Training the mind - challenge

[Andyreas]

The problem itself contained an unsolvable problem.

It's not unsolvable. You just have to find a way on how to incorporate more community cards since you cannot have only lower ones than 3 and 4s

 

[pedrovitorcosta]

if a problem needs to be solved, I solve it in any way I can. The problem itself contained an unsolvable problem. I think the problem is taken from similar content, but if you solved it yourself, my respect .

The problem has a solution, and it came from witnessing it at a poker table, I witnessed it and that's why I decided to post it here.
Once you realize the solution you realize how simple it is, but how hard it is to imagine it happening.
The ridle

 

[pedrovitorcosta]

I'm going to give a tip that I already gave and someone must not have noticed...
How can you incorporate the 4 as the fifth card out of the possible 7 to beat the 3

You just need to know what the sixth and seventh card should be to be able to solve it.

 

[BOXING71]

give me an answer, otherwise today is a sleepless night. The brain exploded, the car boils, Andreas I repeat respect if you yourself decide

 

[bermejoga]

A4 A3
QQAK4
??? Bruh I dont know... That would split woundnt it?

 

[Andyreas]

give me an answer, otherwise today is a sleepless night. The brain exploded, the car boils, Andreas I repeat respect if you yourself decide

I am happy to but I'll hide it, so everyone can still solve, if he wants to:

Spoiler: Solution

The key is to incorporate "useless" community cards which will not be taken into account, i.e. a lower 3rd pair.

For example if the board reads:
A8822, then the 22 are not taking into account and player A has AA883 and player B has AA884

 

[pedrovitorcosta]

A4 A3
QQAK4
??? Bruh I dont know... That would split woundnt it?

wrong, in that situation is a split and 4 and 3 dont get in

 

[bermejoga]

I am happy to but I'll hide it, so everyone can still solve, if he wants to:

Spoiler: Solution

The key is to incorporate "useless" community cards which will not be taken into account, i.e. a lower 3rd pair.

For example if the board reads:
A8822, then the 22 are not taking into account and player A has AA883 and player B has AA884

Thank you so much for letting me sleep today.

 

[pedrovitorcosta]

I am happy to but I'll hide it, so everyone can still solve, if he wants to:

Spoiler: Solution

The key is to incorporate "useless" community cards which will not be taken into account, i.e. a lower 3rd pair.

For example if the board reads:
A8822, then the 22 are not taking into account and player A has AA883 and player B has AA884

So simple and so complicated at the same time
I asked them to put the time, thinking that everyone would solve it and only you did it. Good job

 

[BOXING71]

So simple and so complicated at the same time
I asked them to put the time, thinking that everyone would solve it and only you did it. Good job

I agree, but the task really seemed unsolvable. Thank you for the evening!

 

[iceheart888]

This is simple, the board need to be : A 2 2 +( any pair 55+). Player A make 2 pair AA + 55 ex. and 3 plays as kicker and for player B :4 plays as kicker.

 

[pasha poltava]

I think five cards should be like this A.K.K.3.2, maybe I misunderstood the question

 

[pedrovitorcosta]

I think five cards should be like this A.K.K.3.2, maybe I misunderstood the question

yeah, this is the second possible answer! Well Done!