for the straight flush you are forgetting that once you pick up an Ace, one side of your straight is a dead end.
Jesus Christ....this is such a retarded argument that I'm not going to bother putting my 2 cents in.
Actually I will.
"I didn't know anything was higher than a Royal Flush"
There isn't Sherlock but he did state that he is playing with wild cards.
To put it in the most simplistic way without all the complicated computation.
Someone said it best 5 of a kind is impossible without wild cards so it has to be the better hand
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.
Actually I will.
"I didn't know anything was higher than a Royal Flush"
There isn't Sherlock but he did state that he is playing with wild cards.
To put it in the most simplistic way without all the complicated computation.
Someone said it best 5 of a kind is impossible without wild cards so it has to be the better hand
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.
Oh and for all the people who thinks wild cards are for pussy, give me a break.
Like you have never play poker without wild cards before.
Sure Texas is fun. Ohama is definately fun but throw in a variety is never a bad thing..
Why do you think follow the Queen is such a popular game ?? All those people who plays them are pussy too right ??
Like you have never play poker without wild cards before.
Sure Texas is fun. Ohama is definately fun but throw in a variety is never a bad thing..
Why do you think follow the Queen is such a popular game ?? All those people who plays them are pussy too right ??
Uh, you're all wrong.
first, you pick any card to start with: 13/52
then, there are three other iterations of that card possible, with 51 cards left: *3/51
then, two left, with 50 cards still in the deck: *2/50
one left, 49 on the table: *1/49
then the wild card: if you're going deuces wild, it's the last *4/48
if you're going jokers wild, then it's (13/54)(3/53)(2/52)(1/51)*(2/50)
or if you're going jokers and deuces wild, then it's (13/54)(3/53)(2/52)(1/51)*(6/50)
Unless, of course, your base 4-of-a-kind is deuces (in deuces wild), in which case the odds would be
(13/52)(3/51)(2/50)(1/49)*(48/48) (because any card will make a 5 of a kind. change values accordingly is jokers are in play)
If you got both jokers, then your odds would be:
(2/54)(1/53)(13/52)(3/51)(2/50)
first two for the jokers, and the last three for a three of a kind.
So, the values for those?
5 of a kind with deuces wild: 1/999,600
5 of a kind with jokers wild: 1/2,432,700
5 of a kind with deuces and jokers wild: 1/810,900
5 of a kind with 4 deuces: 1/83,300
5 of a kind with 4 deuces (jokers in play): 1/97,308
5 of a kind with 2 jokers: 1/12,432,700
The end.
Oh right, and for a natural straight flush (no jokers in play):
Start out with any card: 13/52
There are now 12 cards left with that same suit, but only the ones +/-4 in either direction will do: 8/51
Now the odds split depending on what the second card was.
If the second card was 4 off in either direction, then you have three cards left to get: (3/50)(2/49)(1/48)
If the second card was 3 off in either direction, you have two cards between the first and second to get, but two end cards to choose from for the last one: (2/50)(1/49)(2/48)
If the second card was 2 off in either direction, you have one card in between to get, one card to choose from the ends, and then one card to choose from those ends: (1/50)(2/49)(2/48)
If the second card was 1 off in either direction, you have the ends open to fill no matter what, so it's (2/50)(2/49)(2/48)
Odds for final card being:
4 off: 1/499,800
3 off: 1/749,700
2 off: 1/749,700
1 off: 1/374,850
Okay, I'm done. You get the gist.
Start out with any card: 13/52
There are now 12 cards left with that same suit, but only the ones +/-4 in either direction will do: 8/51
Now the odds split depending on what the second card was.
If the second card was 4 off in either direction, then you have three cards left to get: (3/50)(2/49)(1/48)
If the second card was 3 off in either direction, you have two cards between the first and second to get, but two end cards to choose from for the last one: (2/50)(1/49)(2/48)
If the second card was 2 off in either direction, you have one card in between to get, one card to choose from the ends, and then one card to choose from those ends: (1/50)(2/49)(2/48)
If the second card was 1 off in either direction, you have the ends open to fill no matter what, so it's (2/50)(2/49)(2/48)
Odds for final card being:
4 off: 1/499,800
3 off: 1/749,700
2 off: 1/749,700
1 off: 1/374,850
Okay, I'm done. You get the gist.
