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Mike draws five cards from a standard 52-card deck. What is the probability that he draws a card from at least three of the four suits? Express your answer as a simplified fraction.

RektTheNoob  Jan 29, 2018
 #1
avatar+92624 
+2

Mike draws five cards from a standard 52-card deck. What is the probability that he draws a card from at least three of the four suits? Express your answer as a simplified fraction.

 

I'll give it a go :)

 

There are 52*51*50*49*48 =  311875200  permutationss of 5 cards altogether (with no restrictions)

 

I'm going to try and work out how many of these do not include at least 1 card from 3 different suits.

 

5 hearts

1*(13*12*11*10*9) 

4 hearts and one other card

5*(13*12*11*10*39)

3 hearts and 2 from one other suits

5C2* (13*12*11*13*12)*3  

And this can be multiplied by 4 beacuse there are 4 suits 

 

so

 Number of combinations without 3 different suits represented

= [1*(13*12*11*10*9)   +  5*(13*12*11*10*39)   +  5C2* (13*12*11*13*12)*3 ]    *4

=(13*12*11*4)[ 90  +  5*(390)   +  10* (156)*3  ]

= [(13*12*11*4)[ 90  +  1950   +  4680  ]

= [(13*12*11*10*4)[ 9  +  195  +  468  ]

= 13*12*11*10*4* 672 

 

So the number of combinations with at least 3 different suites represented 

=  52*51*50*49*48 - 13*12*11*10*4* 672 

= 12*4   [ 52*51*50*49   -   13*11*10* 672 ]

= 12*4*10*13   [ 4*51*5*49   -   11* 672 ]

= 12*4*10*13*4   [ 51*5*49   -   11* 168 ]

= 12*4*10*13*4 *7  [ 51*5*7   -   11* 24 ]

= 12*4*10*13*4 *7 *3 [ 17*5*7   -   11* 8 ]

= 3*4*4*7 *10*12*13 [ 595   -   88 ]

= 3*4*4*7 *10*12*13 *507

 

So the prob of not getting at least 3 suites 

 

\(=  \dfrac{3*4*4*7 *10*12*13 *507}{52*51*50*49*48}\\ =  \dfrac{507}{17*5*7}\\ =  \dfrac{507}{595}\\\)

Melody  Jan 29, 2018
edited by Melody  Jan 29, 2018
 #7
avatar+376 
+2

THanks Melody!

RektTheNoob  Feb 2, 2018
 #3
avatar+86890 
+1

Here's my take on this....like Melody....not sure of the outcome !!!

 

Number of ways  of drawing all 5 cards from one suit  =  C(13,5)

And there are four ways of doing this =  C(4,1)

 

Number of ways of drawing 5 cards from two suits

We want to first choose any  one of four suits = C(4,1)  and from one of these we want to choose 3 cards  = C(13.3)

And then we want to choose one of the three remaining suits = C(3,1)  and from this suit we want to choose any 2 cards  =  C(13,2)

 

And the number of ways of choosing any 5 cards is C (52, 5)

 

So the probability of choosing 5 cards from two suits is

 

[ C(4,1) * C(13,5)  +  C(4,1) * C(13,3) * C(3,1) * C(13,2) ]  / C(52,5)  ⇒ 

1749 / 16660

 

So...the probability of drawing 5 cards from at  least 3 suits  = 

 

1 -

[ P(drawing 5 cards of all the same suit) + P( drawing 5 cards from two suits )] =

 

1  -  1749 / 16660  =    14911/ 16660   ≈  89.5%

 

 

 

cool cool cool

CPhill  Jan 29, 2018
 #4
avatar+92624 
+1

Hi Chris,

Your answer is better than mine in that I certainly did not need to get into permutations.

My logic did lead to a correct answer I think (as verified below) but it made it unnecessarily messy and there was far too much room to make a numeric error.   

 

However:

Your answer is not correct because you have not included all the possibilities.

 

I'll try again. Using your ideas.

