A Poker Quiz Question

I was watching poker the other day and i thought of this situation and thought i could make it into a question.

Can anyone think of a situation where 2 players are in the pot, 1 has the 2nd nuts based on their cards, the other has the 2nd nuts based on their cards. But 1 hand is superior over the other?

Give your answer like this

Board: X X X X X
Hole Cards: X X X X

I have thought of 1 and think it is the only possible answer but we will see

Yes. Both flop straights but one has flush draw.
Player 1:
7c3c
Player 2:
7d3s
Flop:
4c5h6c


Unless you mean a situation in which the entire board is showing, in which case I am not sure that it is possible.

i think i get it.

here's one?

Board: 4c 5c 6c X X
Hole Cards: player 1 has Ac7c, player 2 has 2c3c

without knowing the other's hand, each player has the 2nd nuts (that is, only one hand can beat them)

well i have thought of 1 situation which i will say after a few more people try. Also i mean when the whole board is showing, since in your case the flush draw hand isnt technically superior yet as it hasnt made a flush yet, so is same superiority

Qh Js Td 2h 9h
Ks 3c for second nut straight (Ax Kx makes nut straight)
Kh xh for second nut flush (Ah yh makes nut flush)

edit. Combuboom's answer is superior.

combuboom said:

i think i get it.

here's one?

Board: 4c 5c 6c X X
Hole Cards: player 1 has Ac7c, player 2 has 2c3c

without knowing the other's hand, each player has the 2nd nuts (that is, only one hand can beat them)

Click to expand...

That sounds as if its correct, congrats, ill post mine and see if anyone else can think of any.

Board. A K K X X

Hole Cards: A A (2nd nuts to pock kings) A K (2nd nuts to pock aces)

pigpen02 said:

Qh Js Td 2h 9h
Ks 3c for second nut straight (Ax Kx makes nut straight)
Kh xh for second nut flush (Ah yh makes nut flush)

Click to expand...

not quite since the player with k,3 doesnt have 2nd nuts, since they are beaten by a,k and many flush combos

Does the guy winner have the nuts and second nuts? If that's the case V is correct if any club hits on the turn or river.

no no, he's saying once the final board is dealt out, one player has the winning hand, but they both have the 2nd nuts based only on their own hands (meaning only one hand in the deck can beat either of them)

This is a devious question.

The game would just about have to be Omaha Hi/Lo, possibly Stud hi/lo. Thats the only place where 2 2nd nuts could exist, but even then we're begging definitions.


There is something real wrong here tho. There could be thousands of hands that fit this incomplete scenario.

example in addition to KaBooms;

board KQ678 rainbow
player 1 kq45
player 2 kq23

the nut hand would be A29T

in any HI/LO game this would be 2 hands with 2nd nuts.

the 2 pair are there to keep the players in the pot to get to the 2nd nuts question?

dj, you're misunderstanding this. this is applying to NL hold em only

look at my example

Board: 4c 5c 6c X X
Hole Cards: player 1 has Ac7c, player 2 has 2c3c

player 1 can only be beat by one hand in the deck: 2c3c. player 2 can only be beat by one hand in the deck. 7c8c. they both have the 2nd nuts but only one hand is the winner

The hand from the movie "Rounders":

A99 on the board, plus other inconsequential stuff

hand 1: A9, only AA has it beat
hand 2: AA, only 99 has it beat

OP looks like a Omaha hand to me, also I edited my post.

here's one other type combining the quads and straight flush elements:

board: 9h9s8d7d6d

hand 1: 9c9d, only 5d4d has it beat
hand 2: 5d4d, only Td9d has it beat

Looks to me like you've nailed it combu.

Interesting fast paced thread, but the OP needs some clarification.

What game.
Pre showdown or post showdown.

combuboom said:

here's one other type combining the quads and straight flush elements:

board: 9h9s8d7d6d

hand 1: 9c9d, only 5d4d has it beat
hand 2: 5d4d, only Td9d has it beat

Click to expand...

damn, you seem to be too good at this game, i thoguht it was really hard, not for you clearly , if anyone else has any quiz questions feel free to post

well if it's hold 'em there's been 3 so far, but as I notice there's XXXX for the hole cards, implying that the game is indeed Omaha. This seems like even easier, adding rags to all previous hold 'em hands gives the same scenarios, so that's at least 3.

zachvac said:

well if it's hold 'em there's been 3 so far, but as I notice there's XXXX for the hole cards, implying that the game is indeed Omaha. This seems like even easier, adding rags to all previous hold 'em hands gives the same scenarios, so that's at least 3.

Click to expand...

no the XX (note the large space between the 2 sets of X's) XX

that represents 2 sets of hole cards for holdem, for 2 players

I don't see a large space, but good question anyway.

I see a space in between each X, but that's irrelevant. 3 answers posted as hold 'em hands that I believe satisfy the requirements.

kcanuck said:

I don't see a large space, but good question anyway.

Click to expand...

touche, i just looked at original post and you are correct, there was a large space when i typed it but it clearly didnt work lol

well we have 3 answers, i think that may be it, hopefully someone else will post a new question to get us thinking

If you bought some powdered water what would you mix it with.

I have another contender!

Board
6h 8d Td Jd Qd

Player 1
4s 9d

Player 2
3c Ad

Player 1 has straight flush Q - 9 so can only be beaten by someone holding Ad Kd.

Player 2 Has the nuts flush, but can be beaten by someone holding 9d. It's true that someone could beat player 2 with either Kd 9d or Kd and anything else. However, player 2 can only lose to someone holding 9d so technically it's irrelevant whether that play also has Kd or not.

The cards NOT highlighted could be pretty much anything, so argueably there are dozens of combinations.