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wtrow
01-24-2011, 11:47 AM
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
01-24-2011, 12:16 PM
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
01-24-2011, 12:39 PM
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
01-24-2011, 12:52 PM
Hello, wtrow!

A deck of 52 cards is dealt among 4 people.
What is the probability that every player gets a Queen?
\text{There are: }\:{52\choose13,13,13,13} \:=\:\frac{52!}{13!\,13!\,13!\,13!}\text{ possible deals.}

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

\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.}

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

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

\text{Don&#39;t panic . . . This can be easily simplified.}

\text{We have: }\;\frac{4!\,48!}{12!\,12!\,12!\,12!}\,\cdot\,\fra c{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{1 3\cdot13\cdot13\cdot13}{1}\,\cdot\,\frac{1}{52\cdo t51\cdot50\cdot49}

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

wtrow
01-24-2011, 08:31 PM

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

Denis
01-25-2011, 01:11 AM
Definitely correct, Soroban; 2197 / 20825 = ~.105498

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

soroban
01-25-2011, 11:02 AM
Nice work, Denis!

. . . Thanks!