Probability- drawing cards from a deck.

In summary: Sorry for the confusion.In summary, the conversation discussed the probability of drawing four of the same rank from a standard deck of 52 cards when drawing 10 cards without replacement. The group came up with different methods of calculating the probability, taking into consideration the number of ways to draw 10 cards and the number of ways to draw four of the same rank. The final answer was 0.0775694%, or 1.01% when considering the number of ways to draw 10 cards.
  • #1
johnsa9
1
0
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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  • #2
Welcome to PF!

johnsa9 said:
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! :smile:

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). :smile:
 
  • #3
[tex]13*1\frac{48!}{42!6!}*\frac{42!10!}{52!}[/tex]=1.01%
 
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  • #4


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

tiny-tim said:
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]

:smile:
 
  • #5
Rogerio said:
[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]

:smile:

Hi Rogerio! :smile:

(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 :rolleyes:)

ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards. :wink:
 
  • #6


tiny-tim said:
Hi Rogerio! :smile:

(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 :rolleyes:)

ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards. :wink:

Hi tiny-tim!
the two things...:wink:

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

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
Rogerio said:
[tex]\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}[/tex]

:smile:
 
  • #7
Hi Rogerio! :smile:

i] oops! :redface:

ii] no, 48C6 is only the number of ways if the first four cards are four aces … but the aces could be anywhere :smile:
 
  • #8
tiny-tim said:
...
ii] no, 48C6 is only the number of ways if the first four cards are four aces … but the aces could be anywhere :smile:

Tiny-tim, the order doesn't matter...
So, you have exactly 48_C_6 combinations.

:smile:
 
  • #9


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


FACT: There is no <CTRL> key on Chuck Norris's keyboard. Chuck Norris is always in control. Believe it.
 
  • #10
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!
 
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  • #11
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%
 
  • #12
Never mind, I misread the problem
 

Related to Probability- drawing cards from a deck.

1. What is the probability of drawing a heart from a standard deck of 52 cards?

The probability of drawing a heart from a standard deck of 52 cards is 13/52 or 25%. This is because there are 13 hearts in a deck and 52 total cards.

2. What is the probability of drawing a face card (jack, queen, king) from a deck of cards?

The probability of drawing a face card from a deck of cards is 12/52 or 23.08%. This is because there are 12 face cards (4 jacks, 4 queens, 4 kings) in a deck and 52 total cards.

3. If you draw two cards from a deck without replacement, what is the probability of getting two aces?

The probability of drawing two aces without replacement is (4/52) * (3/51) = 1.16%. This is because there are 4 aces in a deck of 52 cards and after drawing the first ace, there are only 3 aces left out of 51 cards.

4. What is the probability of drawing a red card or a black card from a deck?

The probability of drawing a red card or a black card from a deck is 1. This is because all cards in a deck are either red or black, so the probability of drawing one or the other is 100%.

5. If you draw three cards from a deck, what is the probability of getting a three of a kind (three cards with the same value)?

The probability of getting a three of a kind when drawing three cards is (52/52) * (3/51) * (2/50) = 0.24%. This is because there are 52 ways to choose the first card, 3 ways to choose the second card (since it must match the first), and 2 ways to choose the third card (since it must match the first two).

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