How many five-card hands can be made with at least one pair of face cards (J, Q, K) from a standard deck of 52 cards?
Quick examples:
- (Q, 7, 6, Q, 2) would count
- (J, 4, K, Q, 4) and (J, A, 3, A, 7) would both not count, the pair is not of face-cards.
- (J, J, J, 7, 10) and (Q, Q, Q, K, K) would both count
My first attempt was to make a decision tree simulating drawing cards, but this got out of hand very quickly.
My second attempt was to create templates for each valid hand (shown below) and count the number of variants that exist for each template. For example, for a hand with a pair and triple of face cards, (F, F, F', F', F'), would have $(12 \times 3 \times 2)(8 \times 3) \times 5!$ configurations, similarly, a single pair of face cards, (F, F, #, #, #), would have $(12 \times 3)(40 \times 39 \times 38) \times 5!$ configurations. My concern with this method is it requires me to come up with a complete, nonoverlapping set of templates. The answer I get is 38,413,440 (or ~12.3% of all possible hands, which passes the sanity check of being less than the probability of a 5-card hand having at least one pair: ~49.29%).
My third attempt was to say "pick a face card, then its pair, then pick whatever until the hand is full, you've already guaranteed at least one face card pair", $(12 \times 3 \times 50 \times 49 \times 48) \times 5!$, but it clearly double counts because I get 508,032,000 (or ~163% of all possible hands).
Note: Since my methods consider each order to be unique, I've been using $\frac{52!}{47!} = 311,875,200$ as my total number of possible hands.
Templates (# – number, F – face card; F and F' are face cards of differing rank):
- (#, #, #, F, F)
- (#, #, F', F, F)
- (#, F'', F', F, F)
- (#, F', F', F, F)
- (F'', F', F', F, F)
- (#, #, F, F, F)
- (#, F', F, F, F)
- (F'', F', F, F, F)
- (F', F', F, F, F)
- (#, F, F, F, F)
- (F', F, F, F, F)
Best Answer
Alternative: count the hands without a such a pair.
We'll count the unordered hands. You can always multiply by $5!$ if you prefer to count the ordered ones.
There are $\binom {52}5$ possible hands (with no restrictions).
There are $\binom {40}5$ hands with no face cards.
There are $12\times \binom {40}4$ hands with one face card.
There are $3\times 4^2 \times \binom {40}3$ hands with two (unmatched) face cards.
There are $4^3\times \binom {40}2$ hands with three unmatched face cards.
This gives $\boxed {320112}$ unordered hands containing at least one pair of face cards.