How many times does Kings run into Aces pre flop

[RioRodent]

right basically P is probability , C is combination"s/combination

There we go 2 out of 4 "simply answered.  

(50,2) is 1225 the number of possible hand combinations

My question here was what does the notation (x,y) mean, not what is the actual result of (50,2).

i hope you see that its not easy to just give a simple answer...but i hope that answers it better  

As I hope you now realise, I wasn"t asking you to explain any advanced mathematical principles... just to clarify what the elements of the formulae are and why the structure of the 1st formula is different to the 2nd two?

[kinboshi]

Can"t say I find probability equations interesting.

However at the Virgin Festival at Dusk Till Dawn in 2008 there where 11 recorded AA v KK confrontations, in the first day. AA was winning 6 to 5, now whats the odds of that?

First bloke knocked out had AA v KK.

He then joined my cash table and told us all about it (in good humour, not angry or anything). 

Then 10 minutes later we get it all-in pre with my  kd kh against his  ac :as:.

:: qd td

He took that quite well to be fair .

[Swinebag]

Can"t say I find probability equations interesting.

However at the Virgin Festival at Dusk Till Dawn in 2008 there where 11 recorded AA v KK confrontations, in the first day. AA was winning 6 to 5, now whats the odds of that?

First bloke knocked out had AA v KK.

He then joined my cash table and told us all about it (in good humour, not angry or anything). 

Then 10 minutes later we get it all-in pre with my  kd kh against his  ac :as:.

:: qd td

He took that quite well to be fair .

That still didn"t stop you whinging about Simon Trumper calling the action in the tourney!!

[kinboshi]

Can"t say I find probability equations interesting.

However at the Virgin Festival at Dusk Till Dawn in 2008 there where 11 recorded AA v KK confrontations, in the first day. AA was winning 6 to 5, now whats the odds of that?

First bloke knocked out had AA v KK.

He then joined my cash table and told us all about it (in good humour, not angry or anything). 

Then 10 minutes later we get it all-in pre with my  kd kh against his  ac :as:.

:: qd td

He took that quite well to be fair .

That still didn"t stop you whinging about Simon Trumper calling the action in the tourney!!

He does it ALL the time.  It gets soooo annoying.  Too loud sir, too loud.

[coprey]

Can"t say I find probability equations interesting.

However at the Virgin Festival at Dusk Till Dawn in 2008 there where 11 recorded AA v KK confrontations, in the first day. AA was winning 6 to 5, now whats the odds of that?

I think thats a routine A level stats question there col. I"ll have a go...but my expertise in teaching maths doesn"t extend to A level stats.

I got 0.039 (3 dps) this is roughly a 1 in 25 chance

I thought it would be much higher so will ask a colleague

I was correct (of course)

to work it out....assuming that the probability of Aces beating Kings is 80% (0.8 ) and therefore kings beating aces is 20% (0.2)

the calculation is

(11 x 10 x 9 x 8 x 7)/(1 x 2 x 3 x 4 x 5) to work out the number of different combinations where Aces could beat kings 6 times in 11 races. This gives 462.

multiply this by the probability of 6 wins and 5 losses..so

462 x (0.8 x 0.8 x 0.8 x 0.8 x 0.8 x 0.8 ) x (0.2 x 0.2 x 0.2 x 0.2 x 0.2)

=0.03875536896

Thanks. Teachers really do have too much time on their hands.

Questions:

0.03875536896 does this equate to 3.86%?

How do you arrive at the following calculation:

(11 x 10 x 9 x 8 x 7)/(1 x 2 x 3 x 4 x 5)

Thanks again. I remember maths used to be fun, but I guess ive become lazy.

[noble1]

probability forum please  

[Swinebag]

0.03875536896 does this equate to 3.86%?

its actually 3.88 % (to 2 d.p.)

How do you arrive at the following calculation:

(11 x 10 x 9 x 8 x 7)/(1 x 2 x 3 x 4 x 5)

It is using the formula for working out how many combination there are from choosing 6 from 11 (which is identical to choosing 5 from 11)

the formula is

number of combinations choosing r (aces winning 6 times) from n (11 races) =
n!/(n-r)! r!

the "!" notation is factorial, e.g. 4! = 1 x 2 x 3 x 4,        7! = 1 x 2 x 3 x 4 x 5 x 6 x 7

the above formula cancels down to give the calculation you quoted. We need to use it because in the case of the Aces winning 6 out of 11 races against Kings, there is no specific race that the aces have to win. For example, Aces could have held up the first 6 times, or the last 6 times or they could have held up in the 1st, 2nd, 4th, 8th, 9th and 10th  race. In fact there are 462 different ways that Aces can beat kings 6 out of 11 times.

An simpler example of this calculation in action is used in lucky 15 or lucky 31 bets at the bookies. If you pick 5 horses in a lucky 31 there are 10 possible doubles, namely

1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5

Using the formula you get 10 by using n=5 and r =2 and this simplifies to (5 x 4)/(1 x 2)

bring your maths exercise book with you to Walsall colin, and you can do a few examples when you are card dead....I"ll mark it in the break

[noble1]

if you working out doubles rob say like in your example of 5 selections you could also work it out in your head by adding 4,3,2,1 =10 doubles

4 selections = 3+2+1=6 doubles
7 selection = 6+5+4+3+2+1=21 doubles

would that be ok?

[Swinebag]

if you working out doubles rob say like in your example of 5 selections you could also work it out in your head by adding 4,3,2,1 =10 doubles

4 selections = 3+2+1=6 doubles
7 selection = 6+5+4+3+2+1=21 doubles

would that be ok?

for doubles only, yes.

[noble1]

can i have £5 e/w on your next 5 selections please?  

[coprey]