1. ## 2 same cards

A pack of cards consists of 52 di fferent cards. A dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each.

What is the probability that one player has two identical cards?

I'm not sure how to do this, do we consider each player getting the same cards seperately then add those probabilities together? And if so, one we are consider each player, do we multiply by the probability of the other players getting normal cards?

2. Hello,l Aquafina!

A pack of cards consists of 52 different cards.
A dealer changes one of the cards for a second copy of another card in the pack
and he then deals the cards to four players, giving thirteen to each.

What is the probability that one player has two identical cards?
First, there are: . ${52\choose13,13,13,13}$ possible ways to deal the cards.

The deck has 2 Twins and 50 Others.

There is a choice of 4 players to get the Twins.

That player must get both Twins and 11 Others.
. . There are: . ${2\choose2}{50\choose11} \:=\:{50\choose11}$ ways.
And the other 39 cards are dealt to the other 3 players: . ${39\choose13,13,13}$ ways.

Hence, there are: . $4{50\choose11}{39\choose13,13,13}$ ways to deal the cards
. . so that one player gets both Twins.

Therefore: . $P(\text{a player gets the Twins}) \;=\;\dfrac{4{50\choose11}{39\choose13,13,13}}{{52 \choose13,13,13,13}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The answer works out to $\frac{4}{17}$ . . . which is a stunning surprise.

I recognize it as the answer to another problem:
. . Two cards are drawn from a deck.
. . What is the probability that the two cards have the same suit?

Are the two problems related?
Could I have found the answer in some streamlined way?

Or is this just a coincidence?

Is a puzzlement!

3. Originally Posted by Soroban
Hello,l Aquafina!

First, there are: . ${52\choose13,13,13,13}$ possible ways to deal the cards.

The deck has 2 Twins and 50 Others.

There is a choice of 4 players to get the Twins.

That player must get both Twins and 11 Others.
. . There are: . ${2\choose2}{50\choose11} \:=\:{50\choose11}$ ways.
And the other 39 cards are dealt to the other 3 players: . ${39\choose13,13,13}$ ways.

Hence, there are: . $4{50\choose11}{39\choose13,13,13}$ ways to deal the cards
. . so that one player gets both Twins.

Therefore: . $P(\text{a player gets the Twins}) \;=\;\dfrac{4{50\choose11}{39\choose13,13,13}}{{52 \choose13,13,13,13}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The answer works out to $\frac{4}{17}$ . . . which is a stunning surprise.

I recognize it as the answer to another problem:
. . Two cards are drawn from a deck.
. . What is the probability that the two cards have the same suit?

Are the two problems related?
Could I have found the answer in some streamlined way?

Or is this just a coincidence?

Is a puzzlement!

Thanks a lot.

In the solutions, the answer is 12/51. I wanted to check what someone else got because I didnt think the approach was right with what they did.

This is their method:

We shall call the four players in this game A, B, C, D, and call the two identical cards a and b. There are many such ways that two cards could appear in the same hand. One such way is that cards a and b are respectively the first and second cards in player A’s hand – the chance of this happening equals

(probability that A’s first card is a) × (probability that A’s second card is b given that A’s first card is a) = 1/52 × 1/51

as there are 52 cards in the deck, and because, once we know the first card to be a, then the second card can be one of the remaining 51 cards.

All these many different ways are equally likely – so to find the total probability we need only count the number of different ways and multiply by the above probability. We note

total number of different ways cards a and b appear in the same hand

equal

4 × total number of different ways cards a and b appear player A’s hand,

as the four players are all equally likely to get both cards. Then the above equals

4 × number of places a can take in A’s hand × number of places b can take in A’s hand given a is already there,

which equals 4×13×12, because there are 13 cards in A’s hand, and once one of these has been chosen to be a there are 12 remaining cards which can be card b.

Finally then we see the total probability is

4 × 13/52 × 12/51 = 12/51

4. Hello everyone
Originally Posted by Aquafina
A pack of cards consists of 52 di fferent cards. A dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each.

What is the probability that one player has two identical cards?

...
Here's the reason the answer is $\frac{12}{51}=\frac{4}{17}$, and it's very simple.

Call the 'twin' cards 'a' and 'b', as Aquafina has suggested, and imagine that all 52 cards have been placed face down in a line, in the order in which they are about to be dealt. Number the cards 0 through 51 according to their position in the line.

Then the first player will get cards 0, 4, 8, ...; the second 1, 5, 9, ...; and so on. Using modular arithmetic notation, the first player has cards numbered 0 (mod 4), the second has cards numbered 1 (mod 4), the third 2 (mod 4) and the fourth 3 (mod 4). There are, of course, 13 cards of each number (mod 4).

Well, card 'a' will be in one of these sets - in other words it will have a certain number mod 4 - and we simply want the probability that card 'b' has the same number mod 4. Since there are 12 such cards out of the remaining 51, the answer is obviously $\frac{12}{51}$.

