Amateur Poker Association & Tour
Poker Forum => Strategy => Topic started by: LongshanksED on January 26, 2009, 20:26:18 PM
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i"ve just had KK run into AA 3 times in the final table of a 30 seat SnG
went into final table with chip lead of 10k
2 orbits later and im down to 560 with 400 BB
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Isn"t it something like 22/1 that AA will face KK (or the other way round of course!) on a full table?
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so for it to happen 3 times on the bounce would be...
22 x 22 x 22 ????
anyway as duncan bannatyne says - im out and on the bubble too
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Pre-flop AA versus KK is 0.004% or 1 chance in 22,560 heads up play
KK vs. AA [9 handed]
[ 6*8*P(48,14)/2^7 - 6*C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ] = 25.6-to-1
KK vs. AA [10 handed]
9*6/C(50,2) - C(9,2)/C(50,4) = 21.8-to-1
KK vs AA [6 handed]
5*6/C(50,2) - C(5,2)/C(50,4) = 39.9-to-1
ALL GONADS ;D if your on a tight table and you get 3bet 4bet and it comes round to you it is an easy fold ;D
check poker tracker if you have it and if KK is running 80% plus from a decent sample of hands say 30,000 then your running normal...............
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each time i was the opening raise and only had one caller each time and never re raised
to even it out though i cracked aces with 10-10 on the button of another 20 man SnG so what comes around evens it
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Pre-flop AA versus KK is 0.004% or 1 chance in 22,560 heads up play
KK vs. AA [9 handed]
[ 6*8*P(48,14)/2^7 - 6*C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ] = 25.6-to-1
KK vs. AA [10 handed]
9*6/C(50,2) - C(9,2)/C(50,4) = 21.8-to-1
KK vs AA [6 handed]
5*6/C(50,2) - C(5,2)/C(50,4) = 39.9-to-1
ALL GONADS ;D if your on a tight table and you get 3bet 4bet and it comes round to you it is an easy fold ;D
check poker tracker if you have it and if KK is running 80% plus from a decent sample of hands say 30,000 then your running normal...............
I"m probably going to regret asking this, but...
- Why is the equation for 9 handed a different format than 10 & 6 handed?
- What is P?
- What is C?
- What does (50,2) mean?
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I"ve not done probability calculations like that since school, I can"t remember how it all works. Out of interest though, have you calculated the odds that:
Someone has AA AND someone has KK (I think this is what you"ve done)
or
Someone has AA GIVEN someone has KK (this will be almost exactly the same as the odds that someone is dealt KK regardless of anyone elses hand).
Just curious
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Pre-flop AA versus KK is 0.004% or 1 chance in 22,560 heads up play
KK vs. AA [9 handed]
[ 6*8*P(48,14)/2^7 - 6*C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ] = 25.6-to-1
KK vs. AA [10 handed]
9*6/C(50,2) - C(9,2)/C(50,4) = 21.8-to-1
KK vs AA [6 handed]
5*6/C(50,2) - C(5,2)/C(50,4) = 39.9-to-1
ALL GONADS ;D if your on a tight table and you get 3bet 4bet and it comes round to you it is an easy fold ;D
check poker tracker if you have it and if KK is running 80% plus from a decent sample of hands say 30,000 then your running normal...............
I"m probably going to regret asking this, but...
- Why is the equation for 9 handed a different format than 10 & 6 handed?
- What is P?
- What is C?
- What does (50,2) mean?
http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
http://en.wikipedia.org/wiki/Poker_probability_(Texas_hold_%27em)
http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&view=&sb=5&o=&fpart=
these are the chances of someone having AA when you have KK
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Pre-flop AA versus KK is 0.004% or 1 chance in 22,560 heads up play
KK vs. AA [9 handed]
[ 6*8*P(48,14)/2^7 - 6*C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ] = 25.6-to-1
KK vs. AA [10 handed]
9*6/C(50,2) - C(9,2)/C(50,4) = 21.8-to-1
KK vs AA [6 handed]
5*6/C(50,2) - C(5,2)/C(50,4) = 39.9-to-1
ALL GONADS ;D if your on a tight table and you get 3bet 4bet and it comes round to you it is an easy fold ;D
check poker tracker if you have it and if KK is running 80% plus from a decent sample of hands say 30,000 then your running normal...............
