A deck of cards has 24 cards remaining, with 3 spades and 21 non-spades. I am going to deal them equally among 3 people (8 cards each) What is the probability that each person ends up with EXACTLY one of the 3 spades when the cards are dealt?
I know the odds of having at least one spade in each hand are 1/3, but this is different than the odds of each person having exactly one of the three available spades in the remaining 24 cards, and can't find any formulas that would handle this.
This just boils down to caclulating how many ways can three people end up with exactly one of three objects. Lets first calculate all possible outcomes, here I’m assuming you deal 8 at a time to each player:
# outcomes = $$\binom{24}{8}\binom{16}{8}\binom{8}{8} = 9,465,511,770$$
The logic behind this is how many ways can you choose 8 items from 24?, then after you pass those out how many ways can you choose 8 things from 16? And then multiply these together. I’m not using permutations because that would be overcounting. It doesn’t matter in what sequence each player recieves a spade, it only matters that they have it in a group of 8.
Now for the outcome we are interested in. We want each player to have exactly one of the remaining spades. This is the product of the following:
$$\binom{3}{1}\binom{21}{7}\binom{2}{1}\binom{15}{7}\binom{1}{1}\binom{7}{7} = 4,489,570,800$$
Then take the ratio to get the probability:
$P($ all three get exactly one spade $) = \frac{4,489,570,800}{9,465,511,770} = .4743$