blonde poker forum

Playing four card Omaha high:

If you have Kx suited and three of that suit appear on the flop, how likely is it at a full table that the A flush is out ?

Now here's the tricky bit.....

If you have Kxx suited and three of that suit appears on the flop, how much less likely is it that the A flush is out ?

i'll have a bash at answering the tricky bit :)

slightly

9players

8*4 = 32 cards dealt to a full table and you get 4 of them

so theres a 28/45 chance the of Ah is out - since you know its not in your hand or the flop (52 - 4 - 3)

3 cards in the hand with the Ah (if its out)

7 hearts left in 44 remaining cards

P(one or more heart in 3 random cards) -

7/44*37/43*36/42*3 + 7/44*6/43*36/42*3 + 7/44*6/43*5/42

multiply that by 28/45 and you get

**Edit(in case anyone stops here)**
0.256

just over 1 in 4 times the nut flush is out.


similarly if you have 3 hearts in your hand

6 hearts in 44 remaining cards

6/44*38/43*37/42*3 + 6/44*5/43*37/42*3 + 6/44*5/43*4/42

0.225 about 2 in 9 so a reduction of about 1/32th times you will have flopped the 2nd nuts (oops).


(Sorry this is a bit messy but I dont have all day and I think the numbers are correct which is the main thing :) )

 :hello: and ;welcome;

so i was right then? phew!!  :D

Thanks a lot for that Stephen.

 :hello: and  ;welcome; to blonde

Wow - stunning first Post Mr Stephen Sir!

Welcome aboard. Methinks you are gonna be busy......

Just what i was going to say ::)  :D

So you can vouch that the figures stephen has quoted are correct? ???

;)

What you also need to be thinking about is how likely is the Queen or Jack flush going to pay me off if the Ace isn't out there?

Oops first post and i got my sums wrong for the first bit (oops!)

 :blonde:

This place must be affecting me somehow.....


9players

8*4 = 32 cards dealt to a full table and you get 4 of them

so theres a 28/45 chance the of Ah is out - since you know its not in your hand or the flop (52 - 4 - 3)

3 cards in the hand with the Ah (if its out)

7 hearts left in 44 remaining cards

P(one or more heart in 3 random cards) -

7/44*37/43*36/42*3 + 7/44*6/43*36/42*3 + 7/44*6/43*5/42

multiply that by 28/45 and you get

0.256 (not 0.329) *************************

just over 1 in 4 times the nut flush is out.

similarly if you have 3 hearts in your hand

6 hearts in 44 remaining cards

6/44*38/43*37/42*3 + 6/44*5/43*37/42*3 + 6/44*5/43*4/42

0.225 about 2 in 9 so a reduction of actually about 1/32'th of the time ! (not quite so significant  8) )

*************************************************

Also this is all done in a 'vacum' obviously where all starting hands see a flop whatever they are. Whilst all hands are dealt equally some reach the flop with a bit more enthusiasm than others and they definately play differently after it. Copious action having flopped the 2nd nut flush with plenty of chips left to play for is getting more into 'know your enemy' territory than just the baseline numbers although I think they are always good to know anyway.

And cheers for all the  :hello: 's

I think you have subtracted 4 twice?  I.e. if 9 players are dealt in, there are 8 players each with 4 unknown cards, so there are 32 unknown cards not 28.

Anyway, don't use actual numbers at this stage... far better maths to assume an n-handed game.


ITYM 7/44*37/43*36/42*3 + 7/44*6/43*37/42*3 + 7/44*6/43*5/42


The correct calculation here is 1-((1-h)^n), where n = the number of opponents, and h = P(any given opponent holds the Ace-high flush) = 4/45*P(at least one of three cards is a heart).

You are very sloppy for a mathematician.  Your homework is to calculate the corresponding probability for an n-handed game of m-card Omaha in which I hold the King of hearts and k other hearts.

Well you are charming !

Yes I did this in a bit of a rush (my apologies) a 9 handed game does indeed have 9*4 cards dealt out. IMO your 'correct calculation' is likely to be of no help to anyone who can't already work out the answer, I'm actually a statistician (although you wouldn't think it from all the mistakes I managed to make, more haste less speed I guess!) and we are allowed to be sloppy because we are cleverer than mathematicians.  8)

My methodology was just a simple a conditioning argument using,

P(nut flush out) = P(Ah dealt out)*P(nut flush out|Ah dealt out)

In order to make the calculation relatively simple. (and hopefully understandable although my rather rushed layout doesn't help with that much obviously)

If I wished to be a smartar*e I could have used up half an hour of my life working out the probability of the nut flush being out in an N-handed game of M-card Omaha but that wouldn't have been answering the question asked.

welcome Stephen, I once had a p nut flush.....i have never eaten them again.

Thanks for the advice sir  ;topofclass;

 rotflmfao rotflmfao rotflmfao

Stephen...very impressed by your first three posts Sir..welcome.