Training the mind - challenge

pedrovitorcosta · May 2, 2023

I'm proposing a challenge just to see how fast you are thinking about poker hands.

I'll mention a question that came up during a game with my friends and I'll ask it for you, tell me how long it took for you to solve the question.

Question:
"In a Texas Hold'm poker game, "Player A" has A/3 and "Player B" has A/4
What is the possible situation for the 4 in player B's hand to enter as a kicker to decide the hand?
(not counting flush)"

Andyreas · May 3, 2023

Not sure if I correctly understand the question but if the board would read K3567, then the 4 would decide and win the hand? ☺️

But I'm thinking that this is not what you are asking for since 4 isn't a kicker in the above solution.

Somehow I cannot come up with a board that uses the 4 as a kicker to decide the hand since the other player holds a 3, the only lower card available are 2s but there are only 4 of them. 🤣

pedrovitorcosta · May 3, 2023

Andyreas said:
Not sure if I correctly understand the question but if the board would read K3567, then the 4 would decide and win the hand? ☺️

But I'm thinking that this is not what you are asking for since 4 isn't a kicker in the above solution.

Somehow I cannot come up with a board that uses the 4 as a kicker to decide the hand since the other player holds a 3, the only lower card available are 2s but there are only 4 of them. 🤣

I forgot to say, but straight it is not valid either.... now you dont have a kicker... you have a straight...
You need a board for use the 4 like high kicker.
Is hard but have one way i guess... or isnt possible?

A kiravio · May 3, 2023

to a single pair of aces

pedrovitorcosta · May 3, 2023

A kiravio said:
to a single pair of aces

Wrong.... remember the hand have 5 cards in total...
I thought it would be easier for CC people to know how to solve

A kiravio · May 3, 2023

Whatever formation is on the table, they both have it, with a pair you say that it falls then it's ,,nothing "😉, tell me too that I still die stupidly😃

pedrovitorcosta · May 3, 2023

A kiravio said:
Whatever formation is on the table, they both have it, with a pair you say that it falls then it's nothing 😉, tell me too that I still die stupidly😃

They would have a pair plus three cards.... there are 5 on the board and 2 in both hands, in which situation would the 4 be used as a kicker?
I can give the answer, but then it loses the fun

Andyreas · May 3, 2023

pedrovitorcosta said:
Is hard but have one way i guess... or isnt possible?

I don't think it's possible but of course I may be wrong.

Since they both have an Ace, there need to be at least 4 other cards lower than 3 and this is not possible?
Edit:
Nevermind, I got it. Nice riddle. 😎

bowserdon · May 3, 2023

If one of the other cards was a 2

pedrovitorcosta · May 3, 2023

Andyreas said:
I don't think it's possible but of course I may be wrong.

Since they both have an Ace, there need to be at least 4 other cards lower than 3 and this is not possible?
Edit:
Nevermind, I got it. Nice riddle. 😎

bowserdon said:
If one of the other cards was a 2

Both Wrong...
it's possible, but isnt only so simply Bowserdon

Both are on the way, but have not yet reached the goal.

Andyreas · May 3, 2023

pedrovitorcosta said:
Both are on the way, but have not yet reached the goal.

I have but I thought you didn't want to post a solution? So I also didn't do it.

pedrovitorcosta · May 3, 2023

Andyreas said:
I have but I thought you didn't want to post a solution? So I also didn't do it.

Yeah nice!
It looks easy but it's not, right?

Andyreas · May 3, 2023

pedrovitorcosta said:
Yeah nice!
It looks easy but it's not, right?

No, it f*cked my brain LOL

BOXING71 · May 3, 2023

To answer this question, we need to consider the possible community cards that could be dealt in a Texas Hold'em game. If the community cards contain a 4, then Player B would have a pair of Aces with a 4 kicker, which would beat Player A's pair of Aces with a 3 kicker.

Assuming that no 4 appears among the community cards, then Player B's best possible hand would be two pairs, Aces and fours, with the highest kicker available. The highest kicker that can be used in this situation is one of the remaining three cards of the same rank as Player B's kicker (in this case, the remaining 3's).

