Given two decks of cards what is the probability to draw a king, and heart card

Question

I thought maybe to separate it into two options,

1) A heart King and a heart card

2) Not a heart king and a heart card

I figured the probability for option 2 is $$\left(\frac{6}{104} \cdot \frac{26}{104}\right) \cdot 2$$ We have $6$ cards of kings which are not heart kind, and we have $26$ heart kind of cards. Since there is no importance to the order of the draw we multiply it by $2$.

From here, I got stuck. I could not manage to calculate the probability for option 1.

Thanks.

Answer 1

If two standard decks are combined, what is the probability of selecting a heart and a king that is not a heart?

Method 1: There are $$\binom{104}{2}$$ ways to choose two cards from the $2 \cdot 52 = 104$ cards in the combined decks.

To be successful, we must choose one of the $26$ hearts and one of the $6$ kings that is not a heart in the combined deck, which can be done in $$\binom{26}{1}\binom{6}{1}$$ ways.

Thus, $$\Pr(\text{heart and king that is not a heart}) = \frac{\dbinom{26}{1}\dbinom{6}{1}}{\dbinom{104}{2}}$$

Method 2: We correct your method.

The probability of choosing a heart, then choosing a king that is not a heart from the remaining $103$ cards is $$\Pr(H)\Pr(\color{red}{K\diamondsuit}~\text{or}~K\clubsuit~\text{or}~K\spadesuit \mid H) = \frac{26}{104} \cdot \frac{6}{103}$$ The probability of choosing a king that is not a heart, then choosing a heart from the remaining $103$ cards is $$\Pr(\color{red}{K\diamondsuit}~\text{or}~K\clubsuit~\text{or}~K\spadesuit)\Pr(H \mid \color{red}{K\diamondsuit}~\text{or}K\clubsuit~\text{or}~K\spadesuit) = \frac{6}{104} \cdot \frac{26}{103}$$ Hence, $$\Pr(\text{heart and king that is not a heart}) = 2 \cdot \frac{26 \cdot 6}{104 \cdot 103}$$

If two standard decks are combined, what is the probability of selecting a king of hearts and a heart?

There are two types of favorable cases. Both kings of hearts are selected or a king of hearts is selected and another heart is selected.

There is $\binom{2}{2} = 1$ way to select both kings of hearts. There are $\binom{2}{1}\binom{24}{1}$ ways to select one king of hearts and one of the other $24$ hearts in the combined decks. Hence, the number of favorable cases is $$\binom{2}{2} + \binom{2}{1}\binom{24}{1}$$ Thus, $$\Pr(\text{king of hearts and a heart}) = \frac{\dbinom{2}{2} + \dbinom{2}{1}\dbinom{24}{1}}{\dbinom{104}{2}}$$

Answer 2

For the first part, think of it like this: you have the first $26$ natural numbers, and you must pick two such that you have at least one of $1$ and $2$. How many ways can you do this? Once you've done this, just assume that cards $1$ and $2$ are the two kings of hearts. Then add this number to the number of ways in which you can get a king not belonging to hearts, and a hearts card. Divide by the total number of ways in which you can pick any two cards.

PS: I'm assuming that the option of having just one king of hearts with some other random card does not satisfy the condition of a king and a heart card?

Answer 3

use a tree

first draw

A draw King (not hearts) = 6/104

B King hearts = 2 / 104

C Hearts, not king of hearts = 24/104

then thinking in each case what you need to complete

for A - you need any of 26 hearts for B - you need any of 5 kings or 24 hearts (excluding king of hearts already counted) for C - you need any of the 8 kings

A x 26 / 103 + B x (4 + 25) / 103 + C x 8 / 103 = 203/5356