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Poker Forums => The Rail => Topic started by: tikay on June 30, 2007, 12:18:44 AM



Title: Help!
Post by: tikay on June 30, 2007, 12:18:44 AM

I need to answer an awkward question for a column I do elsewhere.

On a (full) 6 seater table,what are the odds of 4 players being dealt Pocket Pairs?


Title: Re: Help!
Post by: efcfan on June 30, 2007, 12:29:21 AM
I'll have a stab at about 50,000 to 1.


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 12:46:51 AM
I may be being amazingly thick here but surely it's the same as any other distribution of any pre-flop four hands at a six seater...


Title: Re: Help!
Post by: tikay on June 30, 2007, 12:48:40 AM
I may be being amazingly thick here but surely it's the same as any other distribution of any pre-flop four hands at a six seater...

...which is.......?


Title: Re: Help!
Post by: tikay on June 30, 2007, 12:50:48 AM
I am sure I'm wrong here, but if the chance of being dealt a PP is, say, 16/1, then is the answer 16x16x16x16? Then again, how would that differ - if at all - from a full table of 9 or 10?


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 12:51:12 AM
...which is.......?


Lots

Well I'm just back from the pub but.....progessive distribution of (oh feck lets just deal with 4 hands), would be

2/52 * 2/50 * 2/48 * 2/46...again I'm not sure but I think the pairs bit clouds the question.


Title: Re: Help!
Post by: tikay on June 30, 2007, 12:53:00 AM
...which is.......?


Lots

That's what I thought. What you reckon, a shed-load to 1?


Title: Re: Help!
Post by: efcfan on June 30, 2007, 12:54:46 AM
After the dealer has dealt the first 6 cards, there are 46 left in the pack and only 3 of them will pair you up, so about 15 to 1.

So four to pair up is roughly 15x15x15x14 ?


Title: Re: Help!
Post by: tikay on June 30, 2007, 12:57:21 AM
After the dealer has dealt the first 6 cards, there are 46 left in the pack and only 3 of them will pair you up, so about 15 to 1.

So four to pair up is roughly 15x15x15x14 ?

Thanks Mate.


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 12:57:35 AM
Tikay,
I do think the pairs bit is irrelevant in establishing the odd. James, however, would argue that it is -EVament unless you had position as well.
Cheers,
FishMKK


Title: Re: Help!
Post by: byronkincaid on June 30, 2007, 12:58:24 AM
I'm really bad at maths so I don't even know why I'm posting this but if 16/1 is roughly 6% then is it not

.06*.06*.06*.06*100 = 0.001296%
 
but I can't work out the 6 people v 10 people thing. The odds must be better that it will happen at a 10 seat table right ???


Title: Re: Help!
Post by: tikay on June 30, 2007, 12:59:20 AM
My head hurts.


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 01:00:05 AM
Yeah - I think the Eastbourne FC fan is right..


Title: Re: Help!
Post by: booder on June 30, 2007, 01:01:11 AM
even money on crypto


Title: Re: Help!
Post by: RED-DOG on June 30, 2007, 01:03:31 AM
It's about evens on the Sky Poker Open.


Title: Re: Help!
Post by: efcfan on June 30, 2007, 01:03:51 AM
Yeah - I think the Eastbourne FC fan is right..

Up The Toffees !!!!!!!!


Title: Re: Help!
Post by: Ironside on June 30, 2007, 01:04:41 AM
the answer to the question is Paris


Title: Re: Help!
Post by: KingPoker on June 30, 2007, 01:13:40 AM
Yeah - I think the Eastbourne FC fan is right..

Up The Toffees !!!!!!!!

Good man!!!!!!!


Title: Re: Help!
Post by: bolt pp on June 30, 2007, 01:14:36 AM
you mangled it


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 01:15:57 AM

6 player...12 cards dealt

1st peep 3/50 to hit pair
2nd peep..ditto
3rd peep ditto
4th peep ditto
5th peep ditto
6th peep ditto

(a) So odds to hit four of =  3/50 * 3/50 * 3/50 * 3/50

But you have 6 peeps taking a shot at this so...

