|
|
#1
02-23-2005, 06:51 AM
|
|||
|
|||
Being dealt A2xx in Omaha/8
How do you go about calculating this probability?
ie. Chance of being dealt both at least one ace and at least one deuce in 4 cards. My real-life results are 6.5% but I can't figure out how to calculate the true probability. And how would you extend this for example, to calculate the odds of being dealt A23x? It sounds like it should be easy but I'm not sure where to begin. 
|
|
#2
02-23-2005, 02:00 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4%
|
|
#3
02-23-2005, 08:50 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
[ QUOTE ]
8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4% [/ QUOTE ] This doesn't seem right... Exactly 1 ace and exactly 1 duece = 4*4*(44choose2)/(52choose4) = 0.055909133
|
|
#4
02-23-2005, 09:21 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
[ QUOTE ]
[ QUOTE ] 8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4% [/ QUOTE ] This doesn't seem right... Exactly 1 ace and exactly 1 duece = 4*4*(44choose2)/(52choose4) = 0.055909133 [/ QUOTE ] typo. change the denominator to 52 choose 2 EDIT: Fukc. I mean, 52 choose 4
|
|
#5
02-23-2005, 09:42 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
ie. Chance of being dealt both at least one ace and at least one deuce in 4 cards.
Odds of being dealt at least one ace: 1 - oddsNoAce = 1 - (( 48*47*46*45 )/(52*51*50*49)) = 0.281263274540585 Given the fact that you have an Ace, odds of having at least one deuce in the remaining 3 cards = 1 - oddsNoDeuce = 1 - (( 47*46*45)/(51*50*49) ) = 0.221368547418968 Odds at least one Ace, at least one Deuce, multiply to get: 0.0622628425273517 about 6.23% or roughly 1 in 16. "Ballpark computation" Ace, one in 13, four tries = 4/13 Deuce, on in 13, three tries = 3/13 Both: 12/169 = roughly .071 I know it is not proper probability, but gives a quick approximation which is usually a bit on the high side. Extended to A23x, Odds of a 3 in the remaining 2 cards = 1 - (46*45)/(50*49) and multiply that result by the A2 result.
|
|
#6
02-23-2005, 10:16 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
Angus,
Whoa! There is no need for such a long answer. The answer is: 4*4*(50 chooose 2)/(52 choose 4)= 0.07239819 Mickey, btw, the 8 in my first post was wrong too. Somehow I got 4*4 =8 when I was doing the math in my head. gm
|
|
#7
02-23-2005, 10:35 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
mouse,
Whoa! Since you do not explain why/where/how you came up with your formula (and since you already admit to one mistake), I will believe my answer. Unless you can find where I made a mistake.
|
|
#8
02-23-2005, 10:40 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
Check out this thread:
http://forumserver.twoplustwo.com/sh...747&Forum=,All_Forums,&Words=&Searchpage=0&Limit =25&Main=1420747&Search=true&where=&am p;Name=18572&daterange=&newerval=&newe rtype=&olderval=&oldertype=&bodyprev=# Post1420747
|
|
#9
02-23-2005, 10:50 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
Fine, but what is your answer to this question?
|
|
#10
02-23-2005, 11:34 PM
|
|||
|
|||
|
Re: Being dealt A2xx in Omaha/8
gaming_mouse's answer is slightly out as it double counts hands with AA2 or A22.
I make it A2xx 4*4*(44*43/2)=15136 A22x or AA2x 6*4*44=1056 times 2 = 2112 All aces or twos (8 choose 4) -2 = 28 Total 17276 Divded by (52 choose 4) = 6.38% PS My very first post what do you think?? [img]/images/graemlins/ooo.gif[/img]
|
 |
|
|
Linear Mode
