Texas HoldEm - Help
I need to know what the chances of this happening are:
In a game with 4 players, what are the chances of the flop being 666 and the turn being the last 6.
(This actually happened to me, i woult like to know how unlikely it is.)
I attempted it and got just under 30,000:1 but i don't think its correct. I know i forgot to account for the cards that had already been dealt and burnt.
Thanks in advance.
because the burned cards and the other hands are unknown to you, they do not count and therefore you do not need to deal with them in your odds dealing. As far as those cards are concerned, they are just part of the deck.
I would suggest we start with the assumption that you do not know your own cards either and just go with the basic formula:
You have four options on the first draw out of 52 cards, 3 remaining sixes out of 51 cards, and so on. The value of this is:
24/6497400 or one out of 270725.
NOW: If we want to be even more specific, we could say something along the lines of "the first nine cards cannot be sixes, the next three must be sixes, the next card not a six, and the last card a six". This would include the hands being dealt, the burn cards and the exact order of the sixes. it would look something more like this:
Simplified, we can cancel out some of the numbers on the top and the bottom: 48 through 39 get removed because they're on the top and the bottom. We are left with:
Thanks alot, the answer i got was 10x too low (Rofl)
how would be display the 24/6497400 as a percentage?
I understand that 1/4 = 25% by dividing 1 by 4 but cant seem to use my brain to apply that same logic to 24/6497400 to get a logical answer on a calculator! (Crying)
(24)/(6497400) x 100
Originally Posted by LovePokerHateMath
Which when you do it out is about .0003%
I just pray you have an ace in your hand when it happens ;)