Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black.

Question

Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you’re guessing are red (and then implicitly you’re guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly j of your guesses are correct, for j = 0, 1, 2, 3, 4. Hint: Some probabilities are 0.

My professor worked out this example in class and I know the answers for j = 1 and j = 3 are zero, j = 0 is 1/6, j = 2 is 2/3, and j = 4 is 1/6 but I do not understand the process or concept behind the question. I do not understand where the numbers are coming from and why only the even js have a probability but not the odd js. The j guesses and pointing is confusing me. Can someone please help explain the question, I would appreciate it.

Answer 1

HINT: with $4$ cards, of which two are red and two are black, there are $\frac{4!}{2!2!}=6$ possible sequences. When you choose two cards, you automatically define the other two cards and then choose a sequence.

Now consider the two cards that you choose: they can be both correct (this implies that the other two are also correct), both wrong (this implies that the other two are also wrong) or one correct and one wrong (this implies that the other two are one correct and one wrong as well). Now you can easily complete the solution.

Answer 2

Here's a general answer that should provide you with insight.

Say we have $2n$ cards where $n$ are red and $n$ are black. There are ${2n \choose n}$ arrangements of those cards (you basically choose which $n$ are red).

Great! So now we need count all the ways we can get $j$ correct. While tempting, we can't just choose $j$ of the $2n$ cards. Why? Getting one card correct guarantees you'll get a card of the opposite color correct.

Here's why: Say you guess 1 red card correctly. You now have $n-1$ red cards and $n$ black cards to guess, which means you could at most guess $n-1$ of the $n$ black cards incorrectly leaving one which you are guaranteed to guess correctly.

Given that, the number of ways to guess $j$ cards correctly is the number of ways we can pick $j/2$ red cards and the number of ways we can pick $j/2$ black cards guaranteed by the red cards.

$$ P(j) = {{n \choose j/2}{n \choose j/2} \over {2n \choose n}} = \frac{n!^4}{(2n)!(j/2)!^2(n-j/2)!^2} $$