Gedankenexperiment (Thought Experiment) #1

[PDosterM]

Ciaffone and Brier in Middle Limit Holdem Poker explain that the flop can dramatically change the relative value of players’ hands. Once the flop comes, they say, “The preflop pecking order may well be history.”

Certainly true, but it might be interesting to consider the upper limits of this transformation. It’s quite easy to contrive a situation where three players see a flop that causes the order of the hands to completely reverse.

Player A – A/images/graemlins/spade.gifA/images/graemlins/heart.gif
Player B – 7/images/graemlins/club.gif7/images/graemlins/spade.gif
Player C – 9/images/graemlins/diamond.gif7/images/graemlins/heart.gif

Here Player A is ahead, B is second, and C trails the field until the flop comes:

9/images/graemlins/spade.gif9/images/graemlins/heart.gif7/images/graemlins/diamond.gif

Now Player C is ahead with nines full of sevens, B is second with sevens full of nines, and A (I know how he feels) is last with a solitary pair of aces.

So without creating any ties, can you create a situation where four players see a flop that completely reverses the order? How about five players? How about six? Better yet, what is the theoretical maximum number of players who can see a flop that completely reverses the order?

Answer later with a comment on why this knowledge has some (but admittedly limited) value.

[ChristinaB]

[ QUOTE ]
and A (I know how he feels) is last with a solitary pair of aces.



[/ QUOTE ]

Doesn't player A have two pair?

He has 2 outs to a win.

[daryn]

correct. but he is in last place on that flop. that is the key to the question i believe

[PDosterM]

Right.

I phrased it poorly; he has two pair -- improving to the worst hand.

[ChristinaB]

Just trying to keep your thread alive until you get some good responses that are on topic, as the answer to your original question will take some thought to compose. /images/graemlins/smirk.gif

[shemp]

I understand this is showdown value, but it seems 77 has the worst hand because he can't win.

[DrSavage]

Here's an example for 4:
A)As Ad (55.92%)
B)7s 7d (15.97%)
C)3c 4c (13.96%)
D)2h 4h (12.56%)

flop 5h 6h 7h

D leads with flush, C is second with straight , B is third with a set and A is last with a pair.
Had to change it slightly, used cardplayer calculator the first time and sure enough the numbers were off.

[Schneids]

Player A - Ac Kc
Player B - 9d 9s
Player C - 6d 6s
Player D - 8s 7s
...this is using twodimes to determine the % of the time these four hands win when faced off against eachother. Is that ok to figure out their relative values? Or, do I have to list AKc as #3 even though it wins against these 36.85% of the time?

Flop:
6h 5d 4s

So, now those hands have totally flipped in relative value (as well as the likelihood they will be the winner when it's finally over).

If I can't have AKc ranked above 99 because it's unpaired, I'm sorry.

[Schneids]

[Schneids]

Ac Kc 28.08%
5h 3h 19.31
Ts 9s 19.14
8s 8c 18.97
9d 6d 13.18

flop: 8d 3d Td
flush vs set vs pair vs lower pair vs high card. Immediate showdown value and likelihood of having best hand at the river both come out this way, in this example.

[PDosterM]

Dr. Savage,

Very good answer. Not only are the original hands listed in absolute value order, but in probability of winning order as well. I did not have in mind worrying about the probabilities of winning, but only their absolute poker ranks.

With that in mind, can you come up with a five-player solution? If the probabilities happen to come out in the correct order -- so much the better.

[DrSavage]

[ QUOTE ]
Dr. Savage,

Very good answer. Not only are the original hands listed in absolute value order, but in probability of winning order as well. I did not have in mind worrying about the probabilities of winning, but only their absolute poker ranks.

With that in mind, can you come up with a five-player solution? If the probabilities happen to come out in the correct order -- so much the better.

