Probability of a Straight

[Havie]

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

 

[MarkFL]

Aces can either be 1 or 14. Let's look at from the perspective of building up, from lowest to highest. The lowest can be anything from A to 10, and the suit doesn't matter, so we have 40 choices for the first card, and 4 choices each for the remaining 4 cards in the hand. Thus:

\(\displaystyle N=40\cdot4^4=10240\)

Note, this includes the 4 royal flushes and 36 straight flushes.

Does this make sense?

 

[Havie]

Aces can either be 1 or 14. Let's look at from the perspective of building up, from lowest to highest. The lowest can be anything from A to 10, and the suit doesn't matter, so we have 40 choices for the first card, and 4 choices each for the remaining 4 cards in the hand. Thus:

\(\displaystyle N=40\cdot4^4=10240\)

Note, this includes the 4 royal flushes and 36 straight flushes.

Does this make sense?

hmm this doesn't line up with what i have written down in my notes from class,
i have
\(\displaystyle 9\cdot4^5\)



also yea my bad, totally blanked that A and 1 are the same thing (or 14) so my examples above are a bit wonky.

I still dont understand why you would have 4 choices per card. as you get to the 4th card, dont you only have a more limited amount of options.
lets say you hand is
1,2,4,5
your last card can only be 3.
orrrrrrrrrrr
is it 4 choices because theres 4 suits? so its technically 4 3's??

 

[MarkFL]

hmm this doesn't line up with what i have written down in my notes from class,
i have
\(\displaystyle 9\cdot4^5\)



also yea my bad, totally blanked that A and 1 are the same thing (or 14) so my examples above are a bit wonky.

I still dont understand why you would have 4 choices per card. as you get to the 4th card, dont you only have a more limited amount of options.
lets say you hand is
1,2,4,5
your last card can only be 3.
orrrrrrrrrrr
is it 4 choices because theres 4 suits? so its technically 4 3's??

Yes, the 4 choices for cards 2-5 come from there being 4 suits. It is true that only one rank is possible for each of those choices, but there are 4 suits for each rank, giving a total of 4 choices.

 

[JeffM]

hmm this doesn't line up with what i have written down in my notes from class,
i have
\(\displaystyle 9\cdot4^5\)



also yea my bad, totally blanked that A and 1 are the same thing (or 14) so my examples above are a bit wonky.

I still dont understand why you would have 4 choices per card. as you get to the 4th card, dont you only have a more limited amount of options.
lets say you hand is
1,2,4,5
your last card can only be 3.
orrrrrrrrrrr
is it 4 choices because theres 4 suits? so its technically 4 3's??

Do you understand why we got to 10 instead of 9? We have

A2345
23456
34567
45678
56789
6789T
789TJ
89TJQ
9TJQK
TJQKA

There are THREE RANKS that we cannot start from, J, Q, and K.

So the number of ranks that we can start from is 13 - 3 = 10.

Within each of the 5 sequential ranks, we have 4 suits so we get

\(\displaystyle 10 * 4^5\),

which looks identical to what you got from class except that 10 replaces the 9.

But \(\displaystyle 10 * 4^5 = 10 * 4 * 4^4 = 40 * 4^4\),

which is what Mark told you.

EDIT: I must admit I personally find

\(\displaystyle 10 * 4^5\) more intuitive than \(\displaystyle 40 * 4^4.\)

But when an answer does not look like what you expect, then you need to see if it is the expected answer in a different form.

 

[Havie]

Do you understand why we got to 10 instead of 9? We have

A2345
23456
34567
45678
56789
6789T
789TJ
89TJQ
9TJQK
TJQKA

There are THREE RANKS that we cannot start from, J, Q, and K.

So the number of ranks that we can start from is 13 - 3 = 10.

Within each of the 5 sequential ranks, we have 4 suits so we get

\(\displaystyle 10 * 4^5\),

which looks identical to what you got from class except that 10 replaces the 9.

But \(\displaystyle 10 * 4^5 = 10 * 4 * 4^4 = 40 * 4^4\),

which is what Mark told you.

