Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Poker Hand Odds


Date: 08/07/97 at 14:31:42
From: Anonymous
Subject: Poker Hand Odds

I am trying to calculate the odds of getting a hand of exactly one 
pair (any pair) in five card stud poker, using one 52-card deck.  
I have calculated the total number of possible hands as 2,598,960 
(52 C 5), but I get stuck trying to figure out the possibility of 
exact hands.  Help!


Date: 08/07/97 at 15:53:29
From: Doctor Anthony
Subject: Re: Poker Hand Odds

One pair [two of one face value, and 3 cards of different face 
values, not matching the pair]

                13 x 4C2 x 12C3 x 4^3
 Prob(1 pair) = ----------------------  =  0.422569
                        52C5  


The denominator is the number of ways of selecting 5 cards from 52.  
We will explain the numerator, term by term, starting at the lefthand 
end.

13 is the number of ways of choosing the face value of the pair, e.g. 
two 10's or two 4's or whatever.  There are 13 face values 2, 3, ..., 
Q, K, A.

4C2 is the number of ways of choosing the two 10's (say) from the four 
10's in the pack.

12C3 is the number of ways of choosing 3 different face values from 
the 12 still available - we have used up one face value for the pair.  
For argument's sake, suppose the 3 remaining face values are 3, 7, K

The 4^3 term arises in the following way. Suppose the face values are 
to be 3, 7, K as suggested above. We have 4 ways of choosing the 3, 
since there are four 3's in the pack. Similarly, there are 4 ways of 
choosing the 7 and 4 ways of choosing the K, giving 4 x 4 x 4 = 4^3 
ways of selecting the other 3 cards with a given set of face values.

Multiplying all these factors together gives the total number of ways 
that we could choose exactly one pair.  And this divided by 52C5 gives 
the required probability.
 
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   



Date: 09/03/97 at 15:47:50
From: Doctor Naomi
Subject: Re: Poker Hand Odds

Hi there,

Before talking about the cards, I would like to make a clarification.  
You asked for "odds" in your question, but do you mean "probability?"  

The probability of an event occurring is expressed as a fraction, 
e.g. 2/3, which means that on average an event will happen two times 
out of every three trials (so there is a 67 percent (rounded) chance 
of this event happening in any one trial).  

Odds tell you how many times an event will happen and how many times 
it won't, e.g. odds of 2 to 3 mean that on average, an event
will happen 2 times for every 3 times that it doesn't happen (so there 
is a 40 percent chance of this event happening in any one trial 
because 2/(2+3) = 0.4).  

I assume that you are asking for the probability of getting exactly 
one pair in a hand of five cards randomly dealt from a 52-card deck.  
The probability of getting exactly one pair will be the number of 
hands with exactly one pair divided by the total number of 5-card 
hands.  You have already found the denominator, (52 C 5).  Now we must 
find the numerator, which, as you said, is a lot more tricky.

The total number of 5-card hands with exactly one pair can be thought 
of as the following:

  (number of different face values that the pair might have)  x 
  (number of ways to pick two cards of the same face value from the 
     four available in the deck)  x 
  (number of ways three more cards can be chosen such that none of 
     them have the same face value and none share the same face value 
     as the one chosen for the pair).

Now we have to calculate these quantities. First you must find the 
number of different ways of choosing one face value for the pair 
(e.g., there could be two 10's or two queens, etc.).  There are 13 
face values (2,3,...,Q,K,A) to choose from, so we have (13 C 1) 
different face values that the pair might have.

Now that we have chosen a particular face value (let's say we chose 
Jacks) we need to calculate the number of ways to choose two Jacks 
from the four Jacks available in the deck, yielding (4 C 2).

Now comes the hard part: calculating the number of ways three more 
cards can be chosen such that none of them share the same face value 
as the one chosen for the pair. One way to get this number is by first 
choosing 3 different face values (one for each card), none of them 
matching the face value (Jacks) already picked for the pair. With the 
Jacks out of the running, there are 12 choices of face values left 
from which 3 can be chosen, giving us (12 C 3).  

Let's say that we chose Ace, Queen and King as the other three face 
values.  We now need to multiply (12 C 3) by (the number of ways to 
choose 1 Ace from the 4 Aces in the deck) x (the number of ways to 
choose 1 Queen from the 4 Queens in the deck) x (the number of ways to 
choose 1 King from the 4 Kings in the deck), yielding (12 C 3)(4 C1) 
(4 C 1)(4 C 1).

Putting this all back together, we have

                no. of 5-card hands with exactly one pair
 Prob(1 pair) = -----------------------------------------  
                        total no. of 5-card hands


                (13 C 1)x(4 C 2)x(12 C 3)x(4 C 1)^3
              = ---------------------------------------  
                             (52 C 5)

I hope this helps you out!

-Doctors Bill, Anthony, and Naomi,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Probability

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/