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Bridge is played with four players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start? (Ignore the fact that bridge is played in partnership)

Here is my attempt: For the first player I have $52 \choose 13 $. For the second player I have $39\choose 13$. Then third player $26\choose 13$ and lastly $13\choose 13$.

Therefore, I have $52\choose 13$$39\choose 13$$26\choose 13$$13\choose 13$ of ways a bridge game can start.

Am I correct with this ? Is there another way to think this questions?

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    $\begingroup$ This is right for four distinct players (table positions), which is the correct way to look at bridge; who bids when is important. I'm not quite sure what the question means when it specifies ignoring partnership. $\endgroup$ – Joffan Jul 22 '18 at 17:52
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    $\begingroup$ An equivalent solution is that the number of ways is a multinomial coefficient, $\binom{52}{13 \; 13 \; 13 \; 13} = {52!}/(13!)^4$ $\endgroup$ – awkward Jul 22 '18 at 18:46
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I agree with the comment that it's unclear what the prescription to ignore the fact that bridge is played in partnership means. You've correctly calculated the number of ways a bridge game can start if you regard the players as distinct, which is the appropriate way to look at the situation in bridge.

To illustrate how the result can depend on what you regard as different situations, consider the game of Getaway, where the player who holds the ace of spades starts the game. If you want to know how many different situations you might face as a player, the result would again be the same (assuming four players are playing), since each player is in a distinct position relative to you. However, if you're not one of the players and you don't care about their identities and just want to know how many inequivalent situations might arise in this game, you need to take into account the cyclical symmetry among the players. The order of the players still matters, so there's no full permutational symmetry among them, but it doesn't matter whether you number them e.g. $1234$ or $2341$. Thus you'd have to divide your result by $4$ to get the number of inequivalent starting positions. This doesn't apply to bridge, though, since in bridge the dealer opens the auction, so there are distinguished first, second, third and fourth players. Thus, in the case of bridge, your result is right no matter whether you look from the position of a distinguished player or from outside without regard for the identity of the players. Since this is true with or without partnerships, it's not clear what the prescription to ignore the partnerships means.

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I am attaching a link to another way of looking at the problem. http://personal.psu.edu/drh20/250h/spring2002/practice/quiz07.pdf

Basically, what they are saying is that each card can belong to Player 1, 2, 3, or 4. So color all the cards of Player 1 red, Player 2 blue, Player 3 green and Player 4 yellow. Then ultimately you have 13 cards of each color type and you have to put them all in a deck and use permutation to get the number of ways. This gives us

$$\frac{52!}{13!13!13!13!}$$

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