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To determine the percentage that all four players in a game of bridge receive a 4-3-3-3 distribution—meaning each player has four cards in one suit and three cards in each of the other three suits—we need to calculate the probability of this event and convert it to a percentage. In bridge, a standard deck of 52 cards is dealt equally among four players, with each receiving 13 cards. Let’s compute this step-by-step.
Step 1: Understanding the 4-3-3-3 Distribution
A 4-3-3-3 distribution for a player means their 13 cards are distributed across the four suits (spades, hearts, diamonds, clubs) as 4 cards in one suit and 3 cards in each of the remaining three suits. For all four players to have this distribution simultaneously, each player must have a “long suit” with 4 cards, and we must determine how these long suits are assigned and how the cards are distributed accordingly.
Step 2: Key Insight on Suit Distribution
Since there are exactly four suits and four players, and each player has one suit with 4 cards, a natural configuration emerges: each player’s long suit could be a different suit. For example, player 1 could have 4 spades, player 2 could have 4 hearts, player 3 could have 4 diamonds, and player 4 could have 4 clubs. If two players had 4 cards in the same suit (e.g., both in spades), they’d take 8 spades, leaving only 5 spades for the other two players (e.g., 3 and 2), but a 4-3-3-3 distribution requires at least 3 cards in each non-long suit, making this impossible. Thus, each player must have their 4-card suit in a distinct suit, corresponding to a permutation of the four suits among the four players.
Step 3: Total Number of Possible Deals
The total number of ways to deal 52 cards to four players, with each receiving 13 cards (where players are distinguishable but cards within a suit are not), is the number of ways to partition 52 cards into four unordered sets of 13:
\frac{52!}{13! \times 13! \times 13! \times 13!} = \frac{52!}{(13!)^4}
This is the denominator of our probability fraction.
Step 4: Number of Favorable Outcomes
Number of Suit Assignments
Since each player’s 4-card suit must be distinct, we assign one suit to each player as their long suit. With 4 suits and 4 players, the number of ways to permute the suits among the players is:
4! = 24
Ways to Distribute Cards for a Specific Assignment
Consider a specific assignment: player 1 has 4 spades, player 2 has 4 hearts, player 3 has 4 diamonds, and player 4 has 4 clubs. For each player:
Their long suit has 4 cards.
Each of their other three suits has 3 cards.
Now, examine each suit:
Spades: Player 1 has 4, players 2, 3, 4 have 3 each. Total = 4 + 3 + 3 + 3 = 13.
Hearts: Player 2 has 4, players 1, 3, 4 have 3 each. Total = 3 + 4 + 3 + 3 = 13.
Diamonds: Player 3 has 4, players 1, 2, 4 have 3 each. Total = 3 + 3 + 4 + 3 = 13.
Clubs: Player 4 has 4, players 1, 2, 3 have 3 each. Total = 3 + 3 + 3 + 4 = 13.
For each suit (e.g., spades):
Choose 4 cards out of 13 for player 1:
\binom{13}{4}
.
From the remaining 9, choose 3 for player 2:
\binom{9}{3}
.
From the remaining 6, choose 3 for player 3:
\binom{6}{3}
.
The last 3 go to player 4:
\binom{3}{3} = 1
.
This is:
\binom{13}{4} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3} = \frac{13!}{4!9!} \times \frac{9!}{3!6!} \times \frac{6!}{3!3!} \times \frac{3!}{3!0!} = \frac{13!}{4!3!3!3!}
Since the suits are distinct and their distributions are independent (the cards of each suit are dealt separately), the number of ways for this specific assignment is:
\left( \frac{13!}{4!3!3!3!} \right)^4
Total Favorable Ways
Multiply by the number of permutations:
24 \times \left( \frac{13!}{4!3!3!3!} \right)^4
Step 5: Probability Calculation
The probability is:
P = \frac{24 \times \left( \frac{13!}{4!3!3!3!} \right)^4}{\frac{52!}{(13!)^4}} = 24 \times \frac{(13!)^4}{(4!3!3!3!)^4} \times \frac{(13!)^4}{52!} = 24 \times \frac{(13!)^8}{(4!3!3!3!)^4 \times 52!}
Step 6: Numerical Computation
Let’s compute this numerically:
13! = 6,227,020,800
4! = 24
,
3! = 6
, so
4!3!3!3! = 24 \times 6^3 = 24 \times 216 = 5,184
\frac{13!}{4!3!3!3!} = \frac{6,227,020,800}{5,184} = 1,201,200
(since
5,184 \times 1,201,200 = 6,227,020,800
)
\left( 1,201,200 \right)^4
52!
and
(13!)^4
are large, so use logarithms for precision:
\log_{10}(P) = \log_{10}(24) + 4 \log_{10}\left( \frac{13!}{4!3!3!3!} \right) - \log_{10}\left( \frac{52!}{(13!)^4} \right)
\log_{10}(24) \approx 1.3802
\log_{10}(1,201,200) = \log_{10}(1.2012 \times 10^6) \approx 6.0794
, so
4 \times 6.0794 = 24.3176
\log_{10}(52!) \approx 67.906
,
\log_{10}(13!) \approx 9.794
, so
\log_{10}\left( \frac{52!}{(13!)^4} \right) = 67.906 - 4 \times 9.794 = 28.73
\log_{10}(P) = 1.3802 + 24.3176 - 28.73 \approx -3.0322
P \approx 10^{-3.0322} \approx 0.000927
Percentage =
0.000927 \times 100 = 0.0927\%
.
Step 7: Final Answer
Rounding to two decimal places, the percentage is approximately 0.09%.
Thus, the percentage that all four players receive a 4-3-3-3 distribution is approximately 0.09%.

Ja, jag var uppenbarligen trött. Beräkningen verkar vara korrekt: (13!)^8/(24^3 * 6^12 * 52!). Min slöa miniräknare ger lite andra decimaler, men det är i alla fall i samma storleksordning.

Heja Thomas!

Alltså ungefär en gång på 40 spelkvällar.