So the problem is stated below:
A card game of 52 cards are dealt between 4 people (including me), each receiving 13 cards. 4 of the cards are Aces; one is called Ace of Spades and the other Ace of Clubs.
I wish to work out the probability that I receive neither the Ace of Spades nor the Ace of Clubs.
Progress
let S=receiving one Ace of Spades in hand,
$P(S)=1/4$
let C= Ace of Clubs in hand
$P(C)= (1/51)/(1/12)=4/17$
$=> P(S)*P(C)=1/17$ which gives the probability of getting both an Ace of Spades and an Ace of Clubs.
If I let A=getting all 4 Aces in my hand
$P(A)=[(52-4)!13!]/[52!(13-4)!]$
How do I carry on from there to find the probability that I receive neither the Aces of Spades and Clubs? And, what will be the probability that I receive at least one?
Thank you so much! 🙂
Best Answer
A simple picture will give a simple answer:
Visualize 52 ordered slots into which to put the 52 distinct cards; the first thirteen slots will hold your cards.
Put the Ace of Spades in a slot first, For you to not get it, it must land in one of $39$ slots that you don't own, out of the $52$ possible. Next, put out the Ace of Clubs. To miss it, you must put it in one of the $38$ remaining slots you don't own, out of the $51$ remaining slots. So the answer is: $$P_{neither}=\frac{39}{52}\times \frac{38}{51}$$
As to the probability of getting at least one: Either you get none, or you get some. The two probabilities will add to $1$.