# Combinatorial Probability Question - Poker hand of exactly 3 suits

Sasha12

## 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?

## 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.

damabo
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

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

Homework Helper
Dearly Missed

## 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?

## 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

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damabo
better listen to ray, my analysis is a bit wrong