# Cards-game probabilities

We have four cards that are faced down where two of them are black and two of them are red. We can guess where the two red cards are. We are guessing for $j$ correct in terms of where the red and black cards are. We are to guess for $j=0, 1, 2, 3 ,4$.

We let A be the outcome of our first pick and B the outcome of our second pick. A and B are independent events and we can use the multiplication rule $Pr(A∩B)=Pr(A)Pr(B∣A)$. We use the naive definition of probability.

If we want zero successes, we first pick a black card and then again a black card.

$$P(j=0)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$$

If we want two sucesses, we first pick a red card and then a black card

$$P(j=2)=\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}$$

or we pick a black card and then a red card.

$$P(j=2)=\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}$$

If we want four successes, we first pick a red card and then again a red card.

$$P(j=4)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$$

For $j=1,3$ we have zero chance, from the design of the game.

This is my work but, I have no solution sheet. Are there any errors in my work?

• You have an error in assuming that $A$ and $B$ are independent. $Pr(B)=\frac{1}{2}$, not $\frac{1}{3}$ or $\frac{2}{3}$. That being said, you do have $Pr(A\cap B)=Pr(A)Pr(B\mid A)$ and your calculations seem to be matching this corrected statement. You then seem to have mixed up the number $4$ with the number $2$ in your final lines and messed up your calculations for it as well. You wrote $P(j=\color{red}{4})=1/\color{red}{3}*1/3=1/6$ when it should have been $P(j=2)=1/2*1/3=1/6$. – JMoravitz Nov 24 '17 at 19:19
• Lastly, you mention for one success "or vice versa" in the explanation, but then make absolutely no note of this in your calculation, making this result incorrect by a factor of two. – JMoravitz Nov 24 '17 at 19:20
• Thanks for the feedback @JMoravitz. It's true that $P(j=4)=1/2*1/3=1/6$ (typo) but I am sure that it is for $P(j=4)$ and not $P(j=2)$ since we are implicitly guessing the two black cards correctly when we guess the two red cards correctly. – bubububub Nov 24 '17 at 19:25
• The way you have phrased the problem, we are only counting how many guesses of locations of red cards are correct, and are not guessing locations of black cards at all. If you are instead guessing which two are red and which two are black (and you are never overlapping your guesses) then you would have it be impossible to get only one guess correct and that should instead be the $j=2$. – JMoravitz Nov 24 '17 at 19:32
• You still have not corrected the "or vice versa" part of the calculation. You should know that when you add up all probabilities, it should equal $1$. You currently only have your probabilities adding up to $\frac{2}{3}$. – JMoravitz Nov 24 '17 at 19:33

Since you still seem confused by my comments, but seem to understand enough of what's going on, I'll just explicitly correct your mistakes.

First the problem statement. As discussed in the comments, the problem you are interpreting it as is this:

You are shown four face down cards. Two of which are red and two of which are black, but their backs are otherwise indistinguishable. You are told to randomly select two of the cards. You get a point for each of the cards that you chose that are red and you get a point for each of the cards that you didn't choose that are black. Determine a sample space and probability distribution for the random variable $j$, the number of points you receive.

I still disagree with this interpretation, instead interpreting it as getting points only for correctly identifying red cards and not getting any points for black cards and I expect that whoever originally wrote the problem would agree with me. Regardless, I'll let it slide for now and continue with your interpretation. To correct the calculations to work for my interpretation instead, simply replace $P(j=2)$ with $P(j=1)$ and replace $P(j=4)$ with $P(j=2)$.

Next, we recognize that the only possible scores for your game are $0,2,4$. Letting $A$ be the event that the first card we guess turned out to be red and $B$ be the event that the second card we guess turned out to be red, we recognize that $A$ and $B$ are dependent (not independent) events. Regardless, we may still use multiplication principle as a tool for calculating probabilities. Remember that $Pr(A\cap B)=Pr(A)Pr(B\mid A)$.

Note: It should be remembered that $Pr(A\cap B)=Pr(A)Pr(B\mid A)$ is always true for nontrivial events $A,B$, regardless of whether or not they are dependent or independent. Independent events are special in that two events are said to be independent of one another if and only if $Pr(A\cap B)=Pr(A)Pr(B)$ which further can be said to be true if and only if $Pr(B\mid A)=Pr(B)$ and so on... This is however a moot point for this problem.

Next, we recognize that to end with a score of zero, we must have selected a black card followed by a black card. That is to say:

$Pr(j=0)=Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c\mid A^c)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

Next, we recognize that to end with a score of two, we must have selected a red card followed by a black card or vice versa. This "or vice versa" part of the phrase is important and is what you have been missing in your calculations all this time. Phrasing the probability in terms of events, that is:

$Pr(j=2)=Pr((A\cap B^c)\color{red}{\cup}(A^c\cap B))$

Read aloud, $(A\cap B^c)\cup (A^c\cap B)$ is the event that "the first card is red while the second is black" or "the first card is black while the second is red."

Using addition principle, we note that since $A\cap B^c$ is mutually exclusive with $A^c\cap B$, we may split this apart as a summation. We may then split it apart with multiplication principle.

$Pr(j=2)=Pr(A\cap B^c)+Pr(A^c\cap B)=Pr(A)Pr(B^c\mid A)+Pr(A^c)Pr(B\mid A^c)$

$=\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

In your work above you incorrectly implied that $Pr(j=2)=Pr(A\cap B^c)=Pr(A^c\cap B)$

Finally, to get a score of $4$ we need both of our selected cards to be red. That is to say, $Pr(j=4)=Pr(A\cap B)=Pr(A)Pr(B\mid A)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

Indeed, $Pr(j=0)+Pr(j=2)+Pr(j=4)=\frac{1}{6}+\frac{2}{3}+\frac{1}{6}=1$ as expected.

There are other methods to calculate the probability avoiding the multiplication principle of probability, relying instead on the multiplication principle of counting, either picking simultaneously or in sequence.

$Pr(j=0)=\frac{\binom{2}{0}\binom{2}{2}}{\binom{4}{2}}=\frac{1}{6}$

$Pr(j=2)=\frac{\binom{2}{1}\binom{2}{1}}{\binom{4}{2}}=\frac{4}{6}=\frac{2}{3}$

...

$Pr(j=0)=\frac{2\cdot 1}{4\cdot 3}=\frac{1}{6}$

$Pr(j=2)=\frac{2\cdot 2+2\cdot 2}{4\cdot 3}=\frac{2}{3}$

...