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Suppose we have deck of $52$ cards and we deal it fairly to four players. We want to find out the probability that each player gets an ace.

The straightforward solution is simple - we find number of all possible partitions, which is $\frac{52!}{13!13!13!13!}$, and then number of ways to distribute 4 aces multiplied by ways to distribute other cards, which is $4! \cdot \frac{48!}{12!12!12!12!}$. Therefore, the solution is $\frac{4!*48!}{12!12!12!12!} / \frac{52!}{13!13!13!13!}$.

However, instructor from 6.041x course from MIT proposes this trick - suppose we take the $52$ cards and stack it so that four aces are on the top of the deck and we think of each player containing $13$ slots for cards. This way, the calculations are much simpler - we have $52/52$ slots for dealing the first ace, $39/51$ for second, $26/50$ for third and $13/49$ for fourth. Thus we obtain $$P(\text{each person gets an ace}) = \frac{39}{51} \cdot \frac {26}{50} \cdot \frac {13}{49},$$ which is the same as answer before.

The question is: how can one prove that with this way of dealing the cards all permutations are equally likely?

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The deal begins with fifty two slots, into any one from which the top card of the deck is dealt, without bias.   The second card of the deck is then dealt into any one from the fifty one remaining slots, without bias.   And so forth for the remaining cards until all slots are filled.

If the dealer can guarantee that the assignment remains unbiased each time, then the deck will be shuffled into fifty two slots such that there is equitable probability for occurance of each possible premutation.

One way to ensure the assignment is unbiased is to take a second deck of cards marked with numbers 1 to 52, shuffle it well, and thus use that deck to determine the placement for the cards of the first deck.   Indicating that the results of such a task is identical to simply shuffling the first deck and dealing it to the players.

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Maybe this "visualization" will help?

Original problem setting: the deck is (presumed) randomly shuffled and each of the $52!$ is equally likely. In this case you can deal to 4 people by giving the first 13 cards to player 1, the next 13 to player 2, etc.

Now imagine the players have card sleeves (card protectors if you're into CCGs) which are numbered. Player one has sleeves numbered 1 to 13, player 2 has sleeves numbered 14 to 26, etc. And sleeve number $j$ always gets the $j$th card.

New problem setting: the deck is not randomly ordered at all, but instead the 4 aces are on top (and maybe the 4 kings are next, etc). However, the sleeves (numbered 1-52) are randomly shuffled and every player grabs 13. Now deal the first card (an ace) to sleeve #1, the second card (an ace) to sleeve #2, etc.

The two settings are equivalent because shuffling the cards and shuffling the sleeves make no difference.

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