# Probability of winning a game, considering two options

If I am playing a game, and I'm trying to decide whether to choose option A or option B for my next move (neither option has a 100% success rate), given the following probabilities, what is the minimum success probability in order to choose option B?

• Option A Success Probability: 70%

• Win Probability after option A is successful: 56%

• Win Probability after option A is unsuccessful: 45%
• Win Probability after option B is successful: 60%
• Win Probability after option B is unsuccessful: 46%

I was thinking that the total probability of winning given option A would be calculated with:

(0.56*0.7) + (0.45*0.3) = 0.527

And then, try and solve for the success probability of option B by doing the following:

(0.6*x) + (0.46*(1-x)) = 0.528 (having a total win rate higher than option A)

0.6x + 0.46 - 0.46x = 0.528

0.14x = 0.068

x = 0.486

However, I'm not entirely confident this is the correct answer and would like to see if anyone had any other thoughts.

Thanks!

• Include the self study tag. – Michael R. Chernick May 16 '18 at 1:47

## 1 Answer

### Notation

I'll start with some notation to clarify things.
$$P(A)$$: this is the probability of successful A attempt
$$P(B)$$: this is the probability of successful B attempt

### First question

Your answer looks good to me. In the current state wherein you've selected the A route: $$P(W) = P(W|A) + P(W| \neg A)$$ $$P(W) = (0.7*0.56) + (0.3*0.45)$$ $$P(W) = 0.527$$

### Second question

I see what you are doing with the "solve for the success probability of option B". It looks like you want to decide what value of $$P(B)$$ would suffice to make choosing option B a more probable route to victory. Your math works out; but I'd recommend solving an inequality. (Here we'll use $$P(routeA \cap W)$$ to signify the probability that you've gone down route A and won):

$$P(routeB \cap W) > P(routeA \cap W)$$

$$(0.6*P(B)) + (0.46*(1-P(B))) > P(routeA \cap W)$$

$$(0.6*x) + (0.46*(1-P(B))) > 0.527$$

$$0.6*P(B) + 0.46 - 0.46*P(B) > 0.527$$

$$0.14*P(B) > 0.068$$

$$P(B) > 0.486$$

So, you know that for route B to be a more probable route to victory, the probability of a succesful B attempt must be greater than 0.486.

• Thanks for adding the clarification! Definitely helped me understand the problem better. – kstern May 15 '18 at 22:44