Thread: getting 5 cards of clubs, diamonds, hearts, spades out of 20 turns

1. getting 5 cards of clubs, diamonds, hearts, spades out of 20 turns

Hello!

I got one problem.

It says:

Out of deck of 52 cards we choose 20 cards and we return each card back into the deck. Count the probability that there will be 5 cards of clubs, diamonds, hearts, spades.

A - 5 cards of clubs

B- 5 cards of diamonds

C- 5 cards of hearts

let N be the number of ways of getting 5 cards out of 13.

combinations with repetition:
N(A)=Binomial(13+5-1,5) = Binomial(17,5)=N(B)=N(C)=N(D)

Now I do this:

Assuming the experiments are independent (because we return the card in the deck):

I have probability P = Binomial(17,5) ^4 / Binomial(71,20)

These are combinations with repetition.

I get P = 0.0065059

But I tried the experiment with R Commander and it gives me P = 0.010666 with about 1 000 000 times throwing 20 cards.

So what is correct?

2. do u actually mean the probability of 20 cards including 5 clubs, 5 diamonds, 5 hearts, and 5 spades?
if so, maybe you could analyze it as follows:
consider one possible sequence of the cards, say we subsequently drew 5 clubs, 5 diamonds, 5 hearts, and 5 spades, since each card will be replaced, so the probability of drawing each card with specific type is 1/13. Thus the probability of getting this sequence is simply (1/13)^20;
given 20 cards, there will be P(20, 20) sequences, now sum them up, we get the P(20, 20)*(1/13)^20.
You could try to evaluate P(20, 20)*(1/13)^20 using R Commander, see if it is close to P = 0.010666 by the experiment.

3. Sorry for misunderstood.

I meant probability of 20 cards including five cards of clubs, five cards of diamonds, five cards of hearts and five cards of spades.

I evaluated that expression and it is 0.000128014. I do not what is the problem.

Thank you.

4. Originally Posted by p0oint
Hello!

I got one problem.

It says:

Out of deck of 52 cards we choose 20 cards and we return each card back into the deck. Count the probability that there will be 5 cards of clubs, diamonds, hearts, spades.

A - 5 cards of clubs

B- 5 cards of diamonds

C- 5 cards of hearts

let N be the number of ways of getting 5 cards out of 13.

combinations with repetition:
N(A)=Binomial(13+5-1,5) = Binomial(17,5)=N(B)=N(C)=N(D)

Now I do this:

Assuming the experiments are independent (because we return the card in the deck):

I have probability P = Binomial(17,5) ^4 / Binomial(71,20)

These are combinations with repetition.

I get P = 0.0065059

But I tried the experiment with R Commander and it gives me P = 0.010666 with about 1 000 000 times throwing 20 cards.

So what is correct?

Hi POoint,

If we take a card out of the deck, write down its suite, and then return the card to the deck, and we do this 20 times, the number of possible sequences in which there are 5 hearts, 5 clubs, 5 diamonds, and 5 spades is

$\binom{20}{5 \; 5 \; 5 \; 5} = \frac{20!}{(5!)^4}$

Each such sequence has a probability of
$(\frac{1}{4})^{20}$,
so the probability of seeing 5 of each suite is

$\binom{20}{5 \; 5 \; 5 \; 5} (\frac{1}{4})^{20}$

This is an example of the Multinomial Distribution.
Multinomial distribution - Wikipedia, the free encyclopedia

I think if you work this out you will find the result agrees well with your simulation.

5. Thank you for the response.

It agrees with the simulation.

Yeah, I write P=(13/52)^20 that first came to my mind. But the number of ways that these cards could be permuted was missing.

So 20!/(5!*5!*5!*5!) is actually permutation of repetition of indistinguishable objects.