if you want to practice combinatorics: omaha question
 

so I can do basic combinatorics, but my level (or lack) of talent is that I often befuddle myself for a while before I set the problem up right. this one isn't hard, but I started it the wrong way a couple times. so for anyone who wants to practice, let me ask what you get and then I'll show my answers to check myself.

the question, in several parts (idea is that you're playing omaha--what is the chance someone else has a suited ace).

omaha. 10 players. all 10 see the flop for simplicity. remember a hand is four cards in omaha.

1. before the flop: what is the chance someone is dealt the ace of spades?

2. before the flop: if someone gets the ace of spades as their first card, what is the chance they get at least 1 more spade?

3. on the flop: flop has 2 spades. you have no spades. assume someone has the ace. what is the chance it is suited?

4. on the flop: same as above except this time you have 2 spades yourself (KsXs). what is the chance the guy with the ace has it suited.

if anyone can come up with good questions involving more aces or anything, post and I'll try them.

disclaimer: yes these questions are a bit unrealistic, b/c you generally want to ask: what is the chance he has the suited ace,_given_ his flop action? but we can give it a bit more realism by assuming he will play the naked ace (as a bluff) the same way as the suited ace.


edit: if you know this and never screw it up, maybe try to leave in case someone else wants to try.

Re: if you want to practice combinatorics: omaha question
 

(1) is easy. There are 40 cards dealt, the chance is 40/52 = 10/13 that one of them is A[img]/images/graemlins/spade.gif[/img].

(2) Once someone gets A[img]/images/graemlins/spade.gif[/img] there are C(51,3) = 20,285 possible ways to complete their hand. There are C(39,3) = 9,139 ways to complete the hand without another spade. So the chance is 1 - 9,139/20,285 = 56% that the hand has at least one other spade.

(3) Now there are 44 unknown cards (52 - 4 in your hand - 3 on the board - A[img]/images/graemlins/spade.gif[/img]) with 10 spades. So it's 1 - C(34,3)/C(44,3) = 5,984/13,244 = 55%.

(4) 1 - C(36,3)/C(44,3) = 7,140/13,244 = 46%.

Re: if you want to practice combinatorics: omaha question
 

Omaha hi/lo, 9 handed game.

The board reads:

K [img]/images/graemlins/diamond.gif[/img] 5 [img]/images/graemlins/heart.gif[/img] 3 [img]/images/graemlins/spade.gif[/img] Q [img]/images/graemlins/diamond.gif[/img]

Your hand :

A [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/diamond.gif[/img] 4 [img]/images/graemlins/diamond.gif[/img] 6 [img]/images/graemlins/club.gif[/img]


Assuming that the villain holds an A2 combination in his four card hand, what is the probability that he was dealt a higher flush draw?

Re: if you want to practice combinatorics: omaha question
 

the four answers above are the same that I got, except I managed to be more roundabout for #1, doing (51 choose 40) over (52 choose 40)--duh. (I should right it out and confirm how all the terms cancel out for punishment).

Re: if you want to practice combinatorics: omaha question
 

[ QUOTE ]
Omaha hi/lo, 9 handed game.

The board reads:

K [img]/images/graemlins/diamond.gif[/img] 5 [img]/images/graemlins/heart.gif[/img] 3 [img]/images/graemlins/spade.gif[/img] Q [img]/images/graemlins/diamond.gif[/img]

Your hand :

A [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/diamond.gif[/img] 4 [img]/images/graemlins/diamond.gif[/img] 6 [img]/images/graemlins/club.gif[/img]


Assuming that the villain holds an A2 combination in his four card hand, what is the probability that he was dealt a higher flush draw?

[/ QUOTE ]

[deleted]

let me try it this way: he has an ace, a deuce, and two cards. if any 2 or them are suited in diamonds, he has a higher flush draw than yours. his deuce can not be a diamond. so the question is of his ace and two other cards, how many ways can they be suited in diamonds, versus not.

his total number of possible hands is 1 of 3 aces, 1 of 3 deuces, and 2 from the 38 other unknowns (if I've counted the knowns right: 4 on board, 4 my hand, 3 aces, 3 deuces). so his # of possible hands is
3C1 * 3C1 * 38C2 = 6,327. (or 3*3*703)

I think we have to proceed by cases (can't do the one-minus trick) because we have to consider 2 or 3 diamonds versus 0 or 1 diamonds. I'll count the suited ways b/c 3 diamonds is easy:

he can have 3 diamonds w/ Ad & non-d deuce:
1 * 3 * 8 * 7 = 168
he can have 2 diamonds w/ Ad and non-d deuce:
1 * 3 * 8 * 30 = 720
or by non-d ace and deuce and 2 other diamonds:
2 * 3 * 8 * 7= 336

so the chance he is suited in diamonds is:

(168+720+336)/6327= 19.34%

good problem.