(b) Any 4 peeps from 6 = 6x5x4x3/1*2*3*4

Ergo (A)/(B) = ANSWER


Someone get a calculator







Title: Re: Help!
Post by: tikay on June 30, 2007, 01:24:23 AM

6 player...12 cards dealt

1st peep 3/50 to hit pair
2nd peep..ditto
3rd peep ditto
4th peep ditto
5th peep ditto
6th peep ditto

(a) So odds to hit four of =  3/50 * 3/50 * 3/50 * 3/50

But you have 6 peeps taking a shot at this so...

(b) Any 4 peeps from 6 = 6x5x4x3/1*2*3*4

Ergo (A)/(B) = ANSWER


Someone get a calculator







Thank you, someone get me an Anadin!


Title: Re: Help!
Post by: MKKfish on June 30, 2007, 01:25:26 AM
Bollox to my own calcs.......since I've already said it's the same for any other distribution it must be;
2/52 x 2/52 x 2/52 x 2/52

divided by

6 x 5 x 4 x3 / 1 x 2 x 3 x4


Title: Re: Help!
Post by: vegaslover on June 30, 2007, 01:29:27 AM
Yeah - I think the Eastbourne FC fan is right..
Nah, he's missed a b it's eastbourne borough football club


Title: Re: Help!
Post by: tikay on June 30, 2007, 01:31:00 AM
Bollox to my own calcs.......since I've already said it's the same for any other distribution it must be;
2/52 x 2/52 x 2/52 x 2/52

divided by

6 x 5 x 4 x3 / 1 x 2 x 3 x4

15/1? Surely not?


Title: Re: Help!
Post by: bolt pp on June 30, 2007, 01:32:54 AM
15 to the power of 3 FFS!!!!!!


Title: Re: Help!
Post by: tikay on June 30, 2007, 01:36:53 AM
15 to the power of 3 FFS!!!!!!

Which is........?


Title: Re: Help!
Post by: bolt pp on June 30, 2007, 01:37:50 AM
15 to the power of 3 FFS!!!!!!

Which is........?

 :dontask:


Title: Re: Help!
Post by: CelticGeezeer on June 30, 2007, 02:16:42 AM
start\programs\accessories\calculator


Title: Re: Help!
Post by: RobS on June 30, 2007, 03:06:02 AM
I will have a go.....

There are 15 combinations of any 4 from 6. [(6*5)/(1*2)].

Obviously the chance of one person being dealt a pocket pair is 1/17. If it is known that one person has been dealt a pocket pair, then the odds of another person being dealt a pocket pair are not exactly 1/17, too complicated to work out but 1/17 is a good approximation still.

So the approx chance of 4 players being dealt pocket pairs at a 6 handed table are 15 * 1/17 * 1/17 *1/17 * 1/17 = 1 / 5568.

So once in 5,568 deals. Not exactly right but close enough I think....


Title: Re: Help!
Post by: tikay on June 30, 2007, 09:15:18 AM
I will have a go.....

There are 15 combinations of any 4 from 6. [(6*5)/(1*2)].

Obviously the chance of one person being dealt a pocket pair is 1/17. If it is known that one person has been dealt a pocket pair, then the odds of another person being dealt a pocket pair are not exactly 1/17, too complicated to work out but 1/17 is a good approximation still.

So the approx chance of 4 players being dealt pocket pairs at a 6 handed table are 15 * 1/17 * 1/17 *1/17 * 1/17 = 1 / 5568.

So once in 5,568 deals. Not exactly right but close enough I think....

Thanks Rob.


Title: Re: Help!
Post by: bookiebasher on June 30, 2007, 10:40:43 AM
as a bookie i would offer odds of 25-1....any takers ?


