Be the first
July 4th, 2017 at 12:13:08 PM
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Be the first to answer my question because I've searched everywhere but haven't been able to find an answer. Hopefully the Wizard or one of his goons can help.
I would like to know how many possible combinations there are to get 3 cards to a straight in a 5 card poker hand WITHOUT having 3 of a kind or better.
(Standard 52 card deck, no wilds)
I'd also like to ask the same question but with combos of having 4 cards to a straight. I believe all I need to do is subtract 13 from 101808 (from 4 card poker) but I'm not sure. Please help! Thank you
I would like to know how many possible combinations there are to get 3 cards to a straight in a 5 card poker hand WITHOUT having 3 of a kind or better.
(Standard 52 card deck, no wilds)
I'd also like to ask the same question but with combos of having 4 cards to a straight. I believe all I need to do is subtract 13 from 101808 (from 4 card poker) but I'm not sure. Please help! Thank you
Last edited by: JACK on Jul 4, 2017
July 4th, 2017 at 7:57:18 PM
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Questions:
1. Are you trying to make the best three card poker hand from a 5 card hand? Do you want to know how often you will be able to make a 3-card straight without also having a 3 card flush, 3-of-a-kind, or a three-card straight flush?
2. In the first part of your question, if your 5 card hand can make a 4-card straight does that qualify as a 3-card straight - or must it be 3 -and no more than 3 -consecutive cards in rank?
1. Are you trying to make the best three card poker hand from a 5 card hand? Do you want to know how often you will be able to make a 3-card straight without also having a 3 card flush, 3-of-a-kind, or a three-card straight flush?
2. In the first part of your question, if your 5 card hand can make a 4-card straight does that qualify as a 3-card straight - or must it be 3 -and no more than 3 -consecutive cards in rank?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 4th, 2017 at 8:08:39 PM
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Quote: JACKBe the first to answer my question because I've searched everywhere but haven't been able to find an answer. Hopefully the Wizard or one of his goons can help.
I would like to know how many possible combinations there are to get 3 cards to a straight in a 5 card poker hand WITHOUT having 3 of a kind or better.
(Standard 52 card deck, no wilds)
I'd also like to ask the same question but with combos of having 4 cards to a straight. I believe all I need to do is subtract 13 from 101808 (from 4 card poker) but I'm not sure. Please help! Thank you
Would a hand like 2s, 3h, 5d, Ks, Qc count as "3 to a straight"? You don't mention that they have to be in consecutive order (2,3,4; 6,7,8).
DUHHIIIIIIIII HEARD THAT!
July 4th, 2017 at 8:41:47 PM
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Thank you for responding!
1) the best three card hand doesn't matter nor does a three card flush/straight flush.
2) it must be 3 and no more than 3 consecutive cards in rank.
If I'm dealt 5 cards from a standard deck, what is the probability that I'll have only a 3 card straight? So, no 3 of a kind or better. Making a pair or having two pair along with the 3 card straight is fine. Ex. Ace, 2, 3, 2, 3
As for the 2nd part of my question, I think it can be answered by taking the math from the game 4 card poker on how many straight combinations there are (101808) minus the 5 card straight combinations.
1) the best three card hand doesn't matter nor does a three card flush/straight flush.
2) it must be 3 and no more than 3 consecutive cards in rank.
If I'm dealt 5 cards from a standard deck, what is the probability that I'll have only a 3 card straight? So, no 3 of a kind or better. Making a pair or having two pair along with the 3 card straight is fine. Ex. Ace, 2, 3, 2, 3
As for the 2nd part of my question, I think it can be answered by taking the math from the game 4 card poker on how many straight combinations there are (101808) minus the 5 card straight combinations.
July 5th, 2017 at 8:23:43 AM
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I think this question falls under the heading of "We don't do homework."
And then there's the other thing:
And then there's the other thing:
Goons ???Quote: JACKHopefully the Wizard or one of his goons can help.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
July 5th, 2017 at 9:22:06 AM
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So, you don't know the answer?
July 5th, 2017 at 12:26:40 PM
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I calculate that the probability of getting a 5 card poker hand with 3 (and only 3) consecutive ranks without a 3-of-a-kind is 0.248025 and the total number of combinations is 644,608.
More Detail
I calculate that the AKQ and 32A 'straights' are the most frequent, at 0.024157 each (with the proviso of no 3-of-a-kind.) (62,784 combinations each.)
I calculate that the other 10 3-card straights, from KQJ to 432, each have a frequency of 0.019971. (51,904 combinations each.)
And, btw, I am not "one of the Wizard's goons."
More Detail
I calculate that the AKQ and 32A 'straights' are the most frequent, at 0.024157 each (with the proviso of no 3-of-a-kind.) (62,784 combinations each.)
I calculate that the other 10 3-card straights, from KQJ to 432, each have a frequency of 0.019971. (51,904 combinations each.)
And, btw, I am not "one of the Wizard's goons."
