CareerCup

1-P(no_4)-P(only_one_4)= 1-(5/6)^5 - 5*(1/6)*(5/6)^4

if the result is greater than 0.5, just take the bet, otherwise not.

total faces = 30
total faces of 4 = 5
P(no 4)=(30-5)/30= 25/30 = 5/6
P(only one 4) = 1/30

P(at least two 4)=1-P(no 4)-P(only one 4)


P(at least two 4)= 0.13333

no. of dice: 5
no. of faces for each dice: 6
P(at least two 4s)= 1 - (no. of ways of getting exactly one 4)/(total no. of ways)
= 1 - 6^4/6^5
= 5/6

probability of 2 fours = {a = 1/6, b = 5/6, a^5+(a^4 b) 5+a^3 b^2×(5!)/(3! 2!)+a^2 b^3×(5!)/(2! 3!)} = 763/3888

0.1962 so no bet