 

Ways to get all 5 from the same suit = 4*13C5    (Just like Chris said)

Ways to get 4 from one suit and 1 from another = 4*13C4*  39  (you forgot this one Chris)

Ways to get 3 from one suit and 2 from another = 4*13C3*   3*13C2

 

Total number of ways to choose 5 cards from 52 = 52C5

nCr(52,5) = 2598960 

 

Ways to get no more than 2 suits = 4*13C5   +  4*13C4*  39  +  4*13C3*   3*13C2

4*nCr(13,5)+4*nCr(13,4)*3*13+4*nCr(13,3)*3*nCr(13,2) = 384384

 

Number of ways of NOT getting 2 suits = 52C5 - 384384

nCr(52,5)-384384 = 2214576

 

So 

P(drawing at least 3 different suits) = 2214576 / 2598960

 

2214576/2598960​ = 0.8521008403361345 = 507/595

 

Which is exactly the same as I got doing it with permutations !!   

WOW  laugh

Melody  Jan 29, 2018
 #5
avatar+86890 
+1

OK.....thanks for your explanation, Melody  !!!

 

 

 

cool cool cool

CPhill  Jan 29, 2018
 #6
avatar+1048 
+2

Solution:

\(\text {The easy parts first: }\\ \text {There are }\dbinom{52}{5} = 2598960 \text{ ways to select 5 from 52 cards. } \\ \)

 

\(\text{Single suit probability – five cards of the same suit. }\\ \dbinom{4}{1}\dbinom{5}{13} = 5148\\ \rho(1) = \dfrac{5148}{2598960} = 0.19808 \% \\ ------------------- \)

..

\(\text{Four suit probability – five cards four suits. }\\ \text {A five-card hand will have at least two cards of the same suit. } \\ \text {Select two cards with matching suits. }\\ \text{There are }\underbrace {\dbinom{4}{1}}_ {\small \text { ways to choose suit }} * \underbrace {\dbinom{13}{2}}_ {\small \text { ways to choose 2 of 13 }} \\ \text{Then for the 3 remaining cards }\\ \text{ there are } \dbinom{13}{1}*\dbinom{13}{1}*\dbinom{13}{1} \text { ways to choose unique suit for each of remaining cards. } \\ \text {The product of these counts gives the total number of five-card hands with four unique suites. }\\ \dbinom{4}{1} * \dbinom{13}{2} * \dbinom{13}{1} * \dbinom{13}{1}* \dbinom{13}{1} = 685464 \\ \rho(4) = \dfrac {685464}{2598960} \text { 26.3746}\% \\ \)

-------------------------------------------------------------------

\(\text {Two suits. Five cards }\\ \text {There are } \dbinom{4}{2} \text { ways to chooses two suits. Twenty-six (26) cards represent these two suits. }\\ \text{There are } \dbinom{26}{5} \text { ways to choose five cards from the 26. }\\ \text{These selections include single suits, }\\ \text{so subtract the single-suit counts of three for the five-card selections. } -(3)\dbinom{4}{1} * \dbinom{13}{5}\\ \dbinom{4}{2} * \dbinom{26}{5} – (3)\dbinom{4}{1} * \dbinom{13}{5}= 379236\\ \rho(2) = \dfrac{379236}{2598960} = 14.5918\% \)

-------------------------------------------------------------------

\(\text{For three suits, subtract the counts for 1, 2, and 4 suits from } \dbinom{52}{5}\\ \dbinom{52}{5} - 5148 - 379236 - 685464 = 1529112\\ \rho(3) = \dfrac{1529112}{2598960} = 0.14.5918\% \\ \)

-------------------------------------------------------------------..

\(\rho(1) = \dfrac{5148}{2598960}\\ \rho(2) = \dfrac{379236}{2598960}\\ \rho(3) = \dfrac{1529112}{2598960}\\ \rho(4) = \dfrac {685464}{2598960}\\ -----------------\\ \text{The probability of a five-card hand having at least three (3) unique suits is }\\ \dfrac{1529112}{2598960} + \dfrac {685464}{2598960} = \dfrac {2214576}{2598960} = \dfrac {507}{595}\\ \text{ }\\ \text{ }\\ \small \text { Source: Lancelot Link & Co. Solutions for AoPS probability questions. } \)

-----

 

The result matches Melody’s answer.  This proves we monkey around until we get it right. laugh

 

 

 

GA

GingerAle  Jan 31, 2018
edited by GingerAle  Jan 31, 2018
edited by GingerAle  Jan 31, 2018

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