Probability of dice

Question

If 4 dice tossed together , what is the probability of sum of these dice will be exactly 20? (want to see some shortcut procedure)

Comments

i mentioned it in my answer.

we are not comparing 20 with x^4.

we are comparing it with x²⁰ ....we only have to care about coefficient

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(x+x²+x³+x⁴+x⁵+x⁶)⁴ ----->>  {x(1+x+x²+x³+x⁴+x⁵)}⁴ ==  x⁴ (1+x+x²+x³+x⁴+x⁵)⁴

=x⁴(1+x+x²+x³+x⁴+x⁵) // wrong

=x⁴(1−x⁵)⁴  // wrong Recollect GP formula again    a(1-rⁿ)/ (1-r)  a=first term ; n=number of terms ; r= common ratio

(1+x+x²+x³+x⁴+x⁵)⁴  == 1.(1-x⁶)/ (1-x)

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but why r we taking $\left ( 1-x \right )^{4}$ as denominator? For which part we need to include that denominator?that was my question

There are 4 dice and each dice has 6 sides, that we included numerator part of question. rt? then what denominator is doing here?

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Look srestha let suppose we have to calculate coefficient of x² in (8x³-6x⁵+5+3x²)/7x

then what will you say ? 3/7 ? No why are you not considering denominator. the coeff of x² hence would be 8/7 so that same way with our problem.

you have to take care about whole term.we can't ignore denominator.

and about it's significance -->> what is the significance of calculating area under curve by integration ? we divide area into many tiny partitions and then we sum everyone's area. almost same concept is here we have to tell coefficient then by taking care each and every term we can place some final comment about our desired term's coefficient ... it's not the thing that only numerator is playing the role in coefficient calculation. it's mathematical way to solve the problem and its better if you don't try to define some mid step in words

is it clear now ?

Answer 1

x₁+x₂+x₃+x₄ =20

where x₁ , x₂ , x₃ , x₄ ∈ [1,6]

we need to find coefficient of x²⁰ in  (x¹ +x²+x³+x⁴+x⁵+x⁶)⁴

coefficient of x²⁰ in x⁴ *{(1-x⁶)⁴  / (1-x)⁴ }

ultimately coefficient of x¹⁶ in (1-x⁶ )⁴ /(1-x)⁴   // expand (1-x⁶ )⁴ 

using this formula (here replace x with -x)expand (1-x⁶ )⁴ 

so ultimately we have to find 1*coeff of x¹⁶ in (1-x)^-4   - C(4,1)* coeff of x¹⁰   in (1-x)^-4    + C(4,2)* coeff of x⁴ in (1-x)^-4

 coefficient of x^r in (1 – x)^-n is C(n+r-1 , r)

calculate and get answer =35

Comments

:O i did same i think

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Hello sid. here our goal is not just to answer the question. I think it should be quite clear with all possible details so that anybody can understand it , not just we our self.

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yes :)

Answer 2

$\left ( x_{1}+x_{2}+x_{3}+x_{4} \right )=20$

We have to find coefficient of $\left [ x^{20} \right ]$

$\left ( x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right )^{4}$

$=\left [ x^{4} \right ]\left ( 1+x+x^{2}+x^{3}+x^{4}+x^{5}\right )^{4}$

$=\frac{\left ( 1-x^{6} \right )^{4}}{\left ( 1-x \right )^{4}}$

Now,

$\left ( 1-x^{6} \right )^{4}=1-4x^{6}+6x^{12}-4x^{18}+...$

$\left ( 1-x\right )^{4}=1+4x+\binom{5}{2}.x^{2}+\binom{6}{3}.x^{3}+....$

$=\binom{19}{16}-4\times \binom{13}{10}+6\times \binom{7}{4}=35$