Pairing both live cards when you are not holding a pair (e.g., you hold a hand like AQ and your opponent does not have an A or Q, so you want to know the odds of seeing both an A and Q on the flop)
Pairing both live cards when you are not holding a pair (e.g., you hold a hand like AQ and your opponent does not have an A or Q, so you want to know the odds of seeing both an A and Q on the flop) 2.1% (46.3 to 1) Posted by 35suited
I'll stick with my 48.5 to 1. I've queried a few figures on tworags.com (where I think you got this from) and had no response.
I arrived at my figure by saying that the odds are:
6/50 * 3/49 * 44/48 * 3 = 0.0202 or 48.5 to 1.
To explain the figures, you can hit any of the 6 cards that pair one of your hole cards. Then one of the 3 cards that pair your second hole card and then the third card must be one of the remaining 44 cards that match neither of your hole cards. Finally, the cards can appear on the flop in any of 3 sequences.
To get precisely two pairs: 48/1 approx. Posted by MereNovice
HI MERENOVICE--am just doing a favour for a fella (booboo) in area 51--he would love to know the odds on 14 blacks in a row on a roulette table--perhaps you could help--cheers
In Response to Re: What are the odds? : HI MERENOVICE--am just doing a favour for a fella (booboo) in area 51--he would love to know the odds on 14 blacks in a row on a roulette table--perhaps you could help--cheers Posted by oynutter
in reality, nigh-on impossible....
on skyvegas, fairly short-odds assuming you put ur money on red every time....
In Response to Re: What are the odds? : HI MERENOVICE--am just doing a favour for a fella (booboo) in area 51--he would love to know the odds on 14 blacks in a row on a roulette table--perhaps you could help--cheers Posted by oynutter
Actually, my figure's incorrect because there are not just reds and blacks on a roulette table, are there? I'm not familiar with roulette but there is generally a 0 which is neither red or black, right? Also, on some tables there is an 00 too.
Assuming that the table has 37 slots of which 18 are black, the odds are 24,044 to 1. If the table has 38 slots of which 18 are black, the odds are 34,926 to 1.
Actually, my figure's incorrect because there are not just reds and blacks on a roulette table, are there? I'm not familiar with roulette but there is generally a 0 which is neither red or black, right? Also, on some tables there is an 00 too. Assuming that the table has 37 slots of which 18 are black, the odds are 24,044 to 1. If the table has 38 slots of which 18 are black, the odds are 34,926 to 1. Posted by MereNovice
Right!
So, aasume 1-36 are either Red or Black, & ZERO is neither.
So, 37 numbers, not 36, but only 18 Reds & 18 Blacks.
Comments
I've queried a few figures on tworags.com (where I think you got this from) and had no response.
I arrived at my figure by saying that the odds are:
6/50 * 3/49 * 44/48 * 3 = 0.0202 or 48.5 to 1.
To explain the figures, you can hit any of the 6 cards that pair one of your hole cards. Then one of the 3 cards that pair your second hole card and then the third card must be one of the remaining 44 cards that match neither of your hole cards. Finally, the cards can appear on the flop in any of 3 sequences.
I'm happy to be corrected, as always.
This is 2 * 2 * ... * 2 where there are 14 2's.
I'm not familiar with roulette but there is generally a 0 which is neither red or black, right? Also, on some tables there is an 00 too.
Assuming that the table has 37 slots of which 18 are black, the odds are 24,044 to 1.
If the table has 38 slots of which 18 are black, the odds are 34,926 to 1.
So, aasume 1-36 are either Red or Black, & ZERO is neither.
So, 37 numbers, not 36, but only 18 Reds & 18 Blacks.
More chance of Orford winning a Tourney.
Or MereNovice beating my KQ.