[/ QUOTE ]

I'd like some clarification on this.
You are saying that the rank of hands preflop should be made based on probability of them winning against a random field and not the other hands being in the same hand ?
How big is the field? Should we rank the hands on their probability of beating random hand headsup, or beating a field of 9 random hands or what? How exactly do you rank hands without the competition?
Example : AA, KK and JTs. AA destroys both , KK destroys JTs, but JTs has higher equity than KK if AA is also in the pot.
If it's done this way I don't really see any particular value in these examples. I find it more interesting to analyze each hand against the other hands in competition and then have the order reversed on the flop.
Also I still don't see what good does it do to be second on the flop, since B in your example and C in my example despit having made a second best hand have zero equity since they cannot improve while the other hands are drawing live.
It could be theoretically possible to come up with some sort of 6 hands example , I'd expect it to be smth. like str flush over flush over straight over set over pair over high card.

[Schneids]

Yeah, I'd like more clarification too about PF.

Are we supposed to pretend we are showing down the 2 hole cards, thus meaning that 22 beats AKs PF?

Or, the way how I've been determing PF rankings by going to twodimes and entering in all the hands that I'm going to use, then using their rankings to find the values of 'best' to 'worst' hand, in each set.

[PDosterM]

Schneids,

Not what I had in mind (because of the probabilities), but an inspired answer nonetheless. Very nice.

I doubt if we use probability of win as the sort criterion that we could prove what the theoretical maximum numbers of players is, other than by exhaustive enumeration of the (unreasonably large number of) combinations.

Can you come up with a five-player solution that is based on absolute poker ranks (i.e. bigger pairs beat lower pairs, pairs beat non-pairs, higher card beats lower card)?

But now you have me curious. Using your probability ranking instead of strict poker hand ranking, do you find a six-player solution?

[Festus22]

5 was all I could come up with:

Preflop (in ascending order in terms of rank)
6 /images/graemlins/club.gif 2 /images/graemlins/club.gif
7 /images/graemlins/diamond.gif 6 /images/graemlins/heart.gif
A /images/graemlins/spade.gif 2 /images/graemlins/heart.gif
3 /images/graemlins/diamond.gif 3 /images/graemlins/heart.gif
A /images/graemlins/diamond.gif A /images/graemlins/heart.gif

Flop comes 3 /images/graemlins/club.gif 4 /images/graemlins/club.gif 5 /images/graemlins/club.gif

Now in decending order
6 /images/graemlins/club.gif 2 /images/graemlins/club.gif makes a straight flush
7 /images/graemlins/diamond.gif 6 /images/graemlins/heart.gif makes a straight
A /images/graemlins/spade.gif 2 /images/graemlins/heart.gif makes a lower straight
3 /images/graemlins/diamond.gif 3 /images/graemlins/heart.gif makes a set
A /images/graemlins/diamond.gif A /images/graemlins/heart.gif makes a pair

[pudley4]

4 /images/graemlins/club.gif4 /images/graemlins/heart.gif
A /images/graemlins/club.gif7 /images/graemlins/heart.gif
K /images/graemlins/club.gif8 /images/graemlins/heart.gif
Q /images/graemlins/club.gifT /images/graemlins/heart.gif
J /images/graemlins/club.gif9 /images/graemlins/heart.gif
J /images/graemlins/diamond.gif4 /images/graemlins/diamond.gif
9 /images/graemlins/diamond.gif6 /images/graemlins/diamond.gif

flop: T /images/graemlins/diamond.gif8 /images/graemlins/diamond.gif7 /images/graemlins/diamond.gif

44 becomes bottom pair
A7 becomes 3rd pair
K8 becomes 2nd pair
QT becomes top pair
J9 becomes straight
J4 becomes flush
96 becomes straight flush

[PDosterM]

In the spirit of a gedankenexperiment (which does not necessarily have to lead to anything), I was asking the question with the intent that strict poker rankings (pair beats non-pair) would be used preflop. You have, however, convinced me that your interpretation (using probabilities of winning) is certainly valid – and perhaps even more interesting.

So I remain intrigued by answers to either interpretation.

[Schneids]

[ QUOTE ]

But now you have me curious. Using your probability ranking instead of strict poker hand ranking, do you find a six-player solution?


