12 Cards are delath from 52 playing cards

**desire150** · November 13th 2009, 07:09 AM

Suppose that a hand of 12 cards is dealt from a normal deck of 52 playing cards. What is the probability that of the 12 cards, 3 will be diamonds, 3 will be clubs, 3 will be spades, and 3 will be hearts?

This is the answer I came up with... 0.07692

Please let me know if this is correct.

Thank you.

**Plato** · November 13th 2009, 07:35 AM

Originally Posted by desire150

Suppose that a hand of 12 cards is dealt from a normal deck of 52 playing cards. What is the probability that of the 12 cards, 3 will be diamonds, 3 will be clubs, 3 will be spades, and 3 will be hearts?

This is the answer I came up with... 0.07692

Please let me know if this is correct.

Thank you.

Show us how you got that. Then we can answer.

**desire150** · November 13th 2009, 07:58 AM

Here is how I came up with the answer...

12/52= 0.2307
0.2352/3=0.07692

**Plato** · November 13th 2009, 08:09 AM

Originally Posted by desire150

Here is how I came up with the answer...

12/52= 0.2307
0.2352/3=0.07692

That is, unfortunately, not even close to understanding this answer.
Do you know about combinations? $_N\mathcal{C}_k=\frac{N!}{k!(N-k)!}$ .

Here the number of ways to get three hearts: $_{13}\mathcal{C}_3=\frac{13!}{3!(10)!}$ .

Do that for each of the other three suits.

**desire150** · November 13th 2009, 08:16 AM

Dear Plato,
As you can see, my knowledge is extremely basic.

Here is the result using the formula

13
___
3(10) = 0.433 probability for obtaining 3 diamonds, 3 hearts, 3 clubs, 3 spades.

Thank you.

**Plato** · November 13th 2009, 08:27 AM

Originally Posted by desire150

Dear Plato,
As you can see, my knowledge is extremely basic. Here is the result using the formula

13
___
3(10) = 0.433 probability for obtaining 3 diamonds, 3 hearts, 3 clubs, 3 spades.

That being the case, has it occurred to you that you are not ready to even understand to solution?
You apparently do not even know what $(13!)$ means.
Now don’t come back at saying that we should tell how.
The function of this forum is not to teach.
We can only help if you do have the basic ideas.

**Soroban** · November 13th 2009, 08:39 AM

Hello, desire150!

A hand of 12 cards is dealt from a normal deck of 52 playing cards.
What is the probability that there are: $3 \diamondsuit s,\;3 \clubsuit s,\; 3 \spadesuit s,\,\text{and }3 \heartsuit s\:?$

$\begin{array}{ccc}\text{Number of ways to get 3 }\diamondsuit s &=& {13\choose3} \\ \\[-4mm]<br /> \text{Number of ways to get 3 }\clubsuit s &=& {13\choose3} \\ \\[-4mm]<br /> \text{Number of ways to get 3 } \spadesuit s &=& {13\choose3} \\ \\[-4mm]<br /> \text{Number of ways to get 3 }\heartsuit s &=& {13\choose3}\end{array}$

Hence, there are: . ${13\choose3}^4$ ways to get 3 of each suit.

There are: . ${52\choose12}$ possible 12-card hands.

Therefore: . $P(\text{3 of each suit}) \;=\;\frac{{13\choose3}^4}{{52\choose12}}$

**desire150** · November 13th 2009, 10:36 AM

Soroban,
Thank you for explaining it. After doing research and plugging in the numbers into the formula.. this is what I get.

0.231 probability that 3 spades, hears, clubs and diamonds will be chosen.

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