1
$\begingroup$

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.

2 Answers 2

1

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?

0
  • 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.

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

  • 0
    Thank you for your answer, I did consider this approach, but I'm concerned about the j=3 and j=4 cases. - p(j=0) = 1/6 - p(j=1) = 4/6 As the outcome is contingent on guessing red cards, not black cards, you cannot make a guess for j = 3 or j = 4. It's a mental block, rooted in the formality of the issue for me – 2017-02-20
  • 0
    Hint: The probability for $j$ taking certain values *may* be zero. – 2017-02-20
  • 0
    Thank you for your continued support. Please could you confirm my understanding - that i have the correct labels and probabilities:
      - P(j=0) selected zero red cards P = 1/6 - P(j=1) selected one red card, one black card P = 4/6 - P(j=2) selected two red cards P = 1/6 - P(j=3) you can't select three cards, you can only select two cards P = 0 - P(j=4) you can't select four cards, you can only select two cards P= 0
    – 2017-02-20