What Are the Probabilities of Getting Specific Hands from a Deck of Cards?

[Pizzerer]

I'm not very good at probability and I'm just having trouble interpreting and calculating what the questions actually want me to do. Could someone please help to explain this simply?

1.Five cards are dealt at random from a standard pack. Ignoring suit, what is the probability that at least two of them are the same?

For this question I thought it was 1 - 13c5/52c5 = answer, but I'm not sure.

2. In a poker hand consisting of five cards dealt at random, what is the probability of a royal flush (Ten, Jack, Queen, King and Ace of the same suit)?

No idea how to incorporate the different picture cards into this calculation.

3. In a variant of poker each player has seven cards. What is the probability
of a royal flush in this case?

I thought here it would just be 52c7 instead of 52c5 for all the possible samples, but I don't know where to go from there.

I would really appreciate any help that can be given.

 

[mathman]

For 1. I get 1 - (48/51)(45/50)(42/49)(39/48)

For 2. I get (20/52)(4/51)(3/50)(2/49)(1/48)

3. requires more work.

 

[Pizzerer]

For 1. I get 1 - (48/51)(45/50)(42/49)(39/48)

For 2. I get (20/52)(4/51)(3/50)(2/49)(1/48)

3. requires more work.

Could you explain to me how you got those figures?

 

[Ray Vickson]

Could you explain to me how you got those figures?

For 1, I would have gotten something different. The logic is: find the probability that all 5 cards are different. Well, after drawing the first card, the second must be different from it, and there are 48 cards that are different, in a remaining deck of 51 cards. The third card must be different from the first two, and there are 44 such in a deck of 50. The fourth must be different from the first three, and there are 40 such in a deck of 49, etc.

RGV

 

[chiro]

Hey Pizzerer and welcome to the forums.

Are you familiar with probability trees?

The basic idea is to use different branches to represent events. You can represent each layer of the tree as an independent event and then to find the probability of all the events you multiply each element from the root to some leaf event and that becomes your probability.

This method is a common one that is used and will help you build up intuition for probability.

 

[byrnesj1]

problem 1.

you need to specify that all card are different and singular. i.e (13c1 * 4c1) (12c1 * 4c1) ... denominator is good and so is (1 - .. )

problem 2.
choose a suit (4c1), now you need 5 specific cards. (13c1)*(12c1)*(11c1)*(10c1)*(9c1).. denominator is still (52c5). should get a really small probability.

problem 3.
like problem 2, except in the numerator you should include a (47c2) => choose any 2 cards beyond the 5 needed (47 left), and denominator should be (52c7).

hope this helps

 

[Ray Vickson]

problem 1.

you need to specify that all card are different and singular. i.e (13c1 * 4c1) (12c1 * 4c1) ... denominator is good and so is (1 - .. )

problem 2.
choose a suit (4c1), now you need 5 specific cards. (13c1)*(12c1)*(11c1)*(10c1)*(9c1).. denominator is still (52c5). should get a really small probability.

problem 3.
like problem 2, except in the numerator you should include a (47c2) => choose any 2 cards beyond the 5 needed (47 left), and denominator should be (52c7).

hope this helps

Nope! Your computation for P{all different} in (1) is (13*4 * 12*4 * 11*4 * 10*4)/C(52,5) = 1408/833 > 1!

Basically, in this problem, combinations have nothing to offer and just confuse the issue.

For (2): your answer is wrong. You get P = 4*13*12*11*10*9/C(52,5) = 198/833 ≈ 0.2377. However, the exact answer is obtained: For a Royal Flush of Spades, what is the probability? We have 5 'good' cards (the Royal Flush Spade cards) and 52-5 = 47 "bad' cards. We want to draw 5 good cards. So, P{RFSpades} = 5/52*4/51*3/50*2/49*1/48 = 1/2,598,960, and P{RF} = 4*P{RFSpades} = 4/2,598,960 = 1/649,740 ≈ 0.1539e-05. Note: we could have gotten this from the Hypergeometric distribution: P{RFSpades} = P{draw 5 good cards and 0 bad cards from a deck having 5 good and 47 bad cards} = C(5,5)C(47,0)/C(52,5) = 1/C(52,5) = 1/2,598,960.

For (3): P{RFSpades} = P{draw 5 good and 2 bad cards from a deck of 5 good and 47 bad} = C(5,5)*C(47,2)/C(52,5).
Then P{RF} = 4*C(5,5)*C(47,2)/C(52,7) = 1/30940 ≈ 0.3232e-04.

RGV