EDIT: I must admit I personally find

\(\displaystyle 10 * 4^5\) more intuitive than \(\displaystyle 40 * 4^4.\)

But when an answer does not look like what you expect, then you need to see if it is the expected answer in a different form.

hmm sorry,
you cant do BOTH of the 2 highlighted red elements from your quote, only 1 . Ace high or low, doesnt loop. at least the rules in my class stated this somehow,
so it would be
\(\displaystyle 9* 4^5

but thanks,
i think i get it, the 4 is the suits, not choices.
so really its 1 card 5 times. but that 1 card is represented by a 4 due to suits. gotcha\)

 

[pka]

hmm sorry,
you cant do BOTH of the 2 highlighted red elements from your quote, only 1 . Ace high or low, doesnt loop. at least the rules in my class stated this somehow,
so it would be
\(\displaystyle 9* 4^5\)
but thanks,
i think i get it, the 4 is the suits, not choices.
so really its 1 card 5 times. but that 1 card is represented by a 4 due to suits. gotcha

I for one wish that you would drop the concept "looping" it has no application here.

Straights can begin with one of \(\displaystyle \{\mathbf{A,~2,~3,~4,~5,~6,~7,~8,,~9,T}\}\) i.e. the low card in the five card straight.
Straights can end (some times called the rank) with one of \(\displaystyle \{\mathbf{5,~6,~7,~8,~9,~T,~J,~Q,~K,~A}\}\) i.e. the HIGH card in the five card straight.
Please note that straight either begin with an ace or end with an ace, an ace can be the low card or the high card. (Of course not at the same time.)
Thus JeffM's list is correct. And the count is \(\displaystyle 10* 4^5=10240\). I found some poker websites that give the answer as \(\displaystyle 10220\). It took a while to realize that those sites were removing any straight flush from the count some other hands as well.

 

[Havie]

I for one wish that you would drop the concept "looping" it has no application here.

Straights can begin with one of \(\displaystyle \{\mathbf{A,~2,~3,~4,~5,~6,~7,~8,,~9,T}\}\) i.e. the low card in the five card straight.
Straights can end (some times called the rank) with one of \(\displaystyle \{\mathbf{5,~6,~7,~8,~9,~T,~J,~Q,~K,~A}\}\) i.e. the HIGH card in the five card straight.
Please note that straight either begin with an ace or end with an ace, an ace can be the low card or the high card. (Of course not at the same time.)
Thus JeffM's list is correct. And the count is \(\displaystyle 10* 4^5=10240\). I found some poker websites that give the answer as \(\displaystyle 10220\). It took a while to realize that those sites were removing any straight flush from the count some other hands as well.

hmm
(10 choices - 1 of the red = 9)
A2345
23456
34567
45678
56789
6789T
789TJ
89TJQ
9TJQK
TJQKA

I am not a poker player but in our class we were told its either Ace high or Ace low. you can not have both A2345 and TJQKA.
If this were allowed, it would imply Ace is High and Also low at the same time.
also this is the gateway to "looping" such as JQKA2 which is what i meant.
I appreciate your help but the teachers answer was 9 * 4^5 so im gonna roll with that, it makes sense to me.

 

[Steven G]

hmm
(10 choices - 1 of the red = 9)
A2345
23456
34567
45678
56789
6789T
789TJ
89TJQ
9TJQK
TJQKA

I am not a poker player but in our class we were told its either Ace high or Ace low. you can not have both A2345 and TJQKA.

Were you told in our class we were told its either Ace high or Ace low and then you assumed that you can not have both A2345 and TJQKA OR were you told both statements? If it was the former, than you can have both A2345 and TJQKA since you are not using the A as a high AND low card in any ONE hand

Your teacher is wrong if they said that you can not have both A2345 and TJQKA but that does not really matter. As long as your teacher says you can not have both A2345 and TJQKA, then your answer is correct. The other posters claim (correctly) that a straight can have both A2345 and TJQKA and their answers are also correct.

The thing with word problems is that sometimes they are not very clearly stated (in this case it is clear--your teacher is slightly mistaken) and can be interpreted in multiple ways. As long as you get the correct probability understanding it your way that is good. Sometimes the way you interpret a problem may be wrong and interpreting it correctly is a learning process.