Probability- drawing cards from a deck.


by johnsa9
Tags: combinations, probability
johnsa9
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#1
Feb12-09, 09:19 PM
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I am having some issues with trying to figure out how to go about solving this problem.

You have a standard deck of 52 cards and you draw 10 without replacement.
What is the probability of drawing four of the same rank? (ex. 4 aces)
Order doesn't matter, so I am thinking I have to do some sort of combination over the number of ways to draw ten cards. I could be completely wrong though...

Please help? :D
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tiny-tim
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Feb13-09, 10:58 AM
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Quote Quote by johnsa9 View Post
You have a standard deck of 52 cards and you draw 10 without replacement.
What is the probability of drawing four of the same rank? (ex. 4 aces)
Hi johnsa9! Welcome to PF!

Is this an exam question, or did you just make it up?

Exam questions are usually less complicated than this (in the sense that they don't involve too much number-slugging).

You have to count all the ways of drawing 10 cards (which is easy!), and all the ways in which 4 of them are the same (which is really complicated).
robert Ihnot
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Feb14-09, 04:03 AM
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[tex]13*1\frac{48!}{42!6!}*\frac{42!10!}{52!}[/tex]=1.01%

Rogerio
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#4
Feb15-09, 04:50 AM
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Probability- drawing cards from a deck.


Quote Quote by tiny-tim View Post
You have to count all the ways of drawing 10 cards...
[tex] {52\choose 10} [/tex]

Quote Quote by tiny-tim View Post
and all the ways in which 4 of them are the same.
Well, there are aces, kings, queens, jacks, etc...So,

#combinations we are interested in =
#combinations with 4 aces (including combinations with 4 kings, etc... )
PLUS
#combinations with 4 kings (including combinations with 4 aces, etc... )
PLUS
etc...
MINUS
#combinations with 4 aces e 4 kings (because they were counted twice )
MINUS
#combinations with 4 aces e 4 queens (because they were counted twice )
MINUS
etc...

That is:
[tex]{13\choose 1}{48\choose 6} - {13\choose 2}{44\choose 2}[/tex]



So, the probability of getting four of same rank is
[tex]\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}[/tex]

tiny-tim
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Feb15-09, 06:59 AM
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Quote Quote by Rogerio View Post
[tex]{13\choose 1}{48\choose 6} - {13\choose 2}{44\choose 2}[/tex]

So, the probability of getting four of same rank is
[tex]\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}[/tex]

Hi Rogerio!

(btw, please don't give too much of the answer away before the OP has had a chance to try it him/herself)

two things

i] you have to add back in the probability of three fours (ok, i know that's pretty small )

ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards.
Rogerio
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Feb16-09, 09:11 AM
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Quote Quote by tiny-tim View Post
Hi Rogerio!

(btw, please don't give too much of the answer away before the OP has had a chance to try it him/herself)

two things …

i] you have to add back in the probability of three fours (ok, i know that's pretty small )

ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards.
Hi tiny-tim!
the two things...

i] only Chuck Norris (and maybe David Copperfield) could get three fours using just 10 cards.

ii] the "number of ways of laying down 4 aces among 10 cards" is already the factor [tex]{48\choose 6}[/tex]

So, the probability of getting four of same rank is
Quote Quote by Rogerio View Post
[tex]\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}[/tex]
tiny-tim
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Feb16-09, 12:04 PM
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Hi Rogerio!

i] oops!

ii] no, 48C6 is only the number of ways if the first four cards are four aces but the aces could be anywhere
Rogerio
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Feb16-09, 12:14 PM
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Quote Quote by tiny-tim View Post
...
ii] no, 48C6 is only the number of ways if the first four cards are four aces but the aces could be anywhere
Tiny-tim, the order doesn't matter...
So, you have exactly 48_C_6 combinations.

Adeimantus
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Feb16-09, 01:30 PM
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Quote Quote by Rogerio View Post
...
i] only Chuck Norris (and maybe David Copperfield) could get three fours using just 10 cards....

FACT: There is no <CTRL> key on Chuck Norris's keyboard. Chuck Norris is always in control. Believe it.
robert Ihnot
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Feb16-09, 04:25 PM
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Rogerio: Tiny-tim, the order doesn't matter...
So, you have exactly 48_C_6 combinations.


I thought the simplist way to look at this was to just take out the four cards that are the same--say aces. Well, then there are only 13 such sets of four cards in the whole deck. Now if we add a 5th card, there are 48 ways of adding that fifth card, as Poker players know. So by extension we arrive at [tex]13{48\choose 6}[/tex]. Order does not matter.

But, I did not consider, as did Rogerio did, that duplicate sets could occur after we draw 8 or more cards!
Brunonj
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#11
Dec16-11, 03:49 PM
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I would like to first stress that I have never taken a statistics course in my life, so my logic may be flawed, but if so it would be helpful if someone explained to me where the flaw is. So as far as I can tell the problem is trying to calculate the probability that 4 aces will be drawn out of a 52 card deck when 10 cards are drawn. I believe this ratio would be equal to:

the number of hands possible not counting for order that can be drawn with 4 aces/the number of hands possible not counting for order that can be drawn

so, when looking at the numerator I would assume that each of those hands contain 4 aces and 6 other cards, and that each grouping of 6 would have at least one different card from each other grouping. which would be equal to 48!/(6!42!) total hands (as if I'm drawing 6 cards from a deck of 48). The 48! in this being the number of ways the deck can be arranged, the 42! accounting for the number of ways the nonviewed cards could be arranged without changing the 6 cards in the hand and the 6! accountinf for the number of ways the viewed cards could be arranged without changing the 6 cards that will be in hand.

Out of similar logic I derive the denominator to be 52!/(10!42!) because the number of ways the deck could be arranged is 52!, the number of ways that the viewed and nonviewed cards could be arranged without changing the hand are 10! and 42! respectively.

at this point I come to (48!/(6!42!))/(52!/(10!42!))
=(48!/6!)/(52!/10!)
=(10!/6!)/(52!/48!)
=(10*9*8*7)/(52*51*50*49)
=5040/6,497,400
=.000775694...
=.0775694%
Brunonj
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#12
Dec17-11, 10:52 AM
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Never mind, I misread the problem


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