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# Statistics with poker cards. watch

1. There's this poker game that is popular among the Asian community.
It is a 4-player game where each player is dealt with 13 cards.

What is the probability of getting the following cards among the 13 cards in their hands:
(a) 4 of a kind (i.e. any set of 4 cards with the same rank);

edit: Sorry. That was a bit confusing. I just copied it from wikipedia and it says '4 of a kind'. Edited now.

(b) A straight flush (i.e. straight and flush).
For example,
A 2 3 4 5 of the same suits,
2 3 4 5 6 and so on ....
......
......
J Q K A 2.

I'm just curious to see which is harder to be dealt.
edit: There is always an argument among the players on which is harder to be dealt. Hence, I would like to see the probabilities from a mathematical point of view.
I can try to calculate it myself but I don't trust my statistics.
I would rather do maths!!
2. (Original post by UglyDuckling)
There's this poker game that is popular among the Asian community.
It is a 4-player game where each player is dealt with 13 cards.

What is the probability of getting the following cards among the 13 cards in their hands:
(a) 4 of a kind (i.e. spades, hearts, clubs and diamonds);

(b) A straight flush (i.e. straight and flush).
For example,
A 2 3 4 5 of the same suits,
2 3 4 5 6 and so on ....
......
......
J Q K A 2.

I'm just curious to see which is harder to be dealt.
4 of a kind defined as standard and not what you wrote?

It's not a very simple question (which I suspect is why it hasn't been answered yet), I'll give it a go later, in a hurry now.

edit: Also that's a non-standard definition of straight flush. It's not normally allowed to 'go around' an A.
3. (Original post by nota bene)
4 of a kind defined as standard and not what you wrote?

It's not a very simple question (which I suspect is why it hasn't been answered yet), I'll give it a go later, in a hurry now.

edit: Also that's a non-standard definition of straight flush. It's not normally allowed to 'go around' an A.
Oh I see. Nevermind. Just change the definition.
Make sure you define the scope of a straight flush before doing the calculation. Thanks.

edit: so I suppose 10 J Q K A should be the last of the straight flush.
4. (Original post by UglyDuckling)
There's this poker game that is popular among the Asian community.
It is a 4-player game where each player is dealt with 13 cards.

What is the probability of getting the following cards among the 13 cards in their hands:
(a) 4 of a kind (i.e. any set of 4 cards with the same rank);

edit: Sorry. That was a bit confusing. I just copied it from wikipedia and it says '4 of a kind'. Edited now.

(b) A straight flush (i.e. straight and flush).
For example,
A 2 3 4 5 of the same suits,
2 3 4 5 6 and so on ....
......
......
J Q K A 2.

I'm just curious to see which is harder to be dealt.
edit: There is always an argument among the players on which is harder to be dealt. Hence, I would like to see the probabilities from a mathematical point of view.
I can try to calculate it myself but I don't trust my statistics.
I would rather do maths!!
Not that i play this poker game but sometimes you can't rely just maths alone to win every hand. Psychology
5. it's all luck and facial expressions
6. (Original post by fourdigit)
Not that i play this poker game but sometimes you can't rely just maths alone to win every hand. Psychology
thank you for your words of wisdom
7. I think they should do a poker statistics degree. I would definitely do that
I think they should do a poker statistics degree. I would definitely do that
LOL!!
Then the act of gambling will be lifted up to a whole new level!!
9. How about (13*4C4*48C9)/(52C13)=399/12495 for the first probability? (I may have made a mistake, so just wanting to check if someone else gets the same answer...) Note that I'm only worrying about getting at least one set of four of a kind here, I haven't discounted hands with e.g. 4 of A, 4 of 2 and 5 random cards.

For the second part, I think this works (again I would like someone to check me);

(4*10*5C5*47C8)/(52C13)=399/4998 (4 as there's symmetry wrt what suit and 10 as there are 10 cards a straight can start with (A through 10))
10. a) Assuming you mean at least one set of a 4-of-a-kind. This is how I worked it out.

N choose R (nCr) gives the number of possible hands, with 52 cards and 13 in your hand:
(52! / (39! * 13!)) = 635013559600

There are 13 possible combinations of 4-of-a-kinds, one for each suit. 13

Using nCr again to calculate the remaining number of possible combinations of the other 9 out of 48 cards to get dealt into your hand: (48! / (9! * 39!)) = 1677106640

(13*1677106640) / 635013559600

= 21802386320 / 635013559600

= 0.0343337335

~3.4%

I've probably gone wrong somewhere there, mind. Can anybody verify??
11. That looks like the same answer and method as I used.

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