Probability: Four cards face down, two are red, two are black. Aim is to find the red ones.

Question

There are four playing cards face down on a table, two are red two are black. You must guess which two are red by pointing to them, and implicitly which two are black. Assuming all configurations are equally likely, and that you do not have psychic abilities, find the probability that exactly j of your guesses are correct for j = 0, 1, 2, 3, 4.

Blitzstein, Introduction to Probability, exercise 1-30.

My thoughts:

• First, I do not fully understand the j guesses part. I assume it means that you play n rounds of the game, and must find the probability of 0, 1, 2, 3, 4 of the n games being won...?

• As all configurations are equally likely, this reads as an ordered problem without replacement when picking cards in each round. Therefore probability of selecting two red cards is 1/6

• I think the general answer is P(j wins of n Games) = (1/6)**j * (5/6)**(n-j) but I have invented the variable n because the initial question appears vague.

Answer 1

This is a very vague and confusing question. However the # of way to choose 2 out of 4 is clear $$\binom{4}{2}=6$$

Let r be red card and b be black card. Assume the cards are in rrbb order.

You guess the red cards, and the rest of 2 cards assigned to black implicitly. All possible outcomes:

rrbb
rbbr
rbrb
brrb
brbr
bbrr

j is equal to 4, 2, 2, 2, 2, 0 respectively. Therefore, $$P(j=4)=1/6$$ $$P(j=2)=4/6$$ $$P(j=1)=P(j=3)=0$$ $$P(j=0)=1/6$$

Answer 2

• First, I do not fully understand the j guesses part. I assume it means that you play n rounds of the game, and must find the probability of 0, 1, 2, 3, 4 of the n games being won...?

You point to two cards, guessing which two are red and the other two black, and $j$ of those guesses actually are right.

$$\begin{array}{c}\blacksquare & \blacksquare & \blacksquare & \blacksquare \\ & \uparrow & \uparrow &\end{array}$$

• As all configurations are equally likely, this reads as an ordered problem without replacement when picking cards in each round. Therefore probability of selecting two red cards is 1/6

Yes.   That is the probability that $j=4$.   You pointed to both red cards and did not point at both black cards.

$$\mathsf P(j{=}4) = \frac 16$$

$$\begin{array}{c}\clubsuit & \color{red}\diamondsuit & \color{red}\heartsuit & \spadesuit \\ & \uparrow & \uparrow &\end{array}$$

• I think the general answer is P(j wins of n Games) = (1/6)j * (5/6)(n-j) but I have invented the variable n because the initial question appears vague.

No, you are looking for the probabilities of guessing the colour of 0,1,2,3,or 4 cards by pointing to the two red cards.   You have done the latter.   Find the others.

$$\mathsf P(j{=}0) = \\ \mathsf P(j{=}1) = \\ \mathsf P(j{=}2) = \\ \mathsf P(j{=}3) = $$

Answer 3

In this problem you are always taking four guesses:

• two explicit ones (the ones you point your finger, believing they are red)
• two implict ones (the ones you do not point, since you know there are 2 blacks and 2 reds)

So, if the configuration is (Red, Black, Red, Black) and you point to the first two cards, you are taking four guesses: the first one is red, the second is red, the third is black and the last is black.

In this case, you were right in your first and fourth guess, so j = 2.

Knowing that, how could you guess right an odd number of times? That is, how could j = 1 or j = 3?