blonde poker forum

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?

I'll have a stab at about 50,000 to 1.

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.......?

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?

...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.

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

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.

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

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 ???

My head hurts.

Yeah - I think the Eastbourne FC fan is right..

even money on crypto

It's about evens on the Sky Poker Open.

Up The Toffees !!!!!!!!

the answer to the question is Paris

Good man!!!!!!!

you mangled it

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!

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

Nah, he's missed a b it's eastbourne borough football club

15/1? Surely not?

15 to the power of 3 FFS!!!!!!

Which is........?

 :dontask:

start\programs\accessories\calculator

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.

as a bookie i would offer odds of 25-1....any takers ?

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

can someone explain this bit

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

thx

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

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 !!!

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.

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.

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?

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:

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.

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%.

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.

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?"

Different.