Date with a Psychic

Let $A$ be the event of Peter's picking the heart card and $B$ be the event of Carrie's, where $P$ stands for probability. The probability of $A$ given that $B$ has already happened (if $B$ then $A$ can be expressed as $P(A|B)$ .

Then $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ .

And by set operation, we know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ or $P(A \cap B) = P(A) + P(B) - P(A \cup B)$ .

Then $P(A|B) = \dfrac{P(A) + P(B) - P(A \cup B)}{P(B)}$ .

From the question, we known that $P(B) = 2\times P(A)$ ; $P(B) = \dfrac{5}{6}\times P(A \cup B)$ .

Hence, $P(A|B) = \dfrac{1}{2} + 1 - \dfrac{6}{5} = 1.5 - 1.2 = 0.3 = 30\%$ .

As a result, the chance of Peter's picking the heart card is $\boxed{30\%}$ given that Carrie already getting the heart card.

Let $P$ be the probability that Peter picks a heart. Then, the probability that Carrie picks a heart is $2P$ . Hence, the probability that at least one of them picks a heart is $1 - ( 1 - P ) ( 1 - 2P )$ . This gives us

$\frac{5}{6} [ 1 - ( 1 - P ) ( 1 - 2P) ] = 2P .$

Solving the quadratic, we get that $- 10P^2 + 3P = 0$ , so the solutions are $P = 0 , P = \frac{3}{10}$ .