Hello, was hoping to get some help on this.

Suppose you hold 2347X in 2-7 single draw where you would draw 1 if X is a pair or if it is a q, k, or ace. How many combinations are there of this holding?

What if X is a pair or a 9 or higher?

Why don't you count them?

Not sure what you mean exactly but there are thousands of every particular hand

I think I know what the answer is because Buzz helped me before but this material is going into articles and/or books so I’m looking to confirm my numbers

Basically I’m looking for two numbers of 2347X under the two conditions above out of 2,598,960 possible

I believe there are 7680 combinations of 2347X where X is 9 or greater or a pair

Want to confirm that number and adjust if X is a queen or greater or pair, would be surprised if it’s a division by two of above number...

These calculations are not difficult.

Let's walk through the scenario of 2347X where X is 9-A or a pair.

Case 1: X is not a pair
= 6*(4^5)
= 6,144

Case 2: X is a pair
= C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1)
= 1,536

TOTAL = 6,144 + 1,536 = 7,680

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Now let's walk through the scenario of 2347X where X is Q-A or a pair.

Case 1: X is not a pair
= 3*(4^5)
= 3,072

Case 2: X is a pair
= C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1)
= 1,536

TOTAL = 3,072 + 1,536 = 4,608

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I think these are correct but someone should confirm/deny.

Awesome thanks man!

Looks right to me as I match that first number and thought the second one would be a little greater than half.

Whosnext

Ok, one follow-up question, say you hold 237T under the second constraint where X is a Q or greater or a pair so there are 4608 combos

How many of those combos are double suited? How many of them have specifically two of a suit where the other two cards have two other suits e.g Td7d3h2c?

This has relevance for 2-7 Drawmaha because a 1cd to a ten may be a marginal open from a certain position so a proper strategy may entail playing it if it is double suited or has specifically two of a suit.

It is distressing that you are not using common sense to realize that the number is not cut in half.

Why portray false modesty for such an easy question? Reminds me of when Newton said he stood on the shoulders of giants.

I find it distressing that you feel the need to be so condescending

And I did say I doubted the number is cut off in half

But sometimes in combinatorics something can easily be missed. At least I think so but hey I apologize for wasting people’s time I guess

Assume 237TX where X is either a pair or Q-A.

1. Yes, as above there are 4,608 combos.

2. How many combos have double-suited 237T?

Case 2A: X is not a pair

= 3*C(4,2)*C(4,2)*C(4,1)
= 432

where the terms are the number of ranks for the 5th card, which 2 suits appear in the 4 key cards, which 2 ranks are in the first suit, and which suit is the 5th card.

Case 2B: X is a pair

= C(4,1)*C(4,2)*C(2,1)*C(3,2)*C(1,1)*C(3,1)
= 432

where the terms are which rank is paired, which 2 suits form the double-suits in the 4 key cards, which suit has the paired card, which 2 singletons have the same suit, which suit is the other singleton, and which suit is the other paired card.

Total = 432 + 432 =864

3. How many combos can form a 237T 2-1-1 suit count (excluding double-suited combos)?

Case 3A: X is not a pair

= 3*C(4,3)*C(3,1)*C(4,2)*C(2,1)*C(4,1)
= 1,728

where the terms are rank of 5th card, which 3 suits appear in the 4 key cards, which suit has 2 of the key cards, which ranks are in the 2-suit, which rank is of the first 1-suit, and which suit of 5th card.

Case 3B: X is a pair

Case 3B1: Suits are 3-1-1 with pair being of 3-suit

= C(4,3)*C(3,1)*C(4,3)*C(3,1)*C(2,1)*C(1,1)
= 288

where the terms are which 3 suits appear, which is the 3-suit, which 3 ranks are of 3-suit, which rank is pair, which other suit is other paired card, and remaining rank in remaining suit.

Case 3B2: Suits are 2-1-1-1 with both pair cards being of 1-suits

= C(4,1)*C(4,2)*C(3,2)*C(2,1)*C(1,1)
= 144

where the terms are which is the 2-suit, which ranks are in 2-suit, which suits are pair cards, which other rank is paired card, and remaining rank in remaining suit.

Case 3B3: Suits are 2-1-1-1 with one of paired cards being in 2-suit

= C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(2,1)*C(1,1)
= 288

where the terms are which is 2-suit, which rank is paired, which other suit is other paired card, which other rank is in 2-suit, which other rank is in 1st other suit, remaining rank in remaining suit.

Case 3B4: Suits are 2-2-1 with both pair cards in a different 2-suit

= C(4,3)*C(3,2)*C(4,1)*C(3,1)*C(2,1)*C(1,1)
= 288

where the terms are which 3 suits appear, which 2 suits are the 2-suits, which rank is paired, which other rank is in 1st 2-suit, which other rank is in 2nd 2-suit, remaining rank in remaining suit.

Subtotal for Case 3B = 288 + 144 + 288 + 288 = 1,008

Total = 1,728 + 1,008 = 2,736


Note: I have derived these in the way that makes sense for me. Clearly there are other ways to derive these figures and undoubtedly some of the other ways may be faster or more efficient.

You are top notch whosnext! Thanks!

It is common sense that 9 t j plus pair is more than half of 9 t j q k a plus pair. It has nothing to do with combinatorics. The "distressing" part is that you, along with 90% of the population try to do math through algorithms rather than through thought. At least the American population.

Doubling down on being a jerk for no reason I see. And in society I think dipchits are more distressing then those that may not see things as clearly as you do regarding math.

Your EQ is at bottom of the barrel levels

But sue me, I was only thinking about the 9tj vs qka as I had forgotten about the pair. I felt like I was forgetting something stated as much and whosnext helped me out.

Just stopped by to apologize, David

I got a little personal while you were mostly commenting on people in general

Funny enough I am a former actuary but some things with combinatorics twists me up sometimes but assuredly I do try and think through things

Because my father taught me math at such a young age he was forced to use simple logic rather than any sort of rules. So maybe I am not sympathetic enough towards all those students who stick to algorithms that bad teachers taught them rather than resorting to common sense when it would make things easy.

That will all start to become rectified after I publish my book Algebra For Ten Year Olds (And Maybe You) next year. I will send you a copy in thanks for the apology I often deserve but rarely get.

Thanks DS, that’s kind of you gent

I have never taken a math class that wasn't taught this way, through differential equations. I do not know if this changes eventually or is dependent on the professor, but I have given up on college education because of it. I feel I am better off self-taught.