Card probability problem: Five cards will be dealt from a well-shuffled deck. Find the chance of getting an ace or a king among the 5 cards.

Question

Five cards will be dealt from a well-shuffled deck. Find the chance of getting an ace or a king among the 5 cards.

I think question means getting at least a king or an ace, and here is my procedure:

$p$(an ace or a king among 5 cards)=$p$(at least an ace)+$p$(at least an king)-$p$(ace and king at the same time)

$p$(at least an ace)=$1-\frac{48}{52}\times\frac{47}{52}\times\frac{46}{52}\times\frac{45}{52}\times\frac{44}{52}$

$p$(at least an king)=$1-\frac{48}{52}\times\frac{47}{52}\times\frac{46}{52}\times\frac{45}{52}\times\frac{44}{52}$

$p$(ace and king at the same time)=...

Here I am stuck because there are many combination that ace and king are together. I feel like I misintepreted the question. Could someone give an insight?

Answer 1

I think @lulu's Hint should enable you to get the correct answer. My purpose here is to illustrate a more general approach to this and similar problems.

More generally, consider a hypergeometric distribution, where there are 8 'successes' (A's and K's) in the deck and 44 'failures' in the deck and $X$ is the number of successes when 5 cards are chosen at random without replacement. You can read about the 'hypergeometric distribution' in your text or on Wikipedia.

Computations in R statistical software:

1 - choose(44,5)/choose(52,5)
## 0.5821375
sum(dhyper(1:8, 8, 44, 5))
## 0.5821375

Finally, here is a simulation of a million deals of 5-card hands. A's and K's are put last in the deck, and are numberd 45 through 52. A million iterations should give about three-place accuracy.

m = 10^6;  nr.ak = numeric(m)
deck = 1:52
for (i in 1:m) {
  hand = sample(deck, 5)
  nr.ak[i] = sum(hand >= 45) }
mean(nr.ak);  mean(nr.ak >=1)
## 0.769624  # aprx E(X)
## 0.582149  # aprx P(X >= 1)
10/13
## 0.7692308

The figure below shows a histogram of the simulated hyergeometric distribution and the red dots show exact hypergeometric probabilities. The vertical dotted line shows the location of the mean of the distribution. The resolution of the plot is not sufficient to show simulation error. [If cards were dealt with replacement, the distribution would be $Binom(5, 2/13)$ with the same mean and slightly larger variance; binomial probabilities are shows by open circles, often nearly coincident with the dots.]

hist(nr.ak, prob=T, br=(0:6)-.5, ylim=c(0,.45), col="skyblue2")
points(0:5, dhyper(0:5, 8, 44, 5), pch=19, col="red")
abline(v = 10/13, col="darkgreen", lwd=2, lty="dashed")
points(0:5, dbinom(0:5, 5, 2/13))

Answer 2

Using 1 -

$$1 - \frac {\binom{44}{5} } { \binom{52}{5} } = 0.5821374704$$

Using 1 + 2 + 3 + 4 + 5

$$\frac {\binom{8}{1} \binom{44}{4} + \binom{8}{2} \binom{44}{3} + \binom{8}{3} \binom{44}{2} + \binom{8}{4} \binom{44}{1} + \binom{8}{5}} { \binom{52}{5} } = 0.5821374704$$