Probability of Getting Two Specific Cards in 7-Card Poker Hand, Using Ordinary Deck of 52 Cards?

Question

If we have an ordinary deck of 52 cards, how many possible 7 card poker hands are there that contain exactly one "2" and 1 "King"?

Originally, I was thinking of it as $$ \frac{\binom{52}{4}*\binom{48}{4}}{\binom{52}{7}}$$. However, I'm not sure that it takes into account all necessary combinations. So now I'm thinking that maybe it is just 52*48*(44*43*42*41*40). Does that seem correct?

Thank you in advance for any help that anyone can provide!

Answer 1

Ways. The number of possible hands with exactly one Deuce and exactly one King is $${4\choose 1}^2 {44 \choose 5},$$ as in the numerator of @Henry's first fraction. Evaluated in R as:

4^2*choose(44,5)
[1] 17376128

Exact probability. Then the probability of getting that outcome is $$\frac{{4\choose 1}^2 {44 \choose 5}}{{52\choose y}} = 0.1298814,$$ as in his Comment. Evaluated in R as follows:

4^2*choose(44,5)/choose(52,7)
[1] 0.1298814

Simulated probability. This probability can be simulated with reasonable accuracy by looking at results from 10 million 7-card hands. For simplicity in coding I have simulated Aces (1) and Deuces (20), instead of Deuces and Kings. [Code numbers assigned to denominations are arbitrary as long as there is no ambiguity in the desired total (here 21).]

With ten million iterations, the probability should be accurate to within $\pm 0.0002$ in 95% of such simulations.

set.seed(1234)       # for reproducibility
deck = c(1,1,1,1,20,20,20,20, rep(0, 52-8))
tot = replicate(10^7, sum(sample(deck,7)))
mean(tot==21)        # mean of logical vector is its prop of TRUEs
[1] 0.1299681        # aprx P(1 K & 1 D) = 0.1298814
2*sd(tot==21)/sqrt(10^7)
[1] 0.0002126748     # 95% margin of simulation error