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Full House in Draw Poker

Date: 7/28/96 at 14:40:22
From: Anonymous
Subject: Probability of Getting Full House in Draw Poker

A player is dealt 5 cards - a pair of aces and three other unpaired 
cards. What is the probability that she will wind up with a full house 
(three cards of one rank and two cards of another rank) by using 3 
cards from her hand and drawing one pair of cards?

I know that there will be two other aces to use that will allow me to 
make the three of a kind.  Since I have three other unpaired cards, 
that gives me a total of 9 other cards with which I might make a pair 
of cards in the final hand. Other than that I am lost! 

Date: 7/28/96 at 16:51:27
From: Doctor Anthony
Subject: Re: Probability of Getting Full House in Draw Poker

To fix ideas, suppose that your hand originally consisted of AAKQJ. 
You keep the two aces and the king, and discard the queen and jack. 
Now you have AAK in your hand, and you will get a full house if you 
pick up AK or KK.

There are 52-5 = 47 cards that are available, amongst which are 2 aces 
and 3 kings.

Probability of 1 ace, 1 king in the pair you pick up is 

   (2 C 1)(3 C 1)/(47 C 2) = 2*3/1081 = 6/1081

Probability of two kings is    (3 C 2)/(47 C 2)  = 3/1081

Total probability of a full house = 6/1081 + 3/1081 = .00833

This is in fact the required probability, because you have the choice 
which extra card (besides the two aces) you keep, whatever the face 
value of that card, the probabilities will be the same; there will be 
3 more of that value in the pack and 2 more aces from a total of 47 
cards to be considered.

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Permutations and Combinations
High School Probability

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