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LetYouDown
03-16-2005, 11:00 AM
Assuming you are dealt two random cards, what are the odds that at the river, you will have an unbeatable hand? I'm sure I could write a program to calculate this, but I'm curious if someone has a more purist method. By unbeatable, I obviously mean split pots are ok.

eg.:

A/images/graemlins/spade.gif K/images/graemlins/heart.gif

counts on a board of:

A/images/graemlins/diamond.gif A/images/graemlins/club.gif K/images/graemlins/spade.gif 9/images/graemlins/heart.gif 5/images/graemlins/diamond.gif

Spicymoose
03-16-2005, 02:15 PM
Any method would take SOOOOOOO long to calculate. You could write a program to do this, but for an exact answer, it would take days (maybe weeks?) to complete. The reason there is no easy answer is because for any given hand, there are so many different situations you could be in where you are unbeatable by the river.

For example, 72o.
777AA, 777AK, 777AK (different suits)
3456A, 3456A, 3456A, 3456A, 3456A, 3456A, 3456A (different suits every time), 3456K
straight flush combos, full house combos

There are trillions (im guessing this is an understatement) of different scenarios you would have to run through, so it would take a VERY long time.

LetYouDown
03-16-2005, 02:24 PM
Hence the question /images/graemlins/grin.gif. I've presented this question to a number of stats majors and general all around bright guys and no one seems to have a plan of attack that reasonably cuts down on the amount of time necessary to come up with a definitive answer.

One could certainly get a rough idea by running through a million hands, each time evaluating how many hands beat you. If 0, then increment. This is an awfully loose method and even at a million runs is likely to be significantly off.

2ndGoat
03-16-2005, 03:22 PM
Clearly naked monter carlo is not going to cut it. Take the different classes of nuts (nut straight flush, quads that are not on the board, quads on the board with an ace in your hand, top full house, nut flush, nut straight, nut trip aces/kings/queens) and write a subroutine for determining the probability of ending up with each. This would take a lot of intelligence built into the program but it's certainly more feasible than trying to go through the whole thing with brute force. To do nut straight, for instance, you need to determine the possibility of having a straight, then subtract out the times there is a pair on the board, 3 or more to a suit, or a higher straight (this one might be nasty).

You'd also have to watch out that you include a board of AcKcQdJhTs as the nuts for 63o.

2nd

Edit- you're looking for a method, not a program, sorry... but at least this way you could do it with math for each of the 169 hands and weight according to odds of being dealt each rather than enumerating every possible board for every possible hand.

dabluebery
03-16-2005, 03:37 PM
I wrote a spreadsheet that bypassed Monte-Carlo simulations to consider the 19,600 possible flops for a given hand. It was a nightmare. Evaluating the over two-million possible boards (50c5?) for a given hand dynamically would blow Excel up. Brute force is not a good idea here.

chapstick
03-16-2005, 03:39 PM
[ QUOTE ]
Any method would take SOOOOOOO long to calculate. You could write a program to do this, but for an exact answer, it would take days (maybe weeks?) to complete. The reason there is no easy answer is because for any given hand, there are so many different situations you could be in where you are unbeatable by the river.

For example, 72o.
777AA, 777AK, 777AK (different suits)
3456A, 3456A, 3456A, 3456A, 3456A, 3456A, 3456A (different suits every time), 3456K
straight flush combos, full house combos

There are trillions (im guessing this is an understatement) of different scenarios you would have to run through, so it would take a VERY long time.

[/ QUOTE ]

I don't think it would take that long. There are 52*51 possible starting hands. That's around 2500 starting hands. (Sorry, no calculator handy). There are significantly less if you include the fact that 7h6s is equivalent to 7s6h, etc... I think there are then 169 combinations.

Given two pocket cards, there are 50 cards left in the deck, so then we have C(50,5) possible flops. That's about 2.5 million. So, we have 2.5 million times 169, or about 400 million combinations. I don't think that's a problem to compute.

LetYouDown
03-16-2005, 03:46 PM
Your methodology looks sound...aside from the 2500 starting hands. Order doesn't matter, so it's really half of that. And the 169 number eliminates the prior issue.

I was really looking for some simple way to figure it out by hand, or a solid way of estimating it.

I was wondering on an average hand, how many combinations of cards qualify as the nuts. Obviously on an A-A-K-9-4 hand, you have only 1. But, as mentioned earlier...you have a TON on a A-10 rainbow board. This might be an easier Omaha question...but then again, the thought of that hurts my brain.

yoshi_yoshi
03-16-2005, 04:07 PM
[ QUOTE ]
One could certainly get a rough idea by running through a million hands, each time evaluating how many hands beat you. If 0, then increment. This is an awfully loose method and even at a million runs is likely to be significantly off.

[/ QUOTE ]

I hope I'm not misreading your question, but I don't think brute force is going to be either tedious or inaccurate.

0) Initialize an array of the 52*51/2 possible starting hands to zero.
1) Run 1 million hands (or more)
2) For each hand find the nuts (this should return a SET of hands)
3) For each hand that is in the set, increment.

Since there are ~1300 different starting hands, the average will be about 800. If that's not good enough, do 100 million for an average of 80,000. I doubt that would take more than a couple of hours, depending on how you implement the nut-finding algorithm.

LetYouDown
03-16-2005, 04:20 PM
Sure, there are all kinds of ways to get to the answer through the use of computers. One could easily write a program to give you the exact answer. My curiousity was two-fold.

1.) Has anyone done this?
2.) Is there a good method of approximating this without using any type of brute force? (margin of error + or - .01%.

I'm assuming that's a reasonable margin of error, as I'd expect the probability to be less than 1% as it is.

gaming_mouse
03-16-2005, 05:49 PM
[ QUOTE ]

1.) Has anyone done this?
2.) Is there a good method of approximating this without using any type of brute force? (margin of error + or - .01%.

[/ QUOTE ]

I see no way to do this by hand, or even to get a reasonable approximation by hand.

1. A brute force method, as others have explained, is perfectly feasible.
2. If you want something even faster than that, use monte carlo. I'd imagine you could get a close approximation within a matter of minutes after you'd written the program.

Unfortunately, there are some problems that just don't have a "pure" solution.

gm

turbo the man
03-16-2005, 11:23 PM
this might add more work for you, but AAK94
will have TWO hands that make the nuts.
AA for quads and
AK for the fullhouse, since no-one else can
be holding AA if you have AK. /images/graemlins/spade.gif

SumZero
03-17-2005, 12:19 AM
As a first pass approximation it is worth noting that every board has at least 1 nut hand for it. Therefore the worst case odds are 1 in the number of hands or about 1 in 1000.

LetYouDown
03-17-2005, 10:00 AM
[ QUOTE ]
this might add more work for you, but AAK94
will have TWO hands that make the nuts.
AA for quads and
AK for the fullhouse, since no-one else can
be holding AA if you have AK. /images/graemlins/spade.gif

[/ QUOTE ]

That was the point of mentioning that example.

When I initially posted this question, I semi-expected that someone had published the percentage. Just found it to be an interesting question. I've found that tight play/ABC poker works best at the lower-mid limits online and offline ($100 NL). I was curious if I only got involved in hands where I had the nuts or 2nd nuts, how often I'd actually be playing a hand. This is obviously an absurd method of play, it was purely an exercise in the hypothetical.

Gator
03-18-2005, 02:53 PM
What if you were to take an average hand (say Q8u) - what's the probability for holding the nuts for that hand after the river? I would think that the answer would be pretty close to holding the nuts for any random hand. What do you think?