I love 5! Sorry for being unfair.

Probability Level pending

I have an unfair 6 6 -sided die. Its faces were numbered with integers 1 1 , 2 2 , 3 3 , 4 4 , 5 5 , and 6 6 . I rolled the die 2 2 times. Let A A be the value of the first roll and b b be the value of the second roll. The probability that A A and B B will be both equal to 5 5 O R OR A A and B B will be distinct integers is 241 245 \frac{241}{245} . If the smallest possible probability (minimum value) of rolling a 5 5 in the die is in the form m n \frac{m}{n} where m m and n n are coprime positive integers. Find m + n m + n

Details and Assumptions:

  • The die in this problem is unfair. Each of the probabilities of rolling a 1 1 , 2 2 , 3 3 , 4 4 , 5 5 , and 6 6 are not all equal to each other.

  • If I get 2 2 in the first roll and 5 5 in the second roll, then A = 2 A = 2 and B = 5 B = 5


The answer is 12.

Rindell Mabunga by Rindell Mabunga , 22, Philippines

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1 solution

Rindell Mabunga
Sep 16, 2014

Let P x P_x be the probability that A = B = 5 A = B = 5

P y P_y be the probability that A ≠ B A \neq B

and P z P_z be the probability that A = B A = B

Since P 1 + P 2 + P 3 + P 4 + P 5 + P 6 = 1 P_1 + P_2 + P_3 + P_4 + P_5 + P_6 = 1 where P n P_n is the probability of rolling the integer n n in the die, then P 1 + P 2 + P 3 + P 4 + P 6 = 1 − P 5 P_1 + P_2 + P_3 + P_4 + P_6 = 1 - P_5 .

P x = ( P 5 ) 2 P_x = (P_5)^2 (Since rolling a 5 5 in both first and second roll has a probability of P 5 P_5 )

Also,

P y = 1 − P z = 1 − [ ( P 1 ) 2 + ( P 2 ) 2 + ( P 3 ) 2 + ( P 4 ) 2 + ( P 5 ) 2 + ( P 6 ) 2 ] P_y = 1 - P_z = 1 - [(P_1)^2 + (P_2)^2 + (P_3)^2 + (P_4)^2 + (P_5)^2 + (P_6)^2]

Therefore

P x + P y = ( P 5 ) 2 + 1 − [ ( P 1 ) 2 + ( P 2 ) 2 + ( P 3 ) 2 + ( P 4 ) 2 + ( P 5 ) 2 + ( P 6 ) 2 ] = 1 − [ ( P 1 ) 2 + ( P 2 ) 2 + ( P 3 ) 2 + ( P 4 ) 2 + ( P 6 ) 2 ] P_x + P_y = (P_5)^2 + 1 - [(P_1)^2 + (P_2)^2 + (P_3)^2 + (P_4)^2 + (P_5)^2 + (P_6)^2] = 1 - [(P_1)^2 + (P_2)^2 + (P_3)^2 + (P_4)^2 + (P_6)^2]

Which leads to,

[ ( P 1 ) 2 + ( P 2 ) 2 + ( P 3 ) 2 + ( P 4 ) 2 + ( P 6 ) 2 ] = 1 − ( P x + P y ) = 1 − 241 245 = 4 245 [(P_1)^2 + (P_2)^2 + (P_3)^2 + (P_4)^2 + (P_6)^2] = 1 - (P_x + P_y) = 1 - \frac{241}{245} = \frac{4}{245}

Here's the twist, using Cauchy's Inequality,

[ ( P 1 ) 2 + ( P 2 ) 2 + ( P 3 ) 2 + ( P 4 ) 2 + ( P 6 ) 2 ] ≥ ( P 1 + P 2 + P 3 + P 4 + P 6 ) 2 5 [(P_1)^2 + (P_2)^2 + (P_3)^2 + (P_4)^2 + (P_6)^2] \ge \frac{(P_1 + P_2 + P_3 + P_4 + P_6)^2}{5}

4 245 ≥ ( P 1 + P 2 + P 3 + P 4 + P 6 ) 2 5 \frac{4}{245} \ge \frac{(P_1 + P_2 + P_3 + P_4 + P_6)^2}{5}

4 245 ≥ ( 1 − P 5 ) 2 5 \frac{4}{245} \ge \frac{(1 - P_5)^2}{5}

4 49 ≥ ( 1 − P 5 ) 2 \frac{4}{49} \ge (1 - P_5)^2

2 7 ≥ 1 − P 5 \frac{2}{7} \ge 1 - P_5 (Taking the square root of both sides is "safe" since P 5 > 0 P_5 > 0

P 5 ≥ 1 − 2 7 P_5 \ge 1 - \frac{2}{7}

P 5 ≥ 5 7 P_5 \ge \frac{5}{7}

Therefore,

m = 5 m = 5 and n = 7 n = 7 . m + n = 5 + 7 = 12 m + n = 5 + 7 = \boxed{12}

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