evening.Finally I got a formula for finding the number of ways in which a sum can be obtained.
Let the sum of the outcomes be s=(n + 6k + x)
Where, 
n=>number of throws
k=>arbitrary constant
ϵ {0,1,2,3,4,5}

Then number of ways the sum can be obtained is given by:


Using this formula you can easily find the probability..

This formula can also be extended to a die with 'f' faces
s=(n+fk+x)
Where, 
n=>number of throws
k=>arbitrary constant
ϵ {0,1,2,3,...,f-1}
No. of ways = $\sum_{i=0}^{k}(-1)^i.\binom{n}{i}.\binom{n+(k-i)f+y-1}{(k-i)f+y}$


Uh.Oh Forgot one important thing.


Proof:
We can represent a die with 6 faces as $x+x^2+x^3+x^4+x^5+x^6$ Where the power of x denote the 6 faces.
Now we can mathematically represent n dice throws as $[x+x^2+x^3+x^4+x^5+x^6]^n=x^n[1+x+x^2+x^3+x^4+x^5]^n$
                                                     $=x^{n}\frac{(1-x^{6})^{n}}{(1-x)^{n}}$
                                                     $=x^{n}(1-x^{6})^{n}(1-x)^{-n}$

Now the coefficient of  $x^s$ in the expansion of the above term should give the number of times the term $x^s$ appear in the expansion of  $[x+x^2+x^3+x^4+x^5+x^6]^n$.
To Understand the above statement look for 'combinatorial proof of binomial theorem'


The term $x^s$ can be interpreted as the event of getting a sum $s$ after $n$ dice throws.

All that is left now is to find the general expression for the coeffecient of $x^s$ in the expansion of $x^{n}(1-x^{6})^{n}(1-x)^{-n}$.For that we need to use binomial theorem for negative integral index(Which can be proved using Taylor series expansion).

$x^{n}(1-x^{6})^{n}(1-x)^{-n}=x^n\left ( 1-\binom{n}{1}x^6+\binom{n}{2}x^{12}-...... \right )\left ( 1+\frac{n}{1!}x+\frac{n(n+1)}{2!}x^2 +......\right )$


In the above expansion we can identify a general trend
The coefficient of $x^{n+6} = -\binom{n}{1}+\frac{n(n+1)(n+2)...(n+5)}{6!}$
                                          $=-\binom{n}{1}+\binom{n+6-1}{6}$

The coefficient of $x^{n+6+1} = -\binom{n}{1}.n+\frac{n(n+1)(n+2)...(n+6)}{7!}$
                                          $=-\binom{n}{1}.\binom{n+1-1}{1}+\binom{n}{0}.\binom{n+6+1-1}{6+1}$


A similar detailed analysis shows that the coefficient of $x^{n+6k+y}$ is given by $\sum_{i=0}^{k}(-1)^i.\binom{n}{i}.\binom{n+(k-i)6+y-1}{(k-i)6+y}$ yϵ{0,1,2,3,4,5}
(The detailed analysis had to be omitted here due to the difficulty in writing in math.)

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