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Suppose you are playing a game where you flip a coin to determine who plays first. You know that when you play first, you win the game 60% of the time and when you play the game second, you lose 52% of the time. A Find the probability that you win the game?

Let $A = \{ \text{Play first}\}$ and $\overline{A} = \text{Play second}$, let $B = \{ \text{win} \}$

We want $P(B)$.

I know that $P(B | \overline{A}) = 0.48$ and $P(B | A) = 0.6$

Actual problem:

I get that $P(B) = P(B | A)P(A) + P(B | \overline{A}) P(\overline{A}) = 0.6P(A) + 0.48P(\overline{A})$

I'm not sure how to move ahead. Can someone give me a hint?

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    Hint: a coin is flipped to decide who plays first. What does this mean for $\mathrm{P}(A)$? – 2017-01-24

3 Answers 3

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Hint -

Probability of getting head = $\frac 12$

And getting tail = $\frac 12$

Then use conditional probability to proceed further.

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You flip a coin to decide who plays first and the probability of getting a favourable outcome in the coin toss is $0.5$. Hope it helps.

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Hint:

$$P(A)=\text{chance you get heads at the coin flip}$$

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    Ah... I see, winning or not winning, so 0.5? – 2017-01-24
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    I computed, so the total probability is 0.54? – 2017-01-24
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    Yup, that looks right :-) – 2017-01-24