Card Probability

Question

Consider the standard 52-card deck (4 suits, each consists 13 ranks)

Jack randomly draws 3 cards from the deck. 

Question:

 

1) what's the probability that he gets all three Aces?

 

2) what's the probability that he gets all three cards spades? 

Answer 1

Problem 1)

On his first pick, the probability of an ace is 4/52 because there are 4 aces and 52 total cards. This can be reduced to 1/13. On his second pick, since he has already chosen an ace, the probability of choosing another ace is 3/51 because there are 3 aces left and now, only 51 total cards. This can be reduced to 1/17. On his third pick, since he has already chosen two aces, the probability of choosing another ace is 2/50 because there are 2 aces left and now, only 50 total cards. This can be reduced to 1/25.

So the probability of all three events occurring is 1/13 • 1/17 • 1/25 = 1/5525 (approx 0.018%)

Problem 2)

Probability of a spade on pick #1: 13/52 = 1/4

Probability of a spade on pick #2: 12/51 = 4/17

Probability of a spade on pick #2: 11/50

Probability of all three events occurring: 1/4 • 4/17 • 11/50 = 11/850 (approx 1.29%)

Answer 2

One way to do this is to multiply the probability for getting the 1st ace, by the probability of getting a 2nd ace, multiplied by the probability of getting the 3rd ace (without replacement).

For the 1st draw there are 52 cards in the deck, and 4 aces to pick from so the probability of drawing an ace is 4/52.

Now there are only 51 cards remaining and 3 aces, so the probability of drawing a 2nd ace is: 3/51.

Finally for the 3rd draw there are 50 cards remaining and 2 aces so: 2/50.

P(3 aces) = (4/52)(3/51)(2/50)

Another way is to use combinations. How many ways can 3 cards be drawn from a deck of 52?

This is 52 choose 3 or ₅₂C₃= 52!/49!3!.

How about the number of combinations of choosing 3 aces out of 4? ₄C₃.

P(3 aces) = ₄C₃/₅₂C₃.

Check it out here:

desmos.com/calculator/d6fn5jbbki

And for part 2) the strategy is the same, but ₁₃C₃ for choosing 3 spades from 13.

Answer 3

Question 1

There are 4 aces in a 52-card deck (one for each suit). This means that the probability of Jack drawing one ace is 4/52. If he tries to draw another ace without replacement, the probability becomes (4/52)(3/51). Therefore, for three aces, the probability would be (4/52)(3/51)(2/50) = (1/13)(1/17)(1/25) = 1/5525

Question 2

Each suit has 13 ranks, so if Jack draws one spade, the probability of that happening would be 13/52. If he tries to draw a second spade without replacement, the probability becomes (13/52)(12/51). Therefore, for three spades, the probability would be (13/52)(12/51)(11/50) = (1/4)(4/17)(11/50) = (1/17)(11/50) = 11/850

I hope the explanations helped!