5. . . $\begin{array}{c}\text{Brilliant, Grandad!} \\ \\
\text{Thank you!} \end{array}$

6. Originally Posted by Soroban
Hello,l Aquafina!

First, there are: . ${52\choose13,13,13,13}$ possible ways to deal the cards.

The deck has 2 Twins and 50 Others.

There is a choice of 4 players to get the Twins.

That player must get both Twins and 11 Others.
. . There are: . ${2\choose2}{50\choose11} \:=\:{50\choose11}$ ways.
And the other 39 cards are dealt to the other 3 players: . ${39\choose13,13,13}$ ways.

Hence, there are: . $4{50\choose11}{39\choose13,13,13}$ ways to deal the cards
. . so that one player gets both Twins.

Therefore: . $P(\text{a player gets the Twins}) \;=\;\dfrac{4{50\choose11}{39\choose13,13,13}}{{52 \choose13,13,13,13}}$
Thank you for your help. I understand that you count up the number of ways by doing, 50C11 for {50\choose11}, but how do we calculate {39\choose13,13,13}? Is this 4 * 39C13?

Sorry, I am not familiar with using combinations, apart from in binomial expansions and probability arrangements.

Hello everyoneHere's the reason the answer is $\frac{12}{51}=\frac{4}{17}$, and it's very simple.

Call the 'twin' cards 'a' and 'b', as Aquafina has suggested, and imagine that all 52 cards have been placed face down in a line, in the order in which they are about to be dealt. Number the cards 0 through 51 according to their position in the line.

Then the first player will get cards 0, 4, 8, ...; the second 1, 5, 9, ...; and so on. Using modular arithmetic notation, the first player has cards numbered 0 (mod 4), the second has cards numbered 1 (mod 4), the third 2 (mod 4) and the fourth 3 (mod 4). There are, of course, 13 cards of each number (mod 4).

Well, card 'a' will be in one of these sets - in other words it will have a certain number mod 4 - and we simply want the probability that card 'b' has the same number mod 4. Since there are 12 such cards out of the remaining 51, the answer is obviously $\frac{12}{51}$.

Thank you Grandad! Would the method you used hold if the cards were dealt by giving 13 to player A, then the next 13 to B etc and then give the fourth player the remaining the 13 cards, rather than handing them out one by one to each of them?

In the solution I posted, doesn’t the method show that they are handed to Player A first, by doing 1/52 * 1/51? i.e. He gets 1 out of 52 cards, then 1 out of the next 51.

Just had some confusion there sorry… Also, in the solution I posted, they do 4 x 13 x 12 to calculate the number of ways, but why is the 13th value included for card “a”, as this would mean that card “b” would not be given to the same player?

Thank you guys

7. Originally Posted by Aquafina
A pack of cards consists of 52 different cards. A dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each.
What is the probability that one player has two identical cards?
Aquafina, lets consider a general problem.
What is the probability that player A will get the ace of clubs and the king of clubs?
Answer: $\frac{\displaystyle \binom{50}{11}}{\displaystyle \binom{52}{13}}=\frac{1}{17}$.
That is the answer regardless how we deal the thirteen cards to each of players A, B, C, & D.
Now, what is the probability that some player will get the ace of clubs and the king of clubs?
Answer: $\frac{4}{17}$

Notice this is exactly the same problem you posted, because you have simply specified two particular cards out of fifty-two cards.

8. Originally Posted by Plato
Aquafina, lets consider a general problem.
What is the probability that player A will get the ace of clubs and the king of clubs?
Answer: $\frac{\displaystyle \binom{50}{11}}{\displaystyle \binom{52}{13}}=\frac{1}{17}$.
That is the answer regardless how we deal the thirteen cards to each of players A, B, C, & D.
Now, what is the probability that some player will get the ace of clubs and the king of clubs?
Answer: $\frac{4}{17}$

Notice this is exactly the same problem you posted, because you have simply specified two particular cards out of fifty-two cards.
Thanks, how do you use . to work out the probability? That is the number of ways to get 11 cards out of 50 over getting 13 out of 52?

This is basically what my question was, Soroban used which I didnt understand...

9. Originally Posted by Aquafina
Thanks, how do you use . to work out the probability? That is the number of ways to get 11 cards out of 50 over getting 13 out of 52?

This is basically what my question was, Soroban used [IMG]file:///C:/Users/Mohammad/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG][IMG]file:///C:/Users/Mohammad/AppData/Local/Temp/moz-screenshot-3.jpg[/IMG] which I didnt understand...
Hi Aquafina,

You seem to be having some trouble with the notation.

$\binom{50}{11}$ is the number of ways to get select 11 cards from a deck of 50, or to put it another way, the number of combinations of 50 objects taken 11 at a time, which is $\frac{50!}{11! \; 39!}$

$\binom{52}{13,13,13,13}$ is a generalization of the number of combinations called a multinomial coefficent. Its formula is $\frac{52!}{13! \; 13! \; 13! \; 13!}$. It is the number of ways to select 52 cards and put them in four bins with 13 cards in each bin. See the following link:

Multinomial theorem - Wikipedia, the free encyclopedia