Title: Re: Help!
Post by: bookiebasher on June 30, 2007, 10:46:50 AM
Off the top of me head i would say 3/46 x 3/45 x 3/44 x 3/43 x 3/42 x 3/41 = a big number


Title: Re: Help!
Post by: byronkincaid on June 30, 2007, 10:50:45 AM
can someone explain this bit

There are 15 combinations of any 4 from 6. [(6*5)/(1*2)].

thx


Title: Re: Help!
Post by: bookiebasher on June 30, 2007, 10:58:02 AM
as usual i didnt read the question right but ...i did it for 6 pairs.

So for 4 pairs assuming they sit 1234 then 3/46 x 3/45 x 3/44 x 3/43 = 0.0000206    at a rough guess


Title: Re: Help!
Post by: bookiebasher on June 30, 2007, 11:06:20 AM
but if they sit 3456 then its 3/44 x 3/43 x 3/42 x 3/41 = 0.0000229

This assumes that the other two players dont get dealt any of the same cards.

Obviously this can happen so a further calculation is needed...i suggest a pm to thewy

to get the proper odds !!!


Title: Re: Help!
Post by: AndrewT on June 30, 2007, 11:09:36 AM
can someone explain this bit

There are 15 combinations of any 4 from 6. [(6*5)/(1*2)].

thx

You're looking for any four of the players to have pairs out of the six.

One combination is players 1,2,3 & 4. Another is 1,2,3,5, then 1,2,3,6, or 1,2,5,6 etc. There are 15 of these combinations, any of which will fulfill the criteria.

RobS's answer is closest, provided the question is 'at least four players having pocket pairs'. If it's 'exactly four players' then there's additional calculations, but I think it's near enough.


Title: Re: Help!
Post by: bookiebasher on June 30, 2007, 11:30:24 AM
The more i think about it the more complicated it gets.

When i input the data into my settling machine at the office it told me

to bog off. Tony just say a really big number...as you know....no one

believes you anyway....or ban the person who asked the question.

This is going to bug me all day.


Title: Re: Help!
Post by: doubleup on June 30, 2007, 12:39:27 PM
Get a bit of paper and write out the 988,560,661,000,000,000 permutations of 12 cards.  Then count the permutations that meet the required criterion.

ps can anyone say that number in words?

Is it nine hundred and eighty-eight thousand five hundred and sixty billion six hundred and sixty one million?


Title: Re: Help!
Post by: kinboshi on June 30, 2007, 02:25:24 PM
Right, the maths to work this out is beyond me.  But I've found a page that has worked out something similar - and suggests an answer to this question for a 10-handed table.

The page is:  http://www.math.sfu.ca/%7Ealspach/mag88/

I might have misinterpreted it, but from what I can gather, the chance for 4 players to hold pocket pairs can be worked out (as an approximation), based on these figures.  They show that if you hold 22 there is a 0.01186 chance that there will be three pairs larger than yours.  If you play around with that, you could state that it's the same figure for saying there are three other pairs if you hold any pair (I think).  It discounts the chances of someone having the same pair as you, so a more accurate figure would need to take that into account.

So, the chance of holding a pair x chance of three other pairs (in other words 4 people holding pairs at a 10-handed table)

0.059 x 0.01186 = 0.00069974

One deal in 1,429 will produce 4 pockets pairs at a 10-handed table.

Is that complete nonsense?  I'm not sure it takes into account the correct odds for the first person to hold a PP...

:dontask:



Title: Re: Help!
Post by: kinboshi on June 30, 2007, 02:31:35 PM
Get a bit of paper and write out the 988,560,661,000,000,000 permutations of 12 cards.  Then count the permutations that meet the required criterion.

ps can anyone say that number in words?

Is it nine hundred and eighty-eight thousand five hundred and sixty billion six hundred and sixty one million?

Isn't it Nine hundred and eighty eight million, five hundred and sixty thousand, six hundred and sixty one billion?

