# Probability of a poker hand

## Homework Statement

What is the probability of getting a 5 card hand where the 5 cards are from exactly 2 suits

## Homework Equations

nCr = n!/(r!(n-r)!)

## The Attempt at a Solution

There are two possible cases where the 5 cards are from 2 suits, case one is 4 are from one suite and 1 is from another and case two is that 3 are from one sweet and 2 are from another.

case one:
there are 4 ways to choose a suit and then 13C4 ways to choose 4 cards from that suit. Then there are 39 ways to choose the last card because we must pick from the remaining 3 suites

4(13C4)39

case two:
There are 4 ways to choose a suit and then 13C3 ways to pick 2 cards from it. Then there are 3 ways to pick another suit and 13C2 ways to pick 2 cards from it

4(13C3)3(13C2)

since the total number of 5 card hands is 52C5 the probability where the 5 are from 2 suits is

[4(13C4)39 + 4(13C3)3(13C2)]/52C5

is this correct?

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
At face value, the method looks correct. However, I would do it slightly different and you should end up with the same result.

1. Take the number of combinations where you draw five cards out of 26.
2. Remove two times the number of combinations where you draw five cards out of 13. These are the ones where you only draw from one suite, given that you are drawing from two suites.
3. Multiply this number by the number of ways of selecting two out of four suites.
4. Divide by 52C5.

Zondrina
Homework Helper
I'll tell you one thing, there's more to poker than just math :P.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

What is the probability of getting a 5 card hand where the 5 cards are from exactly 2 suits

## Homework Equations

nCr = n!/(r!(n-r)!)

## The Attempt at a Solution

There are two possible cases where the 5 cards are from 2 suits, case one is 4 are from one suite and 1 is from another and case two is that 3 are from one sweet and 2 are from another.

case one:
there are 4 ways to choose a suit and then 13C4 ways to choose 4 cards from that suit. Then there are 39 ways to choose the last card because we must pick from the remaining 3 suites

4(13C4)39

case two:
There are 4 ways to choose a suit and then 13C3 ways to pick 2 cards from it. Then there are 3 ways to pick another suit and 13C2 ways to pick 2 cards from it

4(13C3)3(13C2)

since the total number of 5 card hands is 52C5 the probability where the 5 are from 2 suits is

[4(13C4)39 + 4(13C3)3(13C2)]/52C5

is this correct?
Yes, it matches what I get.

I used another way:
(1) Pick a first card---any card.
(2) All four remaining cards must either match the first card or come from another suit, (with at least one from the other suit).
There are 3 ways to pick the other suit. The event we want chooses $(i,j)$---meaning $i$ additional cards from the first suit and $j$ cards from the other suit, for $i \geq 0, j \geq 1$ and $i+j = 4$. Calling $p(i,j)$ the probability of choice $(i,j)$, we have
$$p(i,j) = \frac{{12 \choose i} {13 \choose j}}{{51 \choose 4}}$$
and
$$\text{Answer} = 3\,[p(0,4) + p(1,3) + p(2,2) + p(3,1)] = \frac{143}{980}$$

Thank you guys. so many different ways to do it