Combinatorial Probability Question - Poker hand of exactly 3 suits

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SUMMARY

The probability of being dealt a five-card hand with exactly three suits from a standard 52-card deck can be calculated using combinatorial methods. The correct approach involves first selecting three suits from the four available, which can be done in 4 ways. Then, for each selected suit, at least one card must be chosen, followed by selecting additional cards from the remaining suits. The total number of valid combinations is derived from the formula (4 C 3) multiplied by the combinations of choosing cards from the selected suits, ultimately divided by the total number of five-card combinations (52 C 5).

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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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