DrPizza
Administrator Elite Member Goat Whisperer
"and not a royal flush"?? Royal flush is just the highest straight flush. That'd be like saying "which is higher, a flush, or a pair - (and not a pair of aces)"
5 of a kind is higher... just google the rules.
You're wrong too. Care to explain why for your first calculation you have (13/52)....Being dealt 5 cards, the first card you get doesn't matter. The next 4 cards however, must be the same, or be a wild card.
DrPizza
Administrator Elite Member Goat Whisperer
Some of you are looking at this the hard way.
Hopefully you have learned how to do combinations and permutations.
The number of different 5 card hands from a deck of cards is 52C5 or 52*51*50*49*48 / (5*4*3*2*1)
equals 2,598,960 different hands.
There are only 4 different royal straight flushes... so the odds of a royal straight flush are 4/2,598,960
But, for a straight flush, there are 4 suits, times the number of possibilities...
A2345, 23456,34567,45678, etc. 10 ways to get a straight flush per suit, for a total of 40/2,598,960 odds of a straight flush (including the royal straight flush)... 36/2,598,960 otherwise.
4 of a kind: there are 13*48 ways to get 4 of a kind (4 of a kind, plus one other card)
So, the odds of 4 of a kind are 624/2,598,960
To find the probability of ANY type of hand from a 52 card deck, it's the number of ways that hand can occur, divided by 2,598,960
Selecting a wild card from the 52 cards simply increases the number of ways to get a particular hand.
example: if one of the cards below 10 is selected as a wild card, then per suit, you have 6 cards from which you need 5 for a royal straight flush. Doing the combination 6C5 gives 6 different ways to have 5 cards making a royal straight flush per suit. Or a total of 24 different hands out of the 2,598,960.
It may be tempting to simply multiply the 24 by 10 to find the number of possible straight flushes, but you have to be careful... there's a problem - suppose the 7 spades is wild. You can't get the 6S, 7S, 8S, 9S, and expect a wildcard for the 10S.
So, For 3 of the suits, there are 10 different straight flushes. For each, there are 6C5 ways to get the straight flush. For 180 different straight flush hands in the 3 suits that don't include the wild card. Now, the number of possible straight flushes in the suit of the wild card depends on the value of the wild card. That is, if the A is wild, that only affects 2 hands that have the A. But, if the 7 were wild, that effects 5 straights that would include the 7.
Anyway, when one of the 52 cards is wild, that increases the complexity to determine how many actual hands there are out of the 2,598,960 hands. But, if the jokers are used as wild cards, then things get easy again.
The number of hands is now 54C5 = 31,625,510.
The number of royal straight flushes is 7C5 ways to get it per suit, * 4 suits = 21*4 = 84 different ways.
There are 10 times as many straight flushes (including the royal straight flush) = 840 / 31,625,510.
The number of ways to get 5 of a kind is 6C5 * 13. = 78 different ways. Thus, the odds of getting 5 of a kind are only 78 out of 31,625,510.
The chances of getting a 5 of a kind, with 2 wild jokers, are slightly less than the chances of getting a royal straight flush, and quite a bit lower than the odds of a straight flush.
If anyone wants me to calculate any particular odds, just ask.
edit. Since some of you like odds presented as 1 out of ,
the 78/31625,510 works out to 1 in 405454 (odds of 5 of a kind, 2 wild jokers in deck)
and 84/31,625,510 works out to 1 in 376,494 (odds of royal straight flush, 2 wild jokers in deck)
Bow down before my mathematical prowess, ATOT
(God, I hope I don't get owned for that statement)
Hopefully you have learned how to do combinations and permutations.
The number of different 5 card hands from a deck of cards is 52C5 or 52*51*50*49*48 / (5*4*3*2*1)
equals 2,598,960 different hands.
There are only 4 different royal straight flushes... so the odds of a royal straight flush are 4/2,598,960
But, for a straight flush, there are 4 suits, times the number of possibilities...
A2345, 23456,34567,45678, etc. 10 ways to get a straight flush per suit, for a total of 40/2,598,960 odds of a straight flush (including the royal straight flush)... 36/2,598,960 otherwise.
