Is Jqka 2 A Straight In Poker
- An ace-high straight flush, such as A ♦ K ♦ Q ♦ J ♦ 10 ♦, is called a royal flush or royal straight flush and is the best possible hand in high games when not using wild cards. 6 17 18 A five-high straight flush, such as 5 ♥ 4 ♥ 3 ♥ 2 ♥ A ♥, is called a steel wheel and is both the best low hand and usually the best.
- Is JQKA 2 a straight? A straight cannot go “around the corner”, the Ace can only be either the highest or the lowest card, not a card in the middle. So no, J-Q-K-A-2 is no straight in poker.
- A 'bent' straight. Some local rules allow them, but not usually in more traditional versions of poker.
- While the ace may be used as both a high card and a low card, a straight is always counted from the lowest value card to the highest value card, and must contain an uninterrupted sequence in.
Question:In a deck w 52 cards, 4 suits, 13#s per suit.
how many ways are there for a straight?
in class we were taught that you do
13-4 to account for not being able to have 10,J,Q,K roll over to the lowest card val such as like Q,K,A,2,3 would not count.
this =9 .
so then you look at your hand as in 5 spaces
_ _ _ _ _
you have 9 choices for the lowest 1st #
9 _ _ _ _
then here is where i get lost:
we have 4 choices for each remaining slot?
9 4 4 4 4
and the final answer is
9 * 4^5.
I don't understand 2 things
1) if we have 4 choices for the remaining 4 slots, why is it not 4^4 ..
2) how can we have 4 choices for each slot?
shouldn't it be
9 4 3 2 1 ?
Situation 1
i feel like with the 9 4 4 4 4
you could pick your first # as 3.
then 2nd # as 2
then 3rd # as 2
then 4th # as 1
then 5th # as 2
Situation 2
lets say this somehow accounted for the inability to pick the same #.
once again
you could pick your first # as 3.
then you have 4 choices for your next # (1,2,4,5)
then 2nd # as 2
then you have 4 choices for your next # (1,4,5,6) what about Ace if it was low... that would be 5#s or take out 6.
then 3rd # as 6
then you have 4 choices for your next # (1, 4,5,7) ?
then 4th # as 1
then you have 4 choices for your next # (A, 4,5,7) ? this doesnt make any sense
then 5th # as 4
congrats.. the hand is now 1,2,3,4,6... doesnt work out?
Situation 3
lets say you can only pick #s within 2 of your current #s
you could pick your first # as 3.
then you have 4 choices for your next # (1,2,4,5)
then 2nd # as 2
then you have 4 choices for your next # (A, 1,4,5)
then 3rd # as Ace
then you have only 3 choices for your next # (1, 4,5 ) because theres nothing lower than A, and 6 would be 3 spaces away from 3 so it violates assumption rule
then 4th # as 1
then you have 1-2 choices for your next # (4, 5) ? this doesnt make any sense
then 5th # as 5
congrats.. the hand is now A,1,2,3,5... doesnt work out? again and we didnt use 4#s per choice
how many ways are there for a straight?
in class we were taught that you do
13-4 to account for not being able to have 10,J,Q,K roll over to the lowest card val such as like Q,K,A,2,3 would not count.
this =9 .
so then you look at your hand as in 5 spaces
_ _ _ _ _
you have 9 choices for the lowest 1st #
9 _ _ _ _
then here is where i get lost:
we have 4 choices for each remaining slot?
9 4 4 4 4
and the final answer is
9 * 4^5.
I don't understand 2 things
1) if we have 4 choices for the remaining 4 slots, why is it not 4^4 ..
2) how can we have 4 choices for each slot?
shouldn't it be
9 4 3 2 1 ?
Situation 1
i feel like with the 9 4 4 4 4
you could pick your first # as 3.
then 2nd # as 2
then 3rd # as 2
then 4th # as 1
then 5th # as 2
Situation 2
lets say this somehow accounted for the inability to pick the same #.
once again
you could pick your first # as 3.
then you have 4 choices for your next # (1,2,4,5)
then 2nd # as 2
then you have 4 choices for your next # (1,4,5,6) what about Ace if it was low... that would be 5#s or take out 6.
then 3rd # as 6
then you have 4 choices for your next # (1, 4,5,7) ?
then 4th # as 1
then you have 4 choices for your next # (A, 4,5,7) ? this doesnt make any sense
then 5th # as 4
congrats.. the hand is now 1,2,3,4,6... doesnt work out?
Situation 3
lets say you can only pick #s within 2 of your current #s
you could pick your first # as 3.
then you have 4 choices for your next # (1,2,4,5)
then 2nd # as 2
then you have 4 choices for your next # (A, 1,4,5)
then 3rd # as Ace
then you have only 3 choices for your next # (1, 4,5 ) because theres nothing lower than A, and 6 would be 3 spaces away from 3 so it violates assumption rule
then 4th # as 1
then you have 1-2 choices for your next # (4, 5) ? this doesnt make any sense
then 5th # as 5
congrats.. the hand is now A,1,2,3,5... doesnt work out? again and we didnt use 4#s per choice
Is Jqka 2 A Straight In Poker Room
No, a straight cannot be cyclic (e.g. K-A-2-3-4) in any poker variant I have played. The highest straight is 10-J-Q-K-A (also known as Broadway) in most poker games, including Texas Hold'em. Yes, A-2-3-4-5 is the lowest straight in most forms of poker, it is also known as a wheel.