Combinatorial Probability Question - Poker hand of exactly 3 suits

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Homework Help Overview

The problem involves calculating the probability of being dealt a five-card hand from a standard 52-card deck that contains exactly three suits. Participants are exploring combinatorial methods to determine this probability.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to calculate the number of hands with three or fewer suits and questions how to exclude hands with only two or one suit. Another participant suggests a method involving choosing suits and specific card combinations, while also considering the arrangement of suits in the hand.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and questioning the validity of their methods. Some guidance has been offered regarding the selection of suits and the arrangement of cards, but no consensus has been reached on a definitive solution.

Contextual Notes

Participants are grappling with the combinatorial aspects of the problem, including how to account for overlaps in suit combinations and the total number of possible hands. There is an acknowledgment of the complexity involved in ensuring that exactly three suits are represented.

Sasha12
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Homework Statement



A five card hand is dealt from a standard 52-card deck. What is the probability that there will be exactly three suits in the hand?

Homework Equations





The Attempt at a Solution


I started with (4 C 3)(39 C 5), but this gives me all possible hands with three or less suits. How do I get rid of the the possibility of 2 or 1 suits? I tried (4 C 3)(39 C 5) - (4 C 2)(26 C5), but I am not certain that gets rid of all possible overlaps between the sets.
 
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so you do mean for instance club-club-heart-heart-spade?

in this case I would presume these are the options:
1. first choose 3 out of 4 suits. 3 out of 4 gives 4 possibilities.
2. for instance, say we have chosen club,heart and spade.
then there are 6 possibilities: cl-cl-cl-heart spade
cl-cl-heart-heart-spade
cl-heart-heart-heart-spade
cl-cl-heart-spade-spade
cl-heart-heart-spade-spade
cl-heart-spade-spade-spade

we have thus 4*6 possibilites= 24.
now divide this by the total amount of possibilites. I believe this is 4^5.
 
Sasha12 said:

Homework Statement



A five card hand is dealt from a standard 52-card deck. What is the probability that there will be exactly three suits in the hand?

Homework Equations


The Attempt at a Solution


I started with (4 C 3)(39 C 5), but this gives me all possible hands with three or less suits. How do I get rid of the the possibility of 2 or 1 suits? I tried (4 C 3)(39 C 5) - (4 C 2)(26 C5), but I am not certain that gets rid of all possible overlaps between the sets.

I think it is easier to look at what is not in the hand. Say we have no hearts. There must be at least one each of spades, diamonds and clubs; you can represent this as (i) first choose exactly one each of spades, diamonds and clubs; (ii) then, from the remaining deck of 49 cards (13 hearts and 36 others) we must choose two more cards but no hearts. The probability you want is 4 times what you get here, because the 'absent' suit can be selected in 4 ways, and the 4 resulting events are mutually exclusive with the same probability.

RGV
 
Last edited:
better listen to ray, my analysis is a bit wrong
 

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