Number of possible Straights in a deck of 52 cards

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SUMMARY

The discussion focuses on calculating the number of possible straights in a standard 52-card deck. The correct approach identifies that there are 10 distinct starting sequences for straights, with each card in the sequence being able to be of any of the four suits. Therefore, the total number of unique straights is calculated as 10 multiplied by 4 raised to the power of 5, resulting in 10 * 4^5 = 10,240 possible straights. Additionally, it is noted that 40 of these straights can also be classified as straight flushes.

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audifanatic51
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Hi,

I'm a bit thick-headed I guess and I cannot seem to figure out why my answer to this question is wrong.

I need to figure out the number of ways to obtain a straight in a deck of 52 cards (I do not need to ignore straight-flushes or royal flushes).

Note: a hand of 5 cards is used

My answer was to choose 1 card from a deck of 52 cards and then restrict the next 4 cards to one of 4 possible cards (one for each suit).

So, in short:

52C1 * (4C1)^4

which is wrong. Can anybody explain to me why I am wrong. I'm a-Ok with calc 1-4 and graph theory, but counting theories have never been by favorite, even as a kid. Thanks for the help
 
Physics news on Phys.org
"There are 14 effective ranks of cards for a straight (ace can be high or low). The straight can start on anyone of A,K,J,Q,T,9,8,7,6,5 and go down. That makes 10 base straight sequences. Each card in the sequence can be any of the four suits. So the total number of straights is 10 * 4 ^5 = 10240.

40 of those are a straight flush." -Wiki answers.
 
audifanatic51 said:
My answer was to choose 1 card from a deck of 52 cards and then restrict the next 4 cards to one of 4 possible cards (one for each suit).

So, in short:
52C1 * (4C1)^4
You appear to be taking that first card as the top (or maybe the bottom, but not either/or) of the run. As FeynmanIsCool points out, only 40 of the cards can serve as that.
 
haruspex said:
You appear to be taking that first card as the top (or maybe the bottom, but not either/or) of the run. As FeynmanIsCool points out, only 40 of the cards can serve as that.

Ok, you you're basically saying that there really isn't a "first" card since that "first card" can be at the top, end, or somewhere in the middle since order doesn't matter?
 
audifanatic51 said:
Ok, you you're basically saying that there really isn't a "first" card since that "first card" can be at the top, end, or somewhere in the middle since order doesn't matter?
No, I'm saying that if you are taking the first card to define the set of four others then you need to specify where in the sequence that first card comes. You can make it the top, the bottom, or anywhere in between provided you are consistent. But once you have fixed that, there are only forty cards it can be.
 

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