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A poker riddle!!!

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Lukie

01-04-2008, 01:49 AM Post subject: A poker riddle!!!

#1 (permalink)

4-of-a-Kind

Join Date: Jul 2005

Location: back with a vengeance

Posts: 3,307

advice: hit back button before your head asplodes

We've all seen the riddles when we were kids about how if you add and subtract x amount of dollars, you come up short by a dollar or so at the end. I took that brilliant idea and applied it to card combos!

note: scroll down 3 or 4 paragraphs if u already know what i'm talking about.

There are 1326 distinct (meaning suits are of importance) starting hands in hold'em. We take 52 (any card) x 51 (any remaining card after the first is picked) = 2652. Then we divide by 2 because order does not matter. 7d 7s is, for all practical intents and purposes, the same hand as 7s 7d.

We can do the same thing for a specific hand. For example, for all combos of AK (suited and unsuited), we have 8 cards to choose from, then 4 to choose from = 32, divide by 2 since order does not matter. We come up with 16 combos of AK.

If we take this a step further, if we want to find out how many AKo combos there are, we can take any one of the 8 A's or K's, and there will be 3 of the offsuit matching variety. 8x3/2 = 12. It should be simple to figure out that there are only 4 combos of AKs, but the maths are very simple... any A or K can make an AKs combo, but only the matching A or K suited will complete it... so 8x1/2 = 4.

NOW, say we hold exactly Ac Kd, and we are trying to count up all of villain's AK combos. It should be obvious that we have to take away all of the AK combos that contain either Ac or Kd. This is not much of a big deal. There are 4 combos that contain the Ac, so we have to knock all of those out. There are also 4 combos that contain the Kd, so we have to knock those out. There is one combo that contains both, but we don't want to count it twice, so in total we knock out 7 of his possible combos.

So know we want to figure out all of his. He has 6 remaining A's and K's to choose from. First we want to figure out all of his AKo combos-- any of those 6, multiplied by 2 (since the 3rd one would make an AKs combo), then divided by 2 since order does not matter = 6. The suited combos are easy to figure out since we have already knocked out AcKc and AdKd by holding one of each suit.

So we knocked out 7 combos, and villain has 6 possible combos of AKo and 2 possible combos of AKs. WHERE DID THE LAST CARD COMBO GO??????

head asplode

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Lukie

01-04-2008, 02:15 AM

#2 (permalink)

4-of-a-Kind

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picture of head asploding, probably nsfw

01-04-2008, 02:33 AM

#3 (permalink)

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EzDuzIt

01-04-2008, 02:49 AM

#4 (permalink)

Full House

Join Date: Mar 2007

Location: Green Bay

Posts: 915

villan has 7 offsuit combos. i kind of cheated by just bringin up the poker stove hand picker thing.

im pretty sure it has something to do with this being wrong:

"So know we want to figure out all of his. He has 6 remaining A's and K's to choose from. First we want to figure out all of his AKo combos-- any of those 6, multiplied by 2 (since the 3rd one would make an AKs combo), then divided by 2 since order does not matter = 6. The suited combos are easy to figure out since we have already knocked out AcKc and AdKd by holding one of each suit. "

like the AdKd was already dropped out before so there are actually 3 ace of diamonds offsuit combos that the villan can have too. in the other ace of a different suit rows the above quote works.

|~|ypermegachi

01-04-2008, 02:52 AM

#5 (permalink)

4-of-a-Kind

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we're holding it.

bigspenda73

01-04-2008, 03:17 AM

#6 (permalink)

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bigspenda73 will become famous soon enough

HYPER WINS

gabe

01-04-2008, 04:01 AM

#7 (permalink)

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gabe is an unknown quantity at this point

is this a quiz or can you just not figure itout

IowaSkinsFan

01-04-2008, 05:03 PM

#8 (permalink)

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IowaSkinsFan will become famous soon enough

I think the ladder

Check out the new blog!!!

jyms

01-04-2008, 05:54 PM

#9 (permalink)

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Posts: 4,837

6 remaining cards times 3 is 18 divided by 2 = 9.
7 offsuit and 2 suited.

this is your mistake
any of those 6, multiplied by 2 (since the 3rd one would make an AKs combo) The first two make a AKs combo, the third one does not

suited

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Warpe

01-04-2008, 06:16 PM

#10 (permalink)

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Quote:

Originally Posted by IowaSkinsFan

I think the ladder

you think, therefore you climb?

Numbr2intheWorld

01-04-2008, 07:42 PM

#11 (permalink)

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I think you're brain needs to be rebuilt danny

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Lukie

01-05-2008, 12:25 AM

#12 (permalink)

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I couldn't figure it out by myself at first, i was up all night thinking about it thats how much it bothered me. In the morning I made a little grid of the AK card combos like so:

AcKc AcKd AcKh AcKs
AdKc AdKd .. ..
AhKc AhKd .. ..
AsKc AsKd .. ..

Then I figured it out.

The reason I posted it anyway is because I think really analyzing exactly how many combos are hitting each specific boards (and how strong), and how many combos are in villain's entire range can be supremely beneficial. I really believe this kind of precise mathematical breakdown is what sets the very very best shorthanded NL players apart from the rest. I actually remember breaking down these situations about a year ago and I felt my game really improved because of it and I had a monster December ('06). Then I got away from it and I kinda hit a wall ever since. Pretty soon, I'm going to have this whole game solved once I start practicing with all this stuff...

noble007

01-05-2008, 07:51 PM

#13 (permalink)

Flush

Join Date: Jun 2006

Location: London

Posts: 531

To find out the number of unique combinations of AK just multiply the amount of available aces by the amount of available kings.

1)There are 4 aces and 4 kings in the deck. 4x4 = 16
(therefore there are 16 combinations of AK possible)

If you hold an AK. There are 3 aces and 3 kings left in deck. 3x3 = 9
(therefore there are 9 combinations of AK still possible)

2) To find out suited combinations of AK. There are 4 Aces in the deck
but they can only be matched with one king. (the one of the same suit)
4x1=4 (therefore there are 4 suited combination of AK)

eg's.

- You have AhQs how many AQs are possible ?
3 aces are left but only 2 of them can be matched with a Queen of
the same suit. (so the answer is clearly 2 - without doing any maths)

- You have KhJh the flop is JdTd4d

1) How many KJ combinations are left? How many AJ combinations?
2) How many of them have tp and a flush draw?

1)Well there are 3kings left in the deck and 2 jacks. 3x2 =6
Therefore there are 6 combinations of KJ villian could hold.
There are 4aces left in the deck and 2 jacks. 4x2 = 8
Therefore there are 8 combinations of AJ villian could hold.

2) With KJ he has to hold Kd & there are 2 remaining Jacks. 1x2 =2
Therefore there are 2 combinations of KJ with a flush draw (KdJs;KdJc)
With AJ he has to hold Ad & there are 2 remaining Jacks. 1x2 = 2
Therefore there are also only 2 combinations of AJ with a flush draw
(AdJs; AdJc)

I dont know Im sure you can already do this Lukie and co it just seemed you were getting to the answer in a longer way.

Currently thinking of a new quote/signature... Some sort of prayer to the Poker gods for enlightment etc..

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