Deck of Cards

wtrow

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Jan 24, 2011
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A deck of 52 cards is dealt among 4 people. What is the probability that every player gets a Queen?

I thought this was an easy (4/52)*(3/51)*.... until I realized the question said that EVERY card was dealt. So each player gets 13 cards. I know it is without replacement and order, so n!/((n-k)!k!) is relevant. After that I really have no clue, we haven't seen anything like this in class yet.
 

Subhotosh Khan

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wtrow said:
A deck of 52 cards is dealt among 4 people. What is the probability that every player gets a Queen?

I thought this was an easy (4/52)*(3/51)*.... until I realized the question said that EVERY card was dealt. So each player gets 13 cards. I know it is without replacement and order, so n!/((n-k)!k!) is relevant. After that I really have no clue, we haven't seen anything like this in class yet.
Start with 52 cards and 4 queens - deal out 13 cards. What is the probability that "exactly one" of those cards is a queen?

Then you have 39 cards and 3 queens - deal out 13 cards. What is the probability that "exactly one" of those cards is a queen?

and so on....
 

wtrow

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Thank you for the reply, it helped me a lot. I used your method to get the probability of exactly one queen going to the first player which ended up like this:

4/52 (chance the first card is a queen) * 48/51 (chance next is not) *47/50 * ..... *37/40

This ended up as: (4/52) * ((48!/37!)/(51!/40!)) = .036495 * 13 (since any one could be a queen) for the first player.

Using the same method, the second player's probability is .038407 * 13, third's is .04308 * 13, and the fourth's is 1.

Now I just multiply these results to get the final result, right?
 

soroban

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Hello, wtrow!

A deck of 52 cards is dealt among 4 people.
What is the probability that every player gets a Queen?

\(\displaystyle \text{There are: }\:{52\choose13,13,13,13} \:=\:\frac{52!}{13!\,13!\,13!\,13!}\text{ possible deals.}\)


\(\displaystyle \text{Give a Queen to each of the four people.\;\;There are: }4!\text{ ways.}\)

\(\displaystyle \text{Deal the remaining 48 cards to the four people.\;\;There are: }\,{48\choose12,12,12,12} \:=\:\frac{48!}{12!\,12!\,12!\,12!}\text{ ways.}\)

\(\displaystyle \text{Hence, the number of ways that each player gets a Queen is: }\:\frac{4!\,48!}{12!\,12!\,12!\,12!}\)


\(\displaystyle \displaystyle \text{Hence: }\;P(\text{each gets a Queen}) \;=\;\frac{\dfrac{4!\,48!}{12!\,12!\,12!\,12}} {\dfrac{52!}{13!\,13!\,13!\,13}}\)



\(\displaystyle \text{Don't panic . . . This can be easily simplified.}\)


\(\displaystyle \text{We have: }\;\frac{4!\,48!}{12!\,12!\,12!\,12!}\,\cdot\,\frac{13!\,13!\,13!\,13!}{52!} \;=\;\frac{4!}{1}\,\cdot\,\frac{13!\,13!\,13!\,13!}{12!\,12!\,12!\,12!}\,\cdot\,\frac{48!}{52!} \;=\;\frac{4\cdot3\cdot2\cdot1}{1}\,\cdot\,\frac{13\cdot13\cdot13\cdot13}{1}\,\cdot\,\frac{1}{52\cdot51\cdot50\cdot49}\)


. . . . \(\displaystyle \text{which reduces to: }\;\frac{2,\!197}{20,\!825}\)

 

wtrow

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Jan 24, 2011
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Thank you soroban, your answer was very descriptive.

I guess my method was wrong. I'm new to this probability stuff.
 

Denis

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Definitely correct, Soroban; 2197 / 20825 = ~.105498

Ran simulation, a million deals 5 times:
105392
105522
104935
105722
105892
 

soroban

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Nice work, Denis!

. . . Thanks!
 
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