Probability to get 13 cards such that none of the 13 cards have the same number

Question

There are $52$ cards in a deck is distributed equally to $4$ people. What is the probability for one of 4 people to get $13$ cards such that none of them have the same value?

Example:

• {2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is correct
• {3♠, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is incorrect

Note: my teacher said result $\frac{4^{13}}{^{52}C_{13}} \approx 0.0001057$ is incorrect and the order of 13 cards on player's hand does not matter.

My programming experiment:

I know a bit about Python programming, so I did a test, my test specific was:

• A deck is distributed equally to $4$ people
• Then I check if any player satisfies 13 cards and none of them have the same number $\Rightarrow$ this is a successful experiment and if none of 4 people matches the condition so this is an unsuccessful experiment.

I do this test 100 000 times, add up all the successful experiments and divide it by 100000, the probability always falls between 0.0003 and 0.0005

Answer 1

You need to count the number of ways in which each person can receive one card of each denomination. Each denomination can be distributed to the $4$ players in $4!$ different ways, so there are $4!^{13}$ different deals that give each player one card of each denomination. There are

$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}=\binom{52}{13,13,13,13}=\frac{52!}{13!^4}$$

possible deals, so the desired probability is

$$\frac{4!^{13}}{\frac{52!}{13!^4}}=\frac{4!^413!^4}{52!}\,,$$

which is a bit more than $6\times 10^{-24}$.

Added: This is the probability that each player receives one card of each denomination. Subsequent edits make it unlikely that this is the intended interpretation of the question. If what is desired is the probability that at least one player receives a card of each denomination, or that exactly one player does so, the calculation is significantly messier.

Answer 2

If the question asks for the probability that a particular person's hand doesn't have two cards of same value, then:

Out of the 13 cards in your hand, the first can be any of the 52 cards.

For the second one, it can be any of 48 cards (excluding the first card and 3 other cards of other suits with the same value)

For the third one, it can be any of 44 cards (excluding the first two cards and the cards with same value but different suits)

Continuing similarly, there are $52\times 48\times\dots\times 8\times 4=4^{13}\times 13!$ favorable configurations out of the $52!$ configurations. This results in the answer you gave, ie, $\dfrac{4^{13}\times 13!}{52!}=\dfrac{4^{13}}{\binom{52}{13}}$, so I think your teacher referred to the latter case in the problem.

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If the probability is that no person gets two cards of same value, then count configurations where each person gets a hand with all value cards but different suits, so $4!^{13}$ configurations out of a total of $\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$ configurations (total number of possible deals) because for each value card, the suits in the 4 player hands can be permuted in $4!$ ways for each of the 13 valued cards.

Answer 3

Calculation based on first card is free. Get probability all remaining cards have to be different. Second card is $\frac{48}{51}$ third card is $\frac{44}{50}$ summary is $\frac{12!\times 4^{12}\times 39!}{51!}=0.000105681$

This is the probability for one particular player getting it right. Since there are four players the question seems to be at least one gets the desired result. That is approximately $4$ times as much, ignoring dependency of the hands.

............ exact calculation. Let $P_1$ be the result of the above calculation. $P_2= \frac{12!\times3^{12}\times26!}{38!}=0.0001962866$ and $P_3=\frac{12!\times2^{12}\times13!}{25!}= 0.0007876469$

The probability at least one good hand $=P_1$, the probability of at least two good hands is $Q=P_1P_2$, and the probability of all four hands are good is $R=P_1P_2P_3$.

One or more good hands occurs four times each includes two or more three times, while each two or more includes one all. Adding it all up and removing duplicates the final probability is $4P_1-6Q+3R$.

Answer 4

You should get $13$ different values. First card has $52$ choices, second has $48$ choices, etc down to $4$. So multiply all multiples of $4$ from $4$ to $52$, and divide by the total, which is ${52} \choose 13$.

When you simplify you should get $\frac{4^{13}\times 13!}{52 \choose 13}$

Answer 5

Imagine arranging the cards in 13 rows and 4 columns. The first row contains all four aces, the second row contains all four kings, the third row contains all four queens, etc. Assign the first column of cards to person #1, the second column of card to person #2, the third column of cards to person #3, and the fourth column of card to Person #4. Since are $4!$ ways to arrange the cards in each row, the total number of such arrangements is $(4!)^{13}$ making the desired probability $\frac{(4!)^{13}}{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}}$