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
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