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**+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

#3**+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%

CPhill Jan 29, 2018

#4**+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

Melody
Jan 29, 2018

#6**+3 **

Solution:

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

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

..

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

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

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

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

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

GA

GingerAle Jan 31, 2018