Probability of getting the same 10 cards after a 104 card shuffle

Consider this situation:

You have two standard decks merged together (104 cards). You give 10 cards to each of 4 players. What is the probability that the next time you deal (after the deck is randomly shuffled), everyone gets the same hand they got last turn? Just one person gets the same hand they got last turn?

My approach was something along 104 choose 40 (for everyone getting the same card) or 104 choose 10 (for one person getting the same card), but in my approach I'm not considering the duplicate cards in my deck. I don't think it's 52 choose 40 or 52 choose 10 because then I wouldn't be considering that there is more than one card.

I am very curious about this question and I would love to get an answer. I know its a combinatorics question, but I am unsure of the details.

After posting this question here I got an answer that I don't understand (the longer one). Perhaps someone from the Brilliant community could try to explain it to me? Thank you :)

Note by David Holcer
6 years, 1 month ago

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I'll try to explain to you how I think you should proceed.

There are 104 cards. Two of each kind. So we can make 54 pairs, each pair containing two identical cards. Now, we are tasked with finding the probability. To do this, we have to first find the total number of possible ways of distributing the cards.

How will we do this? I'm not sure if I'm right, but shuffling a deck results in it being in one of the permutations of its cards. We then distribute the first 40 cards on the top to our players. Hence, I think the total number of outcomes is S=102!(2!)52S=\frac {102!} {(2!)^{52}}

Now, we want one person to get all the same cards that he/she got in the previous turn. For this, we need only worry about the first ten cards. These 10 cards may include identical pairs also and we have to take these things under considerations. Basically what you do is fix the first ten cards(in the sense that the first ten cards can be permuted among themselves, but they cannot be permuted with the other cards). Then we find the number of permutations of the rest of the 92 cards, multiply it with the number of permutations of the first 10 between themselves.

The catch here is that there are many ways to fix the first 10 cards due to the fact that there are identical pairs. The person may have gotten all different cards, or one pair and 8 different cards, or 2 pairs and other cards etc.

I will illustrate a simple sub case, which is when the person gets an identical pair and 8 other cards in the first turn. The contribution of this case to the answer will be:

(probability that the payer gets an identical pair and 8 other cards) X (probability that it happened again)

The probability of getting an identical pair and 8 other cards is:

(52choose1)×(51choose8)×(P)S\frac {(52choose1) \times (51choose8) \times (P)}{S} (continue to understand what P is)

Say the deck before dealing contained the following. (cards are numbered from 1 to 52)

(23,45,4,19,15,15,32,50,3,9),(other cards 43 pairs and 8 unique cards)

The cards in the first bracket are dealt to the player in the first turn. For this to happen again, the first 10 cards should contain a permutation of the same 10 cards above. The rest of the cards can be arranged in any way. Thus, the number of ways a certain "one pair 8 different" arrangement can occur is:

P=10!2!×94!(2!)43P=\frac{10!}{2!} \times \frac{94!}{(2!)^{43}}

Thus, the probability contribution of this case is:

(52choose1)×(51choose8)×(P)S×PS\frac {(52choose1) \times (51choose8) \times (P)}{S} \times \frac {P}{S}

=((52choose1)×(51choose8)×(10!2!×94!(2!)43)2(102!(2!)52)2=(\frac {(52choose1) \times (51choose8) \times (\frac{10!}{2!} \times \frac{94!}{(2!)^{43}})^2}{(\frac {102!} {(2!)^{52}})^2}

This number is extremely small

The final answer will be sum of probabilities of each case.

Hope you got something out of this, it is not an easy problem, if you ask me.

Raghav Vaidyanathan - 6 years, 1 month ago

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I have no clue if this is right... I'm pretty sure it isn't. The probability seems to be too big :(

Raghav Vaidyanathan - 6 years, 1 month ago

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Thank you very much Raghav, I somewhat understand it now, it's quite a small chance, likely to occur only once in 2500 times. I could try to test it with computer science :).

David Holcer - 6 years, 1 month ago

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Actually, the thing is, I am unsure due to the same reason. 1 in 2500 is a very big probability in card distribution terms.

For example, find the probability that person AA gets dealt the same cards again while playing with four other people, given that each person is dealt 55 cards each. (Deck of 52 cards)

Raghav Vaidyanathan - 6 years, 1 month ago

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