:dontask:


Title: Re: Help!
Post by: AndrewT on June 30, 2007, 02:34:44 PM
Get a bit of paper and write out the 988,560,661,000,000,000 permutations of 12 cards.  Then count the permutations that meet the required criterion.

ps can anyone say that number in words?

Is it nine hundred and eighty-eight thousand five hundred and sixty billion six hundred and sixty one million?

Isn't it Nine hundred and eighty eight million, five hundred and sixty thousand, six hundred and sixty one billion?

:dontask:

Nine hundred and eighty-eight quadrillion, five hundred and sixty trillion, six hundred and sixty one billion.

Or a bajillion.


Title: Re: Help!
Post by: tikay on June 30, 2007, 02:39:28 PM
Right, the maths to work this out is beyond me.  But I've found a page that has worked out something similar - and suggests an answer to this question for a 10-handed table.

The page is:  http://www.math.sfu.ca/%7Ealspach/mag88/

I might have misinterpreted it, but from what I can gather, the chance for 4 players to hold pocket pairs can be worked out (as an approximation), based on these figures.  They show that if you hold 22 there is a 0.01186 chance that there will be three pairs larger than yours.  If you play around with that, you could state that it's the same figure for saying there are three other pairs if you hold any pair (I think).  It discounts the chances of someone having the same pair as you, so a more accurate figure would need to take that into account.

So, the chance of holding a pair x chance of three other pairs (in other words 4 people holding pairs at a 10-handed table)

0.059 x 0.01186 = 0.00069974

One deal in 1,429 will produce 4 pockets pairs at a 10-handed table.

Is that complete nonsense?  I'm not sure it takes into account the correct odds for the first person to hold a PP...

:dontask:



So, for 4 PP's to be dealt at a 6 handed table, the answer is......?

I'm going with RobS on this. Bookiebashers answer will be over-round by 40%.


Title: Re: Help!
Post by: tikay on June 30, 2007, 02:43:22 PM
start\programs\accessories\calculator

Google is better, easier to access, & fasterer, imo.

Just type the sum into the google search box (e.g., 345 x 678) & you get the answer in a nanasecond.


Title: Re: Help!
Post by: tikay on June 30, 2007, 02:47:13 PM
Get a bit of paper and write out the 988,560,661,000,000,000 permutations of 12 cards.  Then count the permutations that meet the required criterion.

ps can anyone say that number in words?

Is it nine hundred and eighty-eight thousand five hundred and sixty billion six hundred and sixty one million?

Am reading Tony Benn's Diaries - quite splendid they are too, politically polar opposites to Alan Clark's equally good Diaries.

Anyway, Tony Benn is with his Son & Grandson one day.

The 5 year old Grandson asks his Dad, "is 168,000 the biggest number in the world, Dad?"

Dad replies, "well, no, 168,001 is bigger".

""Gosh", says the laddo, "I was very close then, eh?"


Title: Re: Help!
Post by: kinboshi on June 30, 2007, 02:50:43 PM
Right, the maths to work this out is beyond me.  But I've found a page that has worked out something similar - and suggests an answer to this question for a 10-handed table.

The page is:  http://www.math.sfu.ca/%7Ealspach/mag88/

I might have misinterpreted it, but from what I can gather, the chance for 4 players to hold pocket pairs can be worked out (as an approximation), based on these figures.  They show that if you hold 22 there is a 0.01186 chance that there will be three pairs larger than yours.  If you play around with that, you could state that it's the same figure for saying there are three other pairs if you hold any pair (I think).  It discounts the chances of someone having the same pair as you, so a more accurate figure would need to take that into account.

So, the chance of holding a pair x chance of three other pairs (in other words 4 people holding pairs at a 10-handed table)

0.059 x 0.01186 = 0.00069974

One deal in 1,429 will produce 4 pockets pairs at a 10-handed table.

Is that complete nonsense?  I'm not sure it takes into account the correct odds for the first person to hold a PP...

:dontask:



So, for 4 PP's to be dealt at a 6 handed table, the answer is......?


Different.