[/ QUOTE ]
Qh Jh 25.61%
Ac Kc 22.96
6s 6c 16.52
8s 4s 14.43
Kd Td 12.95
4d 3d 5.86

Flop: 7d 6d 5d
straight flush, flush, straight, set, ace high, queen high.

Note that as soon as you introduce a straight flush into the picture, nearly all other hands are drawing 100% dead, so, we have some ties in the likelihood of them winning at the river (ie 0%). That might be violating one of the principles you defined beforehand, regarding ties.

**Also, keeping this poker related, interesting to see that 84s has more value in this 6-handed pot than KTd. Kind of counter intuitive till you notice it's pretty much drawing for tens and hoping they'll hold up, since so many diamonds are already in play.**

Second-edit: Esssh, doing it 7-handed this way is gonna be a beast.

[Schneids]

Basically, made a few changes to my 6-hand one and added in another pair. Took a few minutes to find a way to get a second pair in here, keep QJh a top the rankings, and then keep the second flush maker lower than a straight maker, PF. Good times /images/graemlins/grin.gif

Qh Jh 22.76%
Ac Tc 21.37
9c 9h 18.60
6s 6c 13.07
8s 4s 7.71
Jd 8d 7.55
4d 3d 5.86

Flop: 7d 6d 5d
straight flush, flush, straight, set, pair, ace high, queen high.

Note that as soon as you introduce a straight flush into the picture, nearly all other hands are drawing 100% dead, so, we have some ties in the likelihood of them winning at the river (ie 0%). That might be violating one of the principles you defined beforehand, regarding ties.

[gunboat]

[ QUOTE ]
correct. but he is in last place on that flop. that is the key to the question i believe

[/ QUOTE ]

Hand one has two outs while hand 2 is drawing dead. Since there is no prize for second best, hand one is clearly better than hand 2 after the flop.

[PDosterM]

This generated a very interesting set of answers. I liked the twist of using probability of win as the initial sort criterion. It turns out that five players is the maximum if absolute rank is used. Schneids came up with the best solution using probability of win (seven-player), but that’s not to say an eight-player solution does not exist. Very nicely done.

The proof that five is the max in the initial problem is quite lengthy, but the general approach is to divide flops into four categories (three of a kind, a pair and a side card, no pair with a possible flush, and no pair with no possible flush) and then analyze each set.

Here is an example solution:

Player A – A/images/graemlins/spade.gifA/images/graemlins/diamond.gif
Player B – Q/images/graemlins/diamond.gifQ/images/graemlins/club.gif
Player C – A/images/graemlins/club.gifK/images/graemlins/diamond.gif
Player D – A/images/graemlins/heart.gif2/images/graemlins/heart.gif
Player E – 9/images/graemlins/heart.gif8/images/graemlins/heart.gif

Flop – Q/images/graemlins/heart.gifJ/images/graemlins/heart.gifT/images/graemlins/heart.gif giving the reverse order of straight-flush, flush, straight, trips, and one pair.

When you try to find a six-player solution, you wind up trying (for example) to make two flushes where the initial lower hand makes the higher flush (no can do) or you might be trying to make two sets where the initial lower hand makes the higher set (again, no can do). Straights pose an interesting possibility because of the wheel. It’s possible to make a higher straight with 7-6 than with A-2, but if you try using this trick in the solution you’ll find you can’t make both a flush and a straight-flush and still preserve the order constraint. Festus used this method to generate his five-handed solution.

This kind of inversion is also possible when a pair is on the board, but you will quickly find out that three players is the maximum when a pair exists in the flop. (See the original post for an example.)

So why is this even remotely useful? If the flop had the power to completely and easily randomize the order, then the any-two-cards-can-win philosophy might have merit. This exercise is just another indication (among many) that initial starting cards do matter and that being selective on the hands you enter is a valid idea.

I’ve always said, I’m not a strong enough player to play bad cards /images/graemlins/wink.gif.

[Schneids]

[PDosterM]

You're absolutely right.

I tip my hat to Pudley, and I will now have to go find the flaw in my "proof." (It's in the analysis of single pair hands using the low card.)

(This is what I love about this site -- there is always someone with a different -- and often better -- take on an issue.)