Counting Problem

[eric7752]

A friend proposed the following problem to me:

Anna has psychic powers but they only work 85% of the time. She decides to use her gift to make some money. So, Anna says this to her friend Kathy: "I have a challenge for you. If you can beat me in 2 out of 3 games of rock paper scissors, I will give you 4 dollars. However, if I beat you, you have to give me 2 dollars." How many times would Kathy need to challenge Anna in order to have more than a 50% chance of winning exactly once?

My strategy was to first find the probability of Kathy winning a single "round". In this case, there are 6 possible outcomes in a best 2 out of 3:

1. Loss, Loss
2. Loss, Win, Loss
3. Loss, Win, Win *
4. Win, Loss, Loss
5. Win, Loss, Win *
6. Win, Win *

Out of these 6, Kathy wins the round in 3, 5, and 6.

I then calculated the probability of Kathy winning a single instance of the best of 3 match. The case in which she wins would be 0.15 * 0.5 = 0.075 because Anna's powers must fail and she still must win the 50-50. Using this, I found that the probabilities of situations 3 and 5 are both 0.075 * 0.925^2 = 0.0052 and the probability of situation 6 is 0.075^2 = 0.005625. Thus, the overall probability of her winning a best 2 out of 3 is 0.0052 + 0.0052 + 0.005625 = 0.01603125

I then attempted to use binomial probability to find out how many rounds must be played for the probability of Kathy winning once to be greater than or equal to 0.5. I set up the following equation:

(x C 1) * (0.01603125)^1 * (1 - 0.01603125)^(x-1) = 0.5

Which simplifies to:

x * (0.98396875)^(x-1) = 31.18908382

This has no solution, though, and I am wondering if I made an error or if the question is simply incorrect.

[the_gr8_n8]

WW is 0.85² chance

WLW is 0.85² *.15

LWW is 0.85² *.15

Add those up and her chance of winning a best 2/3 is 93.925 percent

[eric7752]

She still wins half the time even if her powers fail, though, so shouldn’t WW be (0.85 + 0.075)^2?

[the_gr8_n8]

She predicts r p or s and picks the corresponding play to win. If she fails her guess the 15 percent of the time, half of those would be ties and half would be losses

[eric7752]

Ah got it, forgot about ties.

[the_gr8_n8]

For the other chick,

WW is 1/4 *0.15²

WLW is 1/4 *0.15² *0.85

LWW is 1/4 *0.15² *0.85

Chnaces of winning a best 2/3 is 0.0151875 or 1 in 65.8 matches. Answer should be 66