Help!

[bookiebasher]

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

[bookiebasher]

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

[byronkincaid]

can someone explain this bit

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

thx

[bookiebasher]

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

[bookiebasher]

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 !!!

[AndrewT]

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.

[bookiebasher]

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.

[doubleup]

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?

[kinboshi]

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...

[kinboshi]

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?

[AndrewT]

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?

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

Or a bajillion.

[tikay]

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...

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%.

[tikay]

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.

[tikay]

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?"

[kinboshi]

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...

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

Different.