Whats higher 5 of a kind or a straight flush??

[Pacfanweb]

Originally posted by: EMPshockwave82
strait flush is a legititamately possible hand... 5 of a kind isnt possible unless wilds are involved

therefore in my opinion a hand that is possible beats a hand that is "impossible" under normal rules

Your opinion is flawed. Everyone knows that the only way to have 5 of a kind is with wild cards, so the original question was obviously assuming that's the way to come up with 5 of a kind.

 

[CTrain]

Originally posted by: Pacfanweb

Originally posted by: CTrain
Now I've played with people who will say "NATURAL straight flush beats 5 of a kind". This I will agree with.
But to say Straight flush(with a wild card) beats 5 of a kind is RETARDED. And your friends are retarded also.

Nothing beats 5 of a kind, whether it's a natural straight or one with a wild card.

I totally agree. If you're willing to play wild cards, then 5 of a kind is the best hand period.
I was just saying I an not oppose to people who considered a natural straight flush(no wild) beating 5 of a kind.

Mathematic wise, 5 of a kind is the best hand period. (with wilds of course)

 

[Rapidskies]

five of a kind w/ 2 jokers is 1 in 40,000 according to this site:

Odds

 

[DrPizza]

Originally posted by: HomeBrewerDude

Originally posted by: DrPizza
Bah, HBD, horrible math. I'm not going to even quote you.
Plus, I think you mean "denomination" where you're saying suit. You illustrated 3 aces. I'd like to see you get 3 aces of the same suit.

The odds of drawing 2 jokers (if you have to get 2 jokers) and 3 cards of the same denomination are
13 suits * 2C2 (jokers) * 4C3
= 13 * 1 (there's only 1 way to have 2 jokers) * 4 (there are 4 ways to have 3 cards of any particular denomination)

= 52 out of 54C5 possible hands.

The odds of getting a 5 of a kind, I repeat, are 13 * 6C5 out of 54C5.

by 6C5, I mean combination (6, choose 5) in the style you prefer to write it in.

This equals 78/31,625,510
THOSE are the odds for getting 5 of a kind in a deck containing 2 wild jokers.

EDIT: I see what you were trying to say. But, then your denominator has to be changed, because it would be a permutation since you're saying the order matters.

you got errors buddy

the problem is in how you are counting combinations. the number of hands with 2 jokers and 3 same rank is far greater than you predict. lose the 6 c 5, and count the total number of unique combinations of 2 Joker, 3 same rank TIMES 13 / total hands and you will compute the same number I did.

Your number is completely wrong. Getting Joker Joker A hearts Aspades Aclubs is the EXACT same hand as A spades Aclubs Ahearts Joker Joker. What you have is a permutation where the order matters. In this case, the order does NOT matter. The total number of possible hands is 54C5. There are 6 cards that can be counted as Aces... 2 jokers and 4 aces. Any combination of 5 of these 6 will give you 5 of a kind. So, 6C5 ( = 6 ) is the number of different hands with 5 aces. There are also 6 different hands with 5 kings, 6 different hands with 5 queens, and so on. The total number of hands with 5 of a kind is 13*6 (or, as I stated above, 13 * 6C5 = 78 )

I'm sorry if you do not understand this. This is covered in freshman level high school mathematics. Too bad you don't go to school here.

If you do wish to do it your way, you could use the permutation (which you decided to list) of 6P5. There are 6*5*4*3*2 ways to receive a 5 of a kind in each suit. (but, since there are 5 cards, as I'm trying to point out, the order of those 5 cards does not matter. Thus, there are 5*4*3*2*1 ways to arrange those cards. You could still use the 6*5*4*3*2 in your probability, however, if you decide that there are 54P5 different hands (again, the hands are not all different, because the order doesn't matter. ) You would then give the probablilty as 6P5 / 54P5. Note that both the numerator and denominator should be divided by 5! because there are 5! different arrangements for each set of 5 cards. Thus, your probability would be equal to mine. But, you apparently don't understand the answer.

 

[SoylentGreen]

5 of a kind is gay, because wild cards are for n00bs.

 

[simms]

Originally posted by: whiteboy81
Straight flush for sheezy...the odds are lower.

That's why your role model is Snoop Dogg. Fo' sheezy, yo.

 

[HomeBrewerDude]

Originally posted by: DrPizza

Originally posted by: HomeBrewerDude

Originally posted by: DrPizza
Bah, HBD, horrible math. I'm not going to even quote you.
Plus, I think you mean "denomination" where you're saying suit. You illustrated 3 aces. I'd like to see you get 3 aces of the same suit.

The odds of drawing 2 jokers (if you have to get 2 jokers) and 3 cards of the same denomination are
13 suits * 2C2 (jokers) * 4C3
= 13 * 1 (there's only 1 way to have 2 jokers) * 4 (there are 4 ways to have 3 cards of any particular denomination)

= 52 out of 54C5 possible hands.

The odds of getting a 5 of a kind, I repeat, are 13 * 6C5 out of 54C5.

by 6C5, I mean combination (6, choose 5) in the style you prefer to write it in.

This equals 78/31,625,510
THOSE are the odds for getting 5 of a kind in a deck containing 2 wild jokers.

EDIT: I see what you were trying to say. But, then your denominator has to be changed, because it would be a permutation since you're saying the order matters.

you got errors buddy

the problem is in how you are counting combinations. the number of hands with 2 jokers and 3 same rank is far greater than you predict. lose the 6 c 5, and count the total number of unique combinations of 2 Joker, 3 same rank TIMES 13 / total hands and you will compute the same number I did.

Your number is completely wrong. Getting Joker Joker A hearts Aspades Aclubs is the EXACT same hand as A spades Aclubs Ahearts Joker Joker. What you have is a permutation where the order matters. In this case, the order does NOT matter. The total number of possible hands is 54C5. There are 6 cards that can be counted as Aces... 2 jokers and 4 aces. Any combination of 5 of these 6 will give you 5 of a kind. So, 6C5 ( = 6 ) is the number of different hands with 5 aces. There are also 6 different hands with 5 kings, 6 different hands with 5 queens, and so on. The total number of hands with 5 of a kind is 13*6 (or, as I stated above, 13 * 6C5 = 78 )

I'm sorry if you do not understand this. This is covered in freshman level high school mathematics. Too bad you don't go to school here.

If you do wish to do it your way, you could use the permutation (which you decided to list) of 6P5. There are 6*5*4*3*2 ways to receive a 5 of a kind in each suit. (but, since there are 5 cards, as I'm trying to point out, the order of those 5 cards does not matter. Thus, there are 5*4*3*2*1 ways to arrange those cards. You could still use the 6*5*4*3*2 in your probability, however, if you decide that there are 54P5 different hands (again, the hands are not all different, because the order doesn't matter. ) You would then give the probablilty as 6P5 / 54P5. Note that both the numerator and denominator should be divided by 5! because there are 5! different arrangements for each set of 5 cards. Thus, your probability would be equal to mine. But, you apparently don't understand the answer.

Okay, I see my error. Its been a while since I have used either combination or permutation. However, your graceful way of pointing out other people mistakes leads me to think your social skills are as dusty as my math skills.