4 of a kind: there are 13*48 ways to get 4 of a kind (4 of a kind, plus one other card)
So, the odds of 4 of a kind are 624/2,598,960
To find the probability of ANY type of hand from a 52 card deck, it's the number of ways that hand can occur, divided by 2,598,960
Selecting a wild card from the 52 cards simply increases the number of ways to get a particular hand.
example: if one of the cards below 10 is selected as a wild card, then per suit, you have 6 cards from which you need 5 for a royal straight flush. Doing the combination 6C5 gives 6 different ways to have 5 cards making a royal straight flush per suit. Or a total of 24 different hands out of the 2,598,960.
It may be tempting to simply multiply the 24 by 10 to find the number of possible straight flushes, but you have to be careful... there's a problem - suppose the 7 spades is wild. You can't get the 6S, 7S, 8S, 9S, and expect a wildcard for the 10S.
So, For 3 of the suits, there are 10 different straight flushes. For each, there are 6C5 ways to get the straight flush. For 180 different straight flush hands in the 3 suits that don't include the wild card. Now, the number of possible straight flushes in the suit of the wild card depends on the value of the wild card. That is, if the A is wild, that only affects 2 hands that have the A. But, if the 7 were wild, that effects 5 straights that would include the 7.
Anyway, when one of the 52 cards is wild, that increases the complexity to determine how many actual hands there are out of the 2,598,960 hands. But, if the jokers are used as wild cards, then things get easy again.
The number of hands is now 54C5 = 31,625,510.
The number of royal straight flushes is 7C5 ways to get it per suit, * 4 suits = 21*4 = 84 different ways.
There are 10 times as many straight flushes (including the royal straight flush) = 840 / 31,625,510.
The number of ways to get 5 of a kind is 6C5 * 13. = 78 different ways. Thus, the odds of getting 5 of a kind are only 78 out of 31,625,510.
The chances of getting a 5 of a kind, with 2 wild jokers, are slightly less than the chances of getting a royal straight flush, and quite a bit lower than the odds of a straight flush.
If anyone wants me to calculate any particular odds, just ask.
edit. Since some of you like odds presented as 1 out of ,
the 78/31625,510 works out to 1 in 405454 (odds of 5 of a kind, 2 wild jokers in deck)
and 84/31,625,510 works out to 1 in 376,494 (odds of royal straight flush, 2 wild jokers in deck)
Bow down before my mathematical prowess, ATOT
(God, I hope I don't get owned for that statement)
DrPizza
Administrator Elite Member Goat Whisperer
Sorry, I meant 4 different royal straight flushes. I later added that there were 40 straight flushes.
edit: I fixed it. Thanks for pointing that out. I had edited as I typed it so many times... it'd have saved me a ton of time to just jot the calculations on paper, scan the paper, and upload it.
HomeBrewerDude
Lifer
- Jan 18, 2001
- 14,465
- 1
- 0
The Prolouge.
You're close, but you forgot to figure that there are multiple ways of getting each type of hand. Your odds are correct for drawing an exact hand, (e.g., joker joker ace ace ace).
However, in the case of 5 of a kind with jokers wild you can:
J J A A A
J A J A A
J A A J A
J A A A J
A J J A A
A J A J A
A J A A J
A A J J A
A A J A J
A A A J J
or Combination (5, choose 3) = 5!/(3!*(5-3)! = (5*4*3*2*1)/(3*2*1)*(2*1) = 5*4/2*1 = 10
So the odds of drawing 2 jokers and 3 cards of the same suit are 10*1/2,432,700 OR 1/243,270
DrPizza
Administrator Elite Member Goat Whisperer
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.
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.
EMPshockwave82
Diamond Member
- Jul 7, 2003
- 3,012
- 2
- 0
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
therefore in my opinion a hand that is possible beats a hand that is "impossible" under normal rules
DrPizza
Administrator Elite Member Goat Whisperer
Well, in my opinion, I'll just add emp to the list of imbeciles who don't even read the first post of a thread before replying.
original post:
HomeBrewerDude
Lifer
- Jan 18, 2001
- 14,465
- 1
- 0
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.
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