If we take all the face cards out of standard deck,
and deal them into four 3-card hands,
what is the probability of each hand having 1 K, 1 Q, 1 J (of any suit) ?
Solution using Combinations
The first player can get any of the 220 possible 3 card hands (12 C 3).
Of these 4 × 4 × 4 consist of (K Q J).
So his chance is 64/220 = 16/55.
Nine cards remain, so there are 84 = (9 C 3) possible hands for
the second player, of which 3 × 3 × 3 consist of (K Q J).
His chances are 27/84 = 9/28.
For the third player it's 2 × 2 × 2 (K Q J) hands
out of 20 = (6 C 3) possible hands, or 8/20 = 2/5.
Finally, there are just 3 cards left for the last player.
Overall then, the probability is 16/55 × 9/28 × 2/5 × 1 = 0.0374025974.
That is, the chances that each player gets 1 K, 1 Q, 1 J is less than 4%.
Solution using Permutations
Let's just assign each card a spot in sequence:
Each card will have some number of places it can land:
- King of Spades: 12
- King of Hearts: 9
- King of Diamonds: 6
- King of Clubs: 3
- Queen of Spades: 8
- Queen of Hearts: 6
- Queen of Diamonds: 4
- Queen of Clubs: 2
That leaves the 4 Jacks which can be placed in the 4 remaining places
in 4! ways.
Finally, the total number of permutations is 12! and for the probability,
we have:
12 × 9 × 6 × 3 × 8 × 6 × 4 × 2 × 4! / 12!
Canceling we have:
18 × 8 / (55 × 70) = 0.37402597 ... the same result as above.
Comparing the common fractions we have above:
18 × 16 / (55 × 140)
Removing a common factor of two in the numerator and denominator,
leaves 18 × 8 / (55 × 70) exactly the same thing.
Another way is this: deal the cards in order from left to right.
- First card can be any of 12.
- Second can be any of 8 (different denomination)
- Third can be any of 4
- Fourth can be any of the remaining 9
- Fifth can be any of 6
- Sixth can be any of 3
- Seventh can be any of 6 remaining
- Eighth can be any of 4
- Ninth can be any of 2
- The last 3 can be any of 3, 2, 1.
This is readily seen to generate the same numerator in a different order of factors.
One last way is:
Shuffle the Kings, Queens, Jacks separately into 4! 4! 4! orders.
Then put each KQJ into one of 3! orders.
So the total number of permutations generated that way is
(4!3 × 3!4) = 4! × 34 × 4! × 24 × 4! × 14 which is the same numerator we had in above.