So the possible situation for the 4 in Player B's hand to enter as a kicker to decide the hand (not counting flush) is if the community cards do not contain a 4, and Player B makes two pairs, Aces and fours, with a 3 kicker

istbno · May 3, 2023

I have a question to which there is no answer ))

pedrovitorcosta · May 3, 2023

BOXING71 said:
To answer this question, we need to consider the possible community cards that could be dealt in a Texas Hold'em game. If the community cards contain a 4, then Player B would have a pair of Aces with a 4 kicker, which would beat Player A's pair of Aces with a 3 kicker.

Assuming that no 4 appears among the community cards, then Player B's best possible hand would be two pairs, Aces and fours, with the highest kicker available. The highest kicker that can be used in this situation is one of the remaining three cards of the same rank as Player B's kicker (in this case, the remaining 3's).

So the possible situation for the 4 in Player B's hand to enter as a kicker to decide the hand (not counting flush) is if the community cards do not contain a 4, and Player B makes two pairs, Aces and fours, with a 3 kicker

no, you can make it without a pair of 4... if you have A A 4 4 XYZ (since you have 7 cards, and the best 5 cards is your game...)

The ridle is, how can a 4 become the highest kicker as the fifth card?

Have one way to your 4 beat the 3 in a simple hight kicker way

DaMooca · May 3, 2023

I tried to find out but I couldn't. Is this really possible??

pedrovitorcosta · May 3, 2023

DaMooca said:
I tried to find out but I couldn't. Is this really possible??

yes is

BOXING71 · May 3, 2023

I see, my apologies for misunderstanding the question earlier. If we assume that Player A and Player B have already made their best possible hands using the community cards, and the only difference between their hands is the kicker, then for Player B's 4 to become the highest kicker, it would need to be the highest-ranked card available among the remaining unpaired cards.

For example, if the community cards are 10-J-Q-K-2 and Player A has A-10 and Player B has A-4, then both players have a pair of Aces, but Player A wins with the better kicker of 10. However, if the last community card is a 4, then Player B's hand improves to two pairs, Aces and fours, with a better kicker than Player A's hand.

In summary, for Player B's 4 to become the highest kicker, it would need to be the highest unpaired card among the community cards and the hole cards of both players. The specific situation would depend on the exact cards dealt and the order in which they are revealed

Andyreas · May 3, 2023

@BOXING71 , are you using ChatGPT for your recent posts, buddy?

BOXING71 · May 3, 2023

Andyreas said:
@BOXING71 , are you using ChatGPT for your recent posts, buddy?

Yes, of course. Such tasks can only be solved by programs specially configured for solving poker problems.

pedrovitorcosta · May 3, 2023

Andyreas said:
@BOXING71 , are you using ChatGPT for your recent posts, buddy?

hahahaha

BOXING71 said:
I see, my apologies for misunderstanding the question earlier. If we assume that Player A and Player B have already made their best possible hands using the community cards, and the only difference between their hands is the kicker, then for Player B's 4 to become the highest kicker, it would need to be the highest-ranked card available among the remaining unpaired cards.

For example, if the community cards are 10-J-Q-K-2 and Player A has A-10 and Player B has A-4, then both players have a pair of Aces, but Player A wins with the better kicker of 10. However, if the last community card is a 4, then Player B's hand improves to two pairs, Aces and fours, with a better kicker than Player A's hand.

In summary, for Player B's 4 to become the highest kicker, it would need to be the highest unpaired card among the community cards and the hole cards of both players. The specific situation would depend on the exact cards dealt and the order in which they are revealed

Yeah,
Yes, but you didn't give the answer, you just explained the poker rule.
The ridle is knowing what the board has to bring for the rule to apply

Andyreas · May 3, 2023

BOXING71 said:
Yes, of course. Such tasks can only be solved by programs specially configured for solving poker problems.

Well I solved it myself - just thinking through.

pedrovitorcosta · May 3, 2023

Andyreas said:
Well I solved it myself - just thinking through.

And that's what's fun, after all, it's to know if our brain is trained for the possibilities that can happen in poker

BOXING71 · May 3, 2023

Andyreas said:
Well I solved it myself - just thinking through.

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 .

Training the mind - challenge

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