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

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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!

Best Answer

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
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