CharmedQuark
CharmedQuark
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October 27th, 2016 at 12:15:10 PM permalink
Standard deck of 52 cards with any 4 different ranks of the 4 cards referred to as group 1 - ex: 4 threes 4 kings 4 nines or 4 aces (16 cards) and the remaining 36 cards.
.
You are dealt a five card hand as follows: The first two cards are dealt in any order - one card is from group 1 and the other card is any of the remaining 51 cards other than a card that pairs the card from group 1. The deal of the remaining 3 cards is irrelevant.

How many 5 card combinations can be made?

What is the math for determining the result?
someone
someone
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October 27th, 2016 at 2:12:35 PM permalink
Assuming suites matter.
Number of first pair is 16*48=768
Number of final 3 if Choose(50,3) = 196,000
Total Combinations is 768*196,000 = 15,052,800
ThatDonGuy
ThatDonGuy
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October 27th, 2016 at 2:17:56 PM permalink
Quote: someone

Assuming suites matter.
Number of first pair is 16*48=768
Number of final 3 if Choose(50,3) = 196,000
Total Combinations is 768*196,000 = 15,052,800


Doesn't that total include duplicates?

Example: suppose the "group 1 ranks" are A, K, Q, J
The first two cards are Ace and King, both spades
The next three cards are Queen, Jack, and 9, all spades
The five-card combination is AKQJ9 of spades
However, if the first card is the King and the second card is the Ace, this is also AKQJ9 of spades, but it is counted differently in your total.
(Besides, if order does not matter, there are only 2,598,960 different five-card hands.)
Also, QJ, then AK9 is also AKQJ9, as are a number of other sets where the first card is any of the A,K,Q,J.
PeeMcGee
PeeMcGee
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October 27th, 2016 at 3:09:38 PM permalink
16*48*50*49*48 / 5! = 752,640

Edit: I may have been jumped the gun with this. Thinking about it a little more, I believe it might be incorrect.

So…in this scenario, it seems that the only hands that are not possible are five cards hands without a card from group 1. Therefore, subtract this number from the number of all possible five cards hands. Hopefully this is the correct assumption.

c(52,5) – c(36,5) = 2,221,968
Last edited by: PeeMcGee on Oct 27, 2016
CharmedQuark
CharmedQuark
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October 27th, 2016 at 8:04:04 PM permalink
OP here - - I'm just trying to get a better grip on determining combinations given certain rules. Here suits and order do not matter, just the first two cards (in any order, one from group 1 and the other from the remaining 51 cards that does not pair up) and three random cards. The resulting hand is unimportant.

I came up with two solutions which I do not believe are accurate:

16c1 * 51c1 * 50c3 - 6 giving 15,993,594

52c2 * 50c3 -6 giving 25,989,594

The -6 is to remove the pair combinations on the first two cards. I'm not sure that is correct logic or math.

So thanks for replying.
CharmedQuark
CharmedQuark
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October 27th, 2016 at 9:43:33 PM permalink
Further thought - "someone", I tend to agree with your solution. 15,052,800 combinations.

16c1 * 48c1* 50c3 giving 15,052,800

Makes sense - pick one card from 16 removes one rank or 4 cards from 52 then one card from 48 for the other card and the remaining 3 cards from 50. This solution indicates order for the first two cards but the math is commutative so either 16c1 * 48c1 or 48c1 * 16c1 - - logically I think 16c1 * 48c1 is more proper. I hope this is an accurate analysis.
someone
someone
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October 28th, 2016 at 3:50:03 AM permalink
My answer was assuming that cards being in the first group were different to having cards in the second group (ie something like hole cards and flop cards), and that suites mattered. Unfortunately, as pointed out by Don, I had a mistake in that I counted AH,KH in the first pair as different to KH AH. To correct this I think the equation is
16*(36+12/2)* 50C3 = 13,171,200
PeeMcGee
PeeMcGee
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October 28th, 2016 at 2:34:31 PM permalink
OP…do you want to distinguish from hands such as AKJQ9 and Q9AKJ?
CharmedQuark
CharmedQuark
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October 28th, 2016 at 7:54:06 PM permalink
Quote: PeeMcGee

OP…do you want to distinguish from hands such as AKJQ9 and Q9AKJ?



PeeM - I'm not sure - i don't think so. The idea was to determine the total number of combinations given the rule.
Think of it this way - hypothetically - you are playing 5 card stud with two hole cards and you fold your hand with these two cards after posting the ante bet. Similar to Mississippi Stud folding the first tow cards. The 'wiz' calculates 48,451,200 for this outcome using his optimum strategy.

Someone - - That might be correct.

BTW - I do not know the answer or combination selection.
PeeMcGee
PeeMcGee
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October 29th, 2016 at 10:49:58 AM permalink
Quote: CharmedQuark

PeeM - I'm not sure - i don't think so. The idea was to determine the total number of combinations given the rule.
Think of it this way - hypothetically - you are playing 5 card stud with two hole cards and you fold your hand with these two cards after posting the ante bet. Similar to Mississippi Stud folding the first tow cards. The 'wiz' calculates 48,451,200 for this outcome using his optimum strategy.

Someone - - That might be correct.

BTW - I do not know the answer or combination selection.


The calculation for Mississippi Stud assumes order matters for the three community cards because the player must make a decision after each street.

But if you ask how different many five cards hands are there in Mississippi Stud—well…that’s all of them (2,598,960). This is the question I initially assumed you were asking.

Now, I think you want to distinguish between the hole cards and the community cards, however order does not matter within these two groups.

So basically…how many two hole cards combinations contain at least one of four ranks but is not a pair?
c(52,2) - c(36,2) – 4*c(4,2) = 672

Now if you want to deal the next three cards all at once (so order doesn’t matter):
672 * c(50,3) = 13,171,200

Which is the number someone gives.

And the total number of combinations for such a game would be:
c(52,2) * c(50,3) = 25,989,600

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