All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Poker Hands (Posted on 2004-07-09) Difficulty: 4 of 5
The likelihoods of being dealt various poker hands are widely published (easily found on the internet). A more difficult problem is: what are the likelihoods of being dealt each poker hand, given a 54 card deck (52 card deck + 2 jokers).

The various hands of interest are:
1 pair
2 pair
3 of a kind
straight
flush
full house
4 of a kind
straight flush
5 of a kind

* Jokers can count as any rank card, in any suit.

No Solution Yet Submitted by Thalamus    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips A start, a hint and a caveat. | Comment 1 of 11

A denominatior that will be used in many of these is 54*53*52*51*50 = 54!/49! = 379,501,200; call it d. Its the number of sequences of five cards that can be drawn, and distinguishes between the two jokers.

One pair can happen with either zero or one joker.
With zero jokers you get 52*1*48*44*40*C(5,2)/d = .115756155711761
With one joker you get 2*52*48*44*40*C(5,1)/d = .115756155711761
for a total of .231512311423521

But, in the case of one joker being used for the pair, one must also subtract out the probability that the four non-jokers have the same suit, form a 4-straight or a straight with a gap of 1, that would then be counted by the player as using that joker for a purpose other than forming a pair, as the alternative (flush or straight) would be better.

It's starting to look messy.


  Posted by Charlie on 2004-07-09 10:45:04
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information