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.
cheers--lads--i will point him here---he was playing the double up system and got very disillusioned when 14 blacks in row came--not sure if this knowledge will cheer him up though--lol
heres a good one for you merenovice--how much would you need to have in your bankroll to play the double up system for a fiver a spin and have a million to one chance of losing your stash? (system is to keep betting on blk till you win--doubling your stake each time if you miss)
heres a good one for you merenovice--how much would you need to have in your bankroll to play the double up system for a fiver a spin and have a million to one chance of losing your stash? (system is to keep betting on blk till you win--doubling your stake each time if you miss) Posted by oynutter
I guess this means betting, £5, £10, £20, £40 etc. The answer would be roughly £5 million pounds.
On a 37 slot roulette wheel, the odds of 19 consecutive blacks is 882,379 to 1 and the odds of 20 consecutive blacks is 1,813,779 to 1.
If you repeatedly doubled up you would be betting £2,621,440 on the 20th spin and you would have "invested" a total of £5,242,875 trying to win £5.
In Response to Re: What are the odds? : I guess this means betting, £5, £10, £20, £40 etc. The answer would be roughly £5 million pounds. On a 37 slot roulette wheel, the odds of 19 consecutive blacks is 882,379 to 1 and the odds of 20 consecutive blacks is 1,813,779 to 1. If you repeatedly doubled up you would be betting £2,621,440 on the 20th spin and you would have "invested" a total of £5,242,875 trying to win £5. Posted by MereNovice
brilliant--thanks mate---was gonna start saving up tomoro---might just give it a miss--how would you be feeling on that last spin--lol--lol--pokers a good game though--init!!--lol
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.
came--not sure if this knowledge will cheer him up though--lol
The answer would be roughly £5 million pounds.
On a 37 slot roulette wheel, the odds of 19 consecutive blacks is 882,379 to 1 and the odds of 20 consecutive blacks is 1,813,779 to 1.
If you repeatedly doubled up you would be betting £2,621,440 on the 20th spin and you would have "invested" a total of £5,242,875 trying to win £5.