is it a proper question? … there is no input or output
Dice Expectation
You just have to output the expected number of dice throws to get a 4.
Probability p of getting a 4 in throwing a die is 1/6 and the probability (q) of not getting a 4 is 1−1/6 = 5/6. Let xth throw of the die results in a 4. This means that preceding (x-1) throws do not result in a 4. So, the probability function is p(x) = (q^(x−1))*p for x= 0,1,2,…….
Hence, E(x) = Summation( x * (q^(x-1)) * p), where x =1 to x = ∞
= p Summation(x q^(x−1)), where x =1 to x = ∞
=p(1+2q+3q^2+4q^3+…)
=p ((1−q)^(−2))
=1/6 ((1−5/6)^(−2))
=1/6 ∗ 36 = 6
Hope this helps
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