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Odds of paired flop?
What are the odds of a paired flop?
Seems to be happening pretty frequently.
Seems to be happening pretty frequently.
Comments
12 cards removed for the hands. leaves us a total of forty. Depending whether live or online it may go down to 39. Then we have to disregard the cards which are in the hands because they can not be known.
so we ignore the first card out because it is irrelevant.
the odds of the second pairing the first are 3/39
The odds of the third card pairing either the first or second are 6/38.
this gives us a total that comes to 23.5%.
Admittedly the real figure in each given hand will be lower due to cards having been dealt hitting the board. But it will be different for every hand.
HI Guys,
This thread caught my eye last night & it'd been doing my head in ever since! lol I'm NOT a maths expert but as I'm trying hard to improve my poker I'm desperately trying to learn pot odds/probability etc. Anyhoo, when i read the first post from Pip I thought it would be around 20% and googled to see if I could find the answer. It took a while but I came across a % quoted as 16.1% (5:1). Unfortunately it didn't give the maths in how this was arrived at. As I'm not a maths expert I had difficulty understanding Pat's answer and Talon's answer made sense. However, could I offer my logical slant on how to arrive at the answer? Someone else will have to do the maths tho! There may be errors in this logic but I LOVE a challenge!
There are 78 PP's possible in 52 cards. (Each set of cards (4) has 6 possible pp potential x 13 = 78)
Therefore I'm thinking that after the hole cards are dealt there are 40 cards left with the potential of having 60 pp. This is where my maths let me down tho. There is a possibility of between 0 and 6 PP's being dealt in the hole cards?. So I'm guessing this is the "variable" Talon mentioned? So this changes the potential to between 54-60 chances? The flop will be 3 x cards with the possibility of any 2 being a pair.
I'm not sure where to go from here tho........HELP!
Pad
p.s. Am I over-thinking this?
EDIT: I think this sounds similar to Talons explaination? However, the 3/39 & 6/38 I don't understand?
The simplest way to look at it mathematicly is to work out the chances that the the flop won't be paired;
The second card must not match the first, which is any 48 out of the 51 remaining cards, then the third card must not match either the first or the second, which is any 44 of the 50 remaining cards.
(48/51)x(44/50)=0.828
So there's an 82.8% chance the flop won't be paired, therefore a 17.2% chance it will be paired. Thats the same as 1 in every 5.8 flops being paired.
There are no variables. With each new flop there is exactly a 17.2% chance it will be paired. It makes no difference how many players are dealt in; the flop is always 3 random cards from 52.
Over small samples you can easily see big variations, ie over 100 hands either 27 paired flops, or only 7 paired flops would be nothing to get excited about. As the samples get bigger the proportion should get much closer to 17.2%. Over 1,000 hands it'll probably be within 1% (16.2% to 18.2%). Over say 100,000 hands it'll probably be within 0.1% (17.1% to 17.3%).
Just a thought, if 10 players are prepared to a look at the last 100 flops in their hand history we can quickly build up a 1,000 hand sample.
As for the talon maths ,whats going on there .....
He was the maths guru!
i put it down to the rng which im not sure but i reckon rngs come in differant types being less random cheap but dearest to rent would be more random and i will be contacting alderney to find out
Firstly the question what cards are held make all the difference to the possible outcomes. So the first card out could be impossible to pair up at all. This is a major unknown variable and would take some very complex mathematics to come to a sensible answer using these factors.
Secondly people are putting up calculations based upon 52 cards. This can never be the case because the maximum number of cards in the deck can only be 48 and that is on a HU game. The flop can never be dealt from a full deck.
The maths that i did used a six player table as its basis. But for the sake of simplicity i had to then ignore the possibility of the flop hitting any of the hands available. This meant that my answer was skewed on the high side because of this. I am aware of the error factor in my calculations and it would be too complex to try to eliminate this error totally. My answer is accurate only if no cards coming out match a players holding in any way. So in effect these are the maximum possible odds in this given situation. So as i said the true answer is lower than mine but the maths is way to complicated to be able to put on this forum
'I have 12 outs'
'NO. Those other 8 players could all hold your outs therefore you can't factor that in' ?
In Response to Re: Odds of paired flop?:
Hi Talon,Firstly, Thats what I thought "There is a possibility of between 0 and 6 PP's being dealt in the hole cards?" Am I correct in saying this is a major variable?