I"m probably going to regret asking this, but...
- Why is the equation for 9 handed a different format than 10 & 6 handed?
- What is P?
- What is C?
- What does (50,2) mean?
http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
http://en.wikipedia.org/wiki/Poker_probability_(Texas_hold_%27em)
http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&view=&sb=5&o=&fpart=
these are the chances of someone having AA when you have KK
I see... you copied it from somewhere else and you don"t actually understand it either!
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if thats what you think ;D thats ok lol lol i take it you couldn"t be arsed to read them
;D ;D
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In answer to the OP: Every feckin" time I get KK, that"s how many times :D :D :D
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if thats what you think ;D thats ok lol lol i take it you couldn"t be arsed to read them
;D ;D
Well, I started to... from the first link - I didn"t find P, C appears to indicate a "Complex Number" (but that whooshed me) and I didn"t find any explanation of (50,2).
The second link goes to an "error/holding page".
I started to read the 3rd link but didn"t find that that answered my questions either.
But, if you do actually understand what it all means but can"t be arsed to give a straight answer to 4 simple questions... well that"s OK.
I don"t dispute the results of the equations and I don"t actually need to understand the formulae... was just curious is all.
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if thats what you think ;D thats ok lol lol i take it you couldn"t be arsed to read them
;D ;D
Well, I started to... from the first link - I didn"t find P, C appears to indicate a "Complex Number" (but that whooshed me) and I didn"t find any explanation of (50,2).
The second link goes to an "error/holding page".
I started to read the 3rd link but didn"t find that that answered my questions either.
But, if you do actually understand what it all means but can"t be arsed to give a straight answer to 4 simple questions... well that"s OK.
I don"t dispute the results of the equations and I don"t actually need to understand the formulae... was just curious is all.
lol its not easy explaining probability Rio i hoped BruceZ article would as it is pretty
confusing at first...
right basically P is probability , C is combination"s/combination , (50,2) is 1225 the number of possible hand combinations http://mathworld.wolfram.com/Combination.html but also you start getting into permutations http://mathworld.wolfram.com/Permutation.html
when dealing with combinations and number of people/total cards involved....
But, if you do actually understand what it all means but can"t be arsed to give a straight answer to 4 simple questions... well that"s OK.
i hope you see that its not easy to just give a simple answer...but i hope that answers it better :)
if anyone asks what are the chances of QQ facing AA KK 10 handed its 10.8to1 :P
before you ask....
you can find charts like on wizard of odds to find out all sorts of probability"s which
are pretty near on the answer [or close enough they are not always spot on]
here is a fun puzzle believe it or not is part of probability , have a go but only look at the picture for 5secs then press randomize
http://www.chemical-ecology.net/java/puzzles4.htm
here are some more if your bored ;D
http://www.chemical-ecology.net/java/probjav.htm
or practice your black jack [probability] http://www.chemical-ecology.net/java/21-1.htm question ?? if you make an additional bet after drawing your 1st 2 cards when your probability of busting is less than 50% would you beat the dealer?
I"ve not done probability calculations like that since school, I can"t remember how it all works. Out of interest though, have you calculated the odds that:
Someone has AA AND someone has KK (I think this is what you"ve done)
or
Someone has AA GIVEN someone has KK (this will be almost exactly the same as the odds that someone is dealt KK regardless of anyone elses hand).
Just curious
these are the odds if you are dealt KK THEN what is the chance of an opponent being dealt AA
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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?
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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) ;D
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
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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?
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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:.
:3c: 9d 5d qd td
He took that quite well to be fair :D.
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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:.
:3c: 9d 5d qd td
He took that quite well to be fair :D.
That still didn"t stop you whinging about Simon Trumper calling the action in the tourney!! ;D
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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:.
:3c: 9d 5d qd td
He took that quite well to be fair :D.
That still didn"t stop you whinging about Simon Trumper calling the action in the tourney!! ;D
He does it ALL the time. It gets soooo annoying. Too loud sir, too loud.
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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) ;D
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.
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probability forum please ;D
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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 ;D
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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?
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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.
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can i have £5 e/w on your next 5 selections please? ;D
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