Andrew has ten different two-player board games on his shelf. He has them numbered 1 through 10. When he plays the game i against Diego, he has a i^2/100 chance of winning. If Andrew rolls a fair ten-sided die to determine which game to play, the probability that he will win against Diego can be expressed as a/b where a and b are positive, coprime integers. What is the value of a+b?

Question

Andrew has ten different two-player board games on his shelf. He has them numbered 1 through 10. When he plays the game i against Diego, he has a i^2/100 chance of winning. If Andrew rolls a fair ten-sided die to determine which game to play, the probability that he will win against Diego can be expressed as a/b where a and b are positive, coprime integers. What is the value of a+b?

Answer 1

Use a probability tree to find that the overall probability of winning is

[∑(i²)/100]/10] for i=1 to 10
=385/1000
which should be simplified to an irreduceable fraction a/b.
The sum of a+b is then 277.

Answer 2

Only question left is how Andrew found a fair 10 sided die. Platonic solids have 4,6,8,12 and 20 sides!

Answer 3

To solve this problem, we need to find the probability that Andrew will win against Diego when choosing a game randomly by rolling a fair ten-sided die.

Let's break down the problem into smaller steps:

Step 1: Calculate the probability of selecting each game.
Since Andrew has ten different games, the probability of selecting any particular game is 1/10 or 0.1.

Step 2: Calculate the probability of winning each game.
According to the given information, the probability of Andrew winning game i against Diego is i^2/100.

Step 3: Calculate the overall probability of winning.
To find the overall probability of winning, we need to multiply the probability of selecting each game by the probability of winning that game, and then sum up all the probabilities:

Overall probability of winning = (prob. of selecting game 1) * (prob. of winning game 1) +
(prob. of selecting game 2) * (prob. of winning game 2) + ...
(prob. of selecting game 10) * (prob. of winning game 10)

Let's do the calculations:

Overall probability of winning = (0.1)*(1^2/100) + (0.1)*(2^2/100) + ... + (0.1)*(10^2/100)
= 0.01 + 0.04 + 0.09 + ... + 0.1
= 0.01(1 + 4 + 9 + ... + 100)
= 0.01(385)

So, the overall probability of winning is 0.01(385) = 3.85/100 = 77/200.

Step 4: Find the sum of a and b.
The value of a is 77, and the value of b is 200.

Therefore, a + b = 77 + 200 = 277.

Hence, the value of a + b is 277.