A Game Of Kings

There are two red and two black Kings on the table. The first card you select is irrelevant. If it is a black King, then of the remaining 3 cards, only one is black, and since your selection is random, you have a 1/3 chance of winning.

If the King you selected is red, there is only one red King left of the other three cards, and the same argument applies.

There are ${4 \choose 2}=6$ possible selections, two of which are winning. This leaves $\frac{1}{3}$ as the answer.

Doesn't matter which colour you choose first, of the other three cards remaining there will be two of the opposite color, and 1 of the same colour. Which leaves you with a probability of a 1/3

Let's say the cards are B♤, B♧, R♡, R◇. The winning situation is drawing 2 cards with the same color (B or R). Let's calculate the possibility of picking our first card. Can we pick our first card? Yes, we can, so it's 100% of possibility, that is to say, 1. What is the possibility of drawing the same color as the second card? 3 cards left on the table in which just one card has the same color as one we have in our hand. 1 card is the right one out of 3 cards, so it's 1/3. Now just multiply those two results: 1×1/3=1/3 In a second way (for those who thought the answer should have been 1/6); Let's calculate the possibility of drawing, for example, 2 B cards: Let's draw our first card: 2 Bs and 2 Rs on the table, so 2 vs. 2, 50/50 of possibility, which gives us the result of 1/2. Now 3 cards remain on the table, and we are going to pick the other B from among other 2 Rs. 1/3 of possibility. Multiply those two results: 1/2×1/3=1/6 Is that correct? Yes, actually it is correct unless you are looking also for B possibility. What about R? We still win if we draw 2 Rs. So just double the possibility(1/6×2): it gives us 1/3. Have a great day!

Think about it this way: the first card you choose; now does it really matter? No it doesn’t, because the second card is what determines whether you get the same card or not. For example, if i pick any card, there are three cards left. Only 1 of those three cards leads to the favourable outcome. Since the first card is irrelevant, and whichever card you pick first there is only one in three cards that could give you your favourable outcome, the probability is $\frac{1}{3}$

Once you pick one, the probability of picking the same color is 1/3 (no matter which card you choose)

There are 6 cases and you can win in two of them.

Since the probability P= $\frac{total cases}{valid cases}$ , we can calculate P= $\frac{2}{6}$ = $\frac{1}{3}$

Obtaining both cards of the same colour on consecutive turns , doesn't put any binding to what colour you pick up in the first turn. Irrespective of the first turn being black or red, all you need to know is the probability that the next card will also be black or red respectively. Suppose you get red first; there are 3 more cards left out of which only one is red. That makes the probability of getting two consecutive red as 1/3. Similar argument if you get black first.

This particular scenario is a rather simple one, as the first decision you make is of no consequence whatsoever. No matter which card you pick first, it will always result in 3 remaining cards. 1 of them is the same color, while the other 2 are the different one - a ratio of 1 out of 3 (1/3).

There is a more elaborate way that will work on any probability calculation, but it's best explained in a visual display, rather than only in words. Sadly it doesn't seem to be possible to add any sort of image here, so that's out of option... Oh well, there's always Google, right?

Once the first card is selected, there are three cards left. Two of these remaining cards have a different color from the one selected, and only one has the same color. Hence, the probability that the second selected card has the same color as the first is $\dfrac{1}{3}$ .

One option is red and red, one is black then black.

For the red & red scenario, the probability looks like this: you have a 2/4 (1/2) chance of choosing a red, because half the cards are red. One that card is taken from the mix, the chance of getting a second red is 1/3 (there's only one red within the remaining selection of 3 cards).

These probabilities are multiplied, because for this to be achieved we need BOTH occur. Multiplication is used for these "and" requirements.

The result is 1/6.

Next, the black & black scenario, the probability is exactly the same.

Because we don't mind if we get red&red OR black&black, we ADD these probabilities together. Addition is used for the "or" requirements. Our result is 1/6 + 1/6 = 2/6 = 1/3.

There are 2 cards of both red and black on the table, for a total of 4 cards. It doesn't matter what the first card you pick is, because all we care about is whether the colors match. now there are 3 cards left in play, 1 of the color you just picked, 2 of the other color, so you have a 1 in 3 chance of picking the right card.

There are six potential outcomes - both red kings, both black kings, king of spades and king of hearts, king of spades and king of diamonds, king of hearts and king of clubs and king of diamonds and king of clubs. Four outcomes out of six do not yield a match and two do, so the possibility of drawing a matching pair is 2/6 = 1/3. How have almost 40% solvers contrived to bungle this one?