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 5th, 2017 at 12:41:39 PM
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Quote: JACKBe the first to answer my question because I've searched everywhere but haven't been able to find an answer. Hopefully the Wizard or one of his goons can help.
I would like to know how many possible combinations there are to get 3 cards to a straight in a 5 card poker hand WITHOUT having 3 of a kind or better.
(Standard 52 card deck, no wilds)
I'd also like to ask the same question but with combos of having 4 cards to a straight. I believe all I need to do is subtract 13 from 101808 (from 4 card poker) but I'm not sure. Please help! Thank you
Well, we don't know Jack. And if you keep up the name-calling, we never will. Personal insults not tolerated here. Warning. Remains to be seen if you're a Biff - troll or just kind of humorously aggressive.
If the House lost every hand, they wouldn't deal the game.
July 5th, 2017 at 12:50:49 PM
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I calculate that the number of combinations for a 5 card hand with 4 consecutive ranks (but not 5 consecutive ranks) is 114,688, that's equivalent to a probability of 0.044128.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 5th, 2017 at 3:54:04 PM
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I didn't mean to offend anybody and I apologize to you and anyone else that was offended. In my circle of friends, goon is just another way to say friend.
July 5th, 2017 at 4:07:05 PM
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Gordonm888 you're the man! Thank you so much!
If you have the time, could you show me how you got those numbers? If not, no big deal. Thanks again!
If you have the time, could you show me how you got those numbers? If not, no big deal. Thanks again!
July 5th, 2017 at 7:12:39 PM
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I'll explain the 3-card straight. It was necessary to break the 3-card straights up into two groups:
1) AKQ, A23 (2 straights -gut shots to make a 4-card straight)
2) KQJ ... 432 (10 straights -open-ended to make a 4-card straight)
For group 1, let's do AKQ
AKQ = (1 combo) * 3^4. For the other two cards in the hand, they are defined to not be a Jack (which would turn it into a four card straight) so that is combin(45,2). The 45 is 52 - 3(cards in your hand) - 4 (Jacks). The last two cards also cannot be AA or KK or QQ (because you said no trips), and there are three ways left to make each of those three pairs. So
AKQ = 1*(3^4)*(combin(45,2)-3*3) =62,784.
With 32A, the remaining 2 cards cannot be a 4, so we use the same formula for another 62,784 combinations, so group 1 is 125,568 combinations
Group 2. These straights are the same formula except the last two cards cannot be from either of the two ranks that surround the straight, i.e., for T98, the last two cards cannot be a J or 7, so the formula for each of these straights is 1*(3^4)*(combin(41,2)-3*3), multiply that by 9 (for 9 straights) =519,040 combinations.
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For the four card straights, its the same deal -divide them up into two groups,
Group 1: AKQJ and 432A
Group 2: The other 9 straights: KQJT ... 5432
And there are no 3-of-a-kinds to worry about so the total number of combinations is:
2*(4^4)*44+9*(4^4)*40=114,688.
Your original approach to the four card straight calculation was not right, probably because the number of combinations you were starting with excluded all 4-card (or 5-card) flushes, and you had specified that you did not want flushes and straight flushes to be excluded from the number of combinations.
1) AKQ, A23 (2 straights -gut shots to make a 4-card straight)
2) KQJ ... 432 (10 straights -open-ended to make a 4-card straight)
For group 1, let's do AKQ
AKQ = (1 combo) * 3^4. For the other two cards in the hand, they are defined to not be a Jack (which would turn it into a four card straight) so that is combin(45,2). The 45 is 52 - 3(cards in your hand) - 4 (Jacks). The last two cards also cannot be AA or KK or QQ (because you said no trips), and there are three ways left to make each of those three pairs. So
AKQ = 1*(3^4)*(combin(45,2)-3*3) =62,784.
With 32A, the remaining 2 cards cannot be a 4, so we use the same formula for another 62,784 combinations, so group 1 is 125,568 combinations
Group 2. These straights are the same formula except the last two cards cannot be from either of the two ranks that surround the straight, i.e., for T98, the last two cards cannot be a J or 7, so the formula for each of these straights is 1*(3^4)*(combin(41,2)-3*3), multiply that by 9 (for 9 straights) =519,040 combinations.
*********************
For the four card straights, its the same deal -divide them up into two groups,
Group 1: AKQJ and 432A
Group 2: The other 9 straights: KQJT ... 5432
And there are no 3-of-a-kinds to worry about so the total number of combinations is:
2*(4^4)*44+9*(4^4)*40=114,688.
Your original approach to the four card straight calculation was not right, probably because the number of combinations you were starting with excluded all 4-card (or 5-card) flushes, and you had specified that you did not want flushes and straight flushes to be excluded from the number of combinations.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 5th, 2017 at 7:39:26 PM
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Quote: JACKI didn't mean to offend anybody and I apologize to you and anyone else that was offended. In my circle of friends, goon is just another way to say friend.
All good, Jack, and welcome to the board.
If the House lost every hand, they wouldn't deal the game.
