Median Best Holdem Starting Hand

[PairTheBoard]

Good observations uuDevil. What I'd really like to see is David or Mason express interest in the idea. If they were to do so I'm sure it would stimulate some work on the problem. A solution would require a moderate amount of programming for simulations. I'm convinced it would be a valuable area for research and the results worthy of a CardPlayer Article or Chapter in a good Poker Book. Plus, I would like to know the answers. I think it would be especially interesting using various Card Ranking Systems taylored for different situations, like flops seen by 2,3,4... players - both Limit and No Limit.

PairTheBoard

[astroglide]

here's my answer: it doesn't matter

[Aisthesis]

Ok, here goes. First, I'm going to skirt the ranking issue for the moment and just assume that we have one (Karlsson-Sklansky might be a good start, but it depends on what we're doing with the results).

I'm going to define "median best starting hand" in the following way: GIVEN a particular hand, it is "median best" if the probability at a table of n players is 50% that yours is the best hand among the random hands dealt.

So, assume we have a way of ranking starting hands, of which there are 1326. There are obviously lots of equivalencies, but it won't hurt the solution if we just assume that we can rank them all 1, 2, 3, 4, ...

You are dealt the hand ranked r at a table of n players. Then the probability that no other hand beats you is: [(1326-r)/1325]~(n-1). (Sorry, but "~" is the best I can find on my keyboard to represent an exponent).

So, at a table of 9 players we want .50 = [(1326-r)/1325]~8 or 2 = [1325/(1326-r)]~8. Solving this, my computer calculator puts r at 111.

Hence, we want the hand ranked 111.

Now, as to particulars (will depend on how to rank these hands, obviously): Pairs all have 6 instances. Unsuited non-pairs 12 instances; suited non-pairs 4 instances.

Pairs 88-AA cover 48 cases. ATs-AKs cover 16 instances. ATo-AKo cover 48 cases. That puts the total at 112.

So, assuming ATo is "better" than any hand other than the aforementioned and the worst of those, ATo should be the median best starting hand at a 9-player table.

Side note to PairTheBoard: This means my initial attempt on UPF, where I came up with AJo on the basis of a slightly faulty attempt at assigning probabilties, was indeed a little off.

As to the practical relevance: One fairly obvious case is shortstacked in a tournament looking for an opportunity to double up. What do you need in EP to go all-in? For me, anyway, it would depend also on just how short my stack is relative to the blinds. But it at least provides some orientation in making this decision, I think.

[TomCollins]

Here's why it doesn't matter. You have no criteria for "best hand". Secondly, it is not transitive. You cannot stick hand ranks on a number line and expect since A > B, B > C, A > C. This greatly increases the complexity of the problem. It's very hard to produce a mathematically precise answer of a poorly defined problem. So there is no way to find a "median" of them. Suppose you did three samples, and the best hands were 4s 4c, Ad Kc, Jh Th. Which of these three is the median? Each lose to one and beat another heads up.

If you actually put up a mathematical percise definition of the problem, I'm willing to bet someone will simulate it.

[Aisthesis]

True, but you can put an exact percentage on the frequency with which a given hand wins against a random hand heads-up (or, for that matter, against any number of random hands, but the results will be different). I think that's probably the only meaningful way to define the problem (at least the only way I can think of).

Shouldn't the order of the Karlsson-Sklanksy rankings reflect this (even though they're expressed as stack sizes in SB)?

[Bozeman]

For a table of ten players who play showdown poker with any hand (no betting or folding), the hands are ranked according to showdown percentage against 9 random hands, and the median best is (from pzhon, etc.) best ~7%. There are 89 hands in this set, and for showdown against 9 random hands these are 99-AA,AK,XTs for 88 hands with AQo on the bubble.

For a more reallife answer, you would need to specify a complete ranking of hands. As others have pointed out, this is difficult or not too meaningful. In addition, knowing the median doesn't help much because much of the coolness of poker comes in the variations (like Ak and qq being dealt together).

Craig

[Louie Landale]

Heads up if you have hand 50% then you are even money to have the best hand. That is, the median "best" hand for one player is hand 50%.

Against 2 players your hand h% is "best" when it beats the first player AND the 2nd player. Chances it beats the first player is h. Chances it beats both is h**2. .5 = h**2. h = 2ndRoot of .5; =70.1%. So if you have hand 70.1% you are 50:50 to have the best hand against 2 random hands.

** Average best hand for n opponents is nthRoot of .5. **

- Louie

Be advised that the actual rank of hands changes somewhat as N changes: trouble hands go down in relative value as N increases (A2 ranks higher heads-up than at a full table); suited hands go up.

This is interesting if you are adopting a starting strategy that says play if you are 50:50 to have the best hand or better.

[PairTheBoard]

By golly I think your on it Aisthesis. I think looking at the whole 1326 possible hands is the way to go since all are equally likely.

When I suggested finding the median best hand among 9 dealt I was thinking of sitting at a 10 player table, getting dealt my 2 cards and asking myself, what is the median best hand of the 9 remaining players? But looking at it your way, when sitting at a 10 player table, ask what hand you need for a 50% chance of being best at the table. Using your solution my calculator comes up with the hand ranked 99 out of the 1326 possible. Where do you figure that falls in the Sklansky ranking system?

btw, here's a link to a Heads UP Showdown Ranking system.

http://www.gocee.com/poker/he_ev_pe.html

How does the change in Ranking Systems affect the results?

PairTheBoard

[PairTheBoard]

You refernce "pzhon" Bozeman. What is that?

PairTheBoard

[PairTheBoard]

I'm thinking Aisthesis has the right approach below Louie. He uses your same reasoning but considers all equally likly 1326 possible starting hands. Here is a link to Hand Rankings accourding to Heads Up matchup.

http://www.gocee.com/poker/he_ev_pe.html

Interesting that J5s is on the bubble heads up.

I guess I agree that in a 3 player game the Bubble would be the hand ranked in the top 70% like you say. I'm not suggesting you think this but for others reading here, that's not the same as the hand that wins 70% of the time against a random hand when taken to showdown.

Thanks,

PairTheBoard