Secondly, That's also what I thought, i.e. that the max is 60pp from a 40 deck (i.e. after the hole cards have been dealt)
Thirdly, is there an idiots version? ;-) I'd be interested to see your maths?
As I said in my post this a.m. my maths isn't great BUT I'm keen to learn poker probs. (I need all the help I can get!)
pad
I agree that we can and should ignore the issue about how many other people are playing.
But the chances of the flop having exactly 2 the same rank must be different depending on whether or not we have a pocket pair.
All the times we have a pocket pair, the 50 unknown cards contain 12 ranks with 4 cards, and 1 rank with 2 cards.
Whenever we do not have a pocket pair, it is 11 ranks with 4 cards, plus 2 with 3 cards.
I am too lazy (and probably too incompetent!) to do the maths, but surely the outcome is different?
Of course, if we're ignoring our own hand, and just averaging it out over all the millions of hands we are dealt in our lives, then I agree with your method.
I then looked at a six handed table and then worked out the percentages of the best and worse case secnarios for the hole cards. The 2 answers were 23.5% and 16.2%. So this is the range of percentages for this particular scenario. Each individual hand will have its own odds due to the distribution of the hole cards but will always fall within this range, provided we dont change the number of people in the hand.
Any attempt to narrow this down would require the use of probablilities to determine the most likely distribution of hole cards, which would naturally have its own built in error factor.
So we have the range of a paired flop in a six handed game as being between 1 in 4 and 1 in 6. Probably not as accurate an answer as most would think possible but is as accurate as it can be. The odds would vary of course depending on how many players were playing. But as a site that operates predominantly 6 handed i thought that this would be the best way to go.
We hold a pocket pair, with (P,P). Ie 2 cards each of rank P. And x, y, z are 3 different ranks which are different to P.
In each deck, when we know our hole cards, there are 50 cards left.
2 of these are rank P.
Any of 48 cards can be rank, x. (ie any card in the deck apart from one of the 2 which is rank P)
Once x is chosen, any of 44 cards can be rank, y.
Once both x and y are chosen, any of 40 cards can be rank, z.
The first card on the flop is chosen 1 out of 50 unkowns. The second is 1/49 and the third is 1/48.
So a flop which is (P,x,y) or (x,P,y) or (x,y,P) all give us a pair on an unpaired board, where the order listed in the brackets is the order in which the flop is dealt. (The order of "x" and "y" does not matter as "x" is the first blank to be dealt and "y" is the second).
The percentage chance of each of these 3 is [(2 x 48 x 44)/(50 x 49 x 48)] x 100% = 3.592%.
So, in total, the chance of us having a set on an unpaired board is 3 x 3.592% = 10.78%.
Whereas a flop of (x, y, z) fails to improve our pocket pair, on an unpaired board.
The percentage chance is [48/50 x 44/49 x 40/48] x 100% = 71.84%
So overall, the chance of an unpaired board when we start off with a pocket pair is 82.62%. This means, of course, the chance of a paired board is 17.38%.
Scenario 2
We do not have a pocket pair. If we have 2 different cards, (H,L), then there are 3 of each of these left in the deck.
Any of 44 cards can be rank, x. (ie any card in the deck apart from one of the 6 which are rank H or L)
Once x is chosen, any of 40 cards can be rank, y.
Once both x and y are chosen, any of 36 cards can be rank, z
Any of these outcomes is an unpaired flop that gives us one pair:
(H, x, y); (x, H, y); (x, y, H); (L, x, y); (x, L, y); (x, y, L).
So there are 6 possibilities each of which is [(3 x 44 x 40)/(50 x 49 x 48)] x 100% = 4.490%
So, in total, the chance of us having a pair on an unpaired board is 6 x 4.490% = 26.94%.
Any of these outcomes is an unpaired flop that gives us two pair:
(H, L, x); (H, x, L); (x, H, L); (L, H, x); (L, x, H); (x, L, H).
So there are 6 possibilities each of which is [(3 x 3 x 44)/(50 x 49 x 48)] x 100% = 0.3367%
So, in total, the chance of us having two pair on an unpaired board is 6 x 0.3367% = 2.02%.
Whereas a flop of (x, y, z) fails to give us any pair, on an unpaired board.
The percentage chance is [44/50 x 40/49 x 36/48] x 100% = 53.88%.
So altogether, the chance of an unpaired flop when we start without a pocket pair is [26.94 + 2.02 + 53.88] is 82.84%. Thus the chance of a 17.16%.
(IMHO)