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Officer Kaoko

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Screw Public decency!!

Imma roam around naked making everyone feel uncomfortable! D:

 

I was beaten to what I was going to say. See one post above.

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@Elli: I got that. Still, what's the meaning of AVEN? lol

 

I published chapter 13 of Better Dayz (And modifying the Hades plot for my fanfic.)

  AVEN's the Asexuality Visibility and Education Network. It's a forum site for asexuals. The model's pretty much the same as here, but the topics of discussion are based on the site's purpose.

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How does one calculate the probability of getting a straight in the same suit with a 8 card hand with a deck of cards of three suits and no face cards (but including 10)?

 

(I need somewhere to put my work for this problem, so here, guys, you can has)

We are searching for the probability. To calculate the probability, we need two numbers: the total number of possible eight-card hands, and the number of distinct desired hands. We already can calculate the total number of hands: it's a combination of eight items out of thirty. There are 5852925 distinct possible combinations of eight cards, or hands.

 

Now we must find the total number of distinct desired hands. To have a straight, we must have 5 specific cards, which are in numerical order. (ex. A - 2 - 3 - 4 - 5). There are six ways to get a five-card straight (running from A-5 straight to the 6-10 straight), and there are three suits, so there are 18 ways to get a straight.

 

However, we are talking about eight-card hands. After we have drawn the necessary five cards, three more cards need to be drawn. Each unique combination of three cards creates a new distinct hand (ex. A 2 3 4 5 plus 6 9 10 and A 2 3 4 5 plus 6 8 9 are distinct hands). Since 25 cards remain in the deck after the straight is drawn, we need to find the total number of three-card combinations from a 25 card deck. There are 2300 potential combinations of three cards after the straight is drawn. Each straight paired with a three-card combination creates a distinct hand, so:

 

2300 * 18 = 41400 distinct hands.

 

We have found the number of desired hands, and the total number of hands possible. At this point, simply divide:

 

41400 / 5852925 = 0.0071. The low probability is expected, so the answer is checked as reasonable.

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(I need somewhere to put my work for this problem, so here, guys, you can has)

We are searching for the probability. To calculate the probability, we need two numbers: the total number of possible eight-card hands, and the number of distinct desired hands. We already can calculate the total number of hands: it's a combination of eight items out of thirty. There are 5852925 distinct possible combinations of eight cards, or hands.

 

Now we must find the total number of distinct desired hands. To have a straight, we must have 5 specific cards, which are in numerical order. (ex. A - 2 - 3 - 4 - 5). There are six ways to get a five-card straight (running from A-5 straight to the 6-10 straight), and there are three suits, so there are 18 ways to get a straight.

 

However, we are talking about eight-card hands. After we have drawn the necessary five cards, three more cards need to be drawn. Each unique combination of three cards creates a new distinct hand (ex. A 2 3 4 5 plus 6 9 10 and A 2 3 4 5 plus 6 8 9 are distinct hands). Since 25 cards remain in the deck after the straight is drawn, we need to find the total number of three-card combinations from a 25 card deck. There are 2300 potential combinations of three cards after the straight is drawn. Each straight paired with a three-card combination creates a distinct hand, so:

 

2300 * 18 = 41400 distinct hands.

 

We have found the number of desired hands, and the total number of hands possible. At this point, simply divide:

 

41400 / 5852925 = 0.0071. The low probability is expected, so the answer is checked as reasonable.

:o That's awesome... Btw just a question, I can just calculate the extra three cards after I've drawn the five cards? Because I can have something like this:

S1, S2, S3, S4, C1, C2, C3, S5 for example and it still considers that I have a straight, can I just calculate as if I got S1, S2, S3, S4, S5, then C1, C2, C3 (as with your workings)? I am pretty confused on that part.

 

And the next part of the question involves having 4 sets of this deck D:

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:o That's awesome... Btw just a question, I can just calculate the extra three cards after I've drawn the five cards? Because I can have something like this:

S1, S2, S3, S4, C1, C2, C3, S5 for example and it still considers that I have a straight, can I just calculate as if I got S1, S2, S3, S4, S5, then C1, C2, C3 (as with your workings)? I am pretty confused on that part.

 

And the next part of the question involves having 4 sets of this deck D:

No, I'm lazy so I didn't do it right. :V

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@Elli: Sure, you will have it.

 

OMM: I got this unique avatar. I just watched some moves to fill my inspiration for my fanfic and my drawing (Need to be scanned before being published....).

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OMM: That 'Gravity' Movie seems interesting. Still, No one really recommend it to me.

 

My aunt didn't like it at all. She was talking about how much of a rip-off it was for half an hour nonstop. But then again, she's too tough to please.

 

OMM: The rp I'm in has its own tv tropes page and apparently has fans. The pressuuuuure.

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Speaking of fanfiction, I'm finally motivated enough to start working on one. I think first chapter'll be done...next week? Maybe? No?

 

Speaking of fanfiction, I'm finally motivated enough to start working on one. I think first chapter'll be done...next week? Maybe? No?

Sounds cool what is going to be about?

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