Conditional Probability - truth/lie The probability that two poker players speak truth is 3/5 and 2/3.  A card is drawn from the pack and they both agree that it is a queen.  What is the probability that it is really a queen ?
My approach:-
4 cases possible for A and B based on one saying Truths and Lies (TT,TF,FT,FF)
Case 1:- Both speaking truth and it is really a queen = 3/5 * 2/3 * 4/52
Case 2 and 3:- One person saying truth and other lie will not be possible as it would lead to an inconsistency because it is given that both agree it to be queen
Case 4:-Both speaking a lie, and it is not a queen = 2/5 * 1/3 * 48/52 = 8/65
Answer= 12/65
Answer given in book is 1/5
Can this be solved using probability tree diagram ?
 A: Let $A$ be the event that the first player says the card is a queen, $B$ be the event that the second player says the card is a queen, and $Q$ be the event that a queen is actually drawn. By the problem statement, we can reasonably assume that $A$ and $B$ are independent of each other, when $Q$ is held fixed.
Then we are given $P(Q),P(A|Q),P(B|Q),P(A|Q^C),P(B|Q^C)$, and we are looking to find $P(Q|AB)$.
By Bayes' Theorem, we know:
$$P(Q|AB)=\frac{P(AB|Q)\cdot P(Q)}{P(AB)}
\\=\frac{P(AB|Q)\cdot P(Q)}{P(AB|Q)\cdot P(Q)+P(AB|Q^C)\cdot P(Q^C)}
\\=\frac{P(A|Q)\cdot P(B|Q)\cdot P(Q)}{P(A|Q)\cdot P(B|Q)\cdot P(Q)+P(A|Q^C)\cdot P(B|Q^C)\cdot P(Q^C)}
\\=\frac{(3/5)(2/3)(1/13)}{(3/5)(2/3)(1/13)+(2/5)(1/3)(12/13)}
\\=\frac{6}{6+24}=\frac{1}{5}$$
A: If, as lulu suggested, A and B choose their lie with a uniform distribution, we wind up with a vastly different answer. The problem can be broken down as follows:
Define $Q, A_Q, B_Q$ to be the events that the drawn card is a queen, $A$ says the card is a queen, and $B$ says the card is a queen respectively.
Then, you have the following truth table:
$$\begin{array}{c|c|c|c}Q & A_Q & B_Q & \text{Prob} \\ \hline T & T & T & \tfrac{1}{13}\cdot \tfrac{3}{5}\cdot \tfrac{2}{3} = \tfrac{2}{65} \\ T & T & F & \tfrac{1}{13} \cdot \tfrac{3}{5} \cdot \tfrac{1}{3} = \tfrac{1}{65} \\ T & F & T & \tfrac{1}{13} \cdot \tfrac{2}{5}\cdot \tfrac{2}{3} = \tfrac{4}{195} \\ T & F & F & \tfrac{1}{13}\cdot \tfrac{2}{5}\cdot \tfrac{1}{3} = \tfrac{2}{195} \\ F & T & T & \tfrac{12}{13}\cdot \left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\cdot \left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{1}{1170} \\ F & T & F & \tfrac{12}{13}\left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{7}{234} \\ F & F & T & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{29}{1170} \\ F & F & F & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{203}{234}\end{array}$$
Note: I am assuming a uniform distribution of lies So, if the card is not a queen, and they say it is a queen, not only is it the probability of them lying, it is also the probability that their lie happened to be a queen. So, $P(Q^C\cap A_Q^C)$ is the probability that it is not a queen and $A$ says a different card. $A$ will tell the truth with probability $\tfrac{3}{5}$. $A$ will say it is a queen with probability $\tfrac{2}{5}\cdot \tfrac{1}{12}$. $A$ will lie and say it is another card (that is not a queen) with probability $\tfrac{2}{5}\cdot \tfrac{11}{12}$. For example, if the actual card is a king, he will say it is not a king with probability $\tfrac{2}{5}$, but the chosen lie (ace through queen) is uniformly distributed over that event. So, each individual card that is not a king has its own $\tfrac{2}{5}\cdot \tfrac{1}{12} = \tfrac{1}{30}$ chance of being told.
Next, we add up the probabilities to ensure we have a total probability space:
$$\dfrac{2}{65}+\dfrac{1}{65}+\dfrac{4}{195}+\dfrac{2}{195}+\dfrac{1}{1170}+\dfrac{7}{234}+\dfrac{29}{1170}+\dfrac{203}{234} = 1$$
And indeed we do. So, unless I made a calculation error, we have:
$$P(Q|A_Q\cap B_Q) = \dfrac{P(Q\cap A_Q\cap B_Q)}{P(A_Q \cap B_Q)} = \dfrac{\left(\dfrac{2}{65}\right)}{\left(\dfrac{2}{65}\right)+\left(\dfrac{1}{1170}\right)} = \dfrac{36}{37}$$
Another note: This also assumes all events are independent (players choose to lie independently of reality and each other). This seems like another valid answer to the original problem since it was not clearly worded.
A: I think first you have to understand the meaning of Bayes's Formula
$$P(Q|D)=\frac{P(D|Q)P(Q)}{P(D)}$$
First you have an event $Q$ with prior probability $P(Q)$, then we got new evidence about $Q$ after observing $D$,that evidence is incorporated as $P(D|Q)$ to finally evaluate the probability of $Q$ given evidence $D$ using Bayes's Formula.
In your case $Q$ is the event of drawing a Queen with prior $P(Q)= 1/13$,then you got evidence provided by players $D = AB$ as hightlighted @ash4fun. Both players are agree but both can lie, that's why you need to apply Total Probability Law on $AB$ condition respect to $Q$. So thw anwer you are looking for was already provided by @ash4fun.
A: Conceptualize the event like this:
A draws the card on top. Now, when he 'says' he sees a queen, it can mean one of two things:

*

*It was a Queen, and he's telling the truth

*It wasn't a Queen, and he's lying

B looks at the card and does the same thing
So what does it mean when they 'agree' on the card being a queen? It means they both 'say' it is a queen, i.e.

*

*If the card is a queen, they're BOTH telling the truth

*If the card isn't a queen, they're BOTH lying


Now, its just an application of Bayes' theorem. (they agree means they agree that it is a Queen)
$$ P(\text{it is Queen} |\text{they agree} ) = \frac { P(\text{they agree}|\text{it is Queen}) \cdot P(\text{it is Queen}) } {P(\text{they agree}|\text{it is Queen})\cdot P(\text{it is Queen}) + P(\text{they agree}|\text{it is not Queen}) \cdot P(\text{it is not Queen})}$$
$$ = \frac{ \frac{6}{15} \cdot \frac{1}{13} }{ \frac{6}{15} \cdot \frac{1}{13}+ \frac{2}{15} \cdot \frac{12}{13} } $$
$$=